For more information, please see full course syllabus of Pre Calculus
For more information, please see full course syllabus of Pre Calculus
Intermediate Value Theorem and Polynomial Division
 There is no general root formula for all polynomials. There are formulas for degree 3 and 4 polynomials, but they're so long and complicated, we wonÕt even look at them. Thus, if we want to find the precise roots of a high degree polynomial, we're stuck factoring.
 If you manage to find a root of a polynomial (say, by pure luck or an educated guess), then you now know one of the factors of the polynomial. Each root of a polynomial automatically implies a factor:
Once you know a factor, it becomes that much easier to find the other factors.f(k) = 0 ⇔ (x−k) is a factor of f(x).  The intermediate value theorem says that If:
 f(x) is a polynomial (or any continuous function),
 a, b are real numbers such that a < b,
 u is a real number such that f(a) < u < f(b);
 This means we can use the intermediate value theorem to help us find roots. If we know that f(a) and f(b) are opposite signs (one +, other −), then we know there must be a root in (a,b). [Note that there they may be more than one. It guarantees the existence of at least one root, but does not say the precise number.] Knowing there is a root in some interval makes it that much easier to guess it, check to make sure, then use it to find a factor.
 The division algorithm states that if f(x) and d(x) are polynomials where the degree of f is greater than or equal to d and d(x) ≠ 0, then there exist polynomials q(x) and r(x) such that
Alternatively, we can write this asf(x) = d(x) · q(x) + r(x) . f(x) d(x)
= q(x) + r(x) d(x)
.  In the above, d(x) is what we divide by, q(x) is the quotient (what comes out of the division), and r(x) is the remainder. Thus if after we use the division algorithm we get that r(x) = 0, we know that d(x) divides evenly into f(x), or, in other words, d(x) is a factor of f(x).
 We can use the division algorithm through polynomial long division. [This is a rather difficult technique to describe with words, but much easier to see in action. Check out the video if you haven't already to understand how to do this.] The process for long division goes like this:
1. Divide the first term of the dividend (thing being divided) by the first term of the divisor (thing doing the division). Write the result above.
2. Multiply the entire divisor by the result, then subtract that from the dividend (so that the first term of one lines up with first term of the other).
3. Bring down the next term from the dividend.
4. Repeat the process (divide first terms, multiply, subtract, bring down) until finished with the entire dividend.  There is also a shortcut method that goes a bit faster if the divisor is in the form (x−k). This is called synthetic division. [Again, this is a rather difficult technique to describe with words, but much easier to see in action. Check out the video if you haven't already to understand how to do this.] The process for synthetic division goes like this:
1. Write out the coefficients of the polynomial in order.
2. Write k in the topleft corner [from the factor you're dividing by, (x−k)].
3. Start on the left of the coefficients and work towards the right. Every step will have a "vertical" part and a "diagonal" part.
4. On the "vertical" part, add the two numbers above and below each other to produce another number. On the "diagonal" part, multiply the result from the "vertical" by k.
5. The final number produced by the process is the remainder. The other numbers are the resulting coefficients of the quotient.  If you need to do division on a polynomial that is "missing" a variable raised to a certain exponent, make sure to fill that space in with a 0 multiplying the appropriate variable with exponent. For example, if you wanted to divide x^{4} + 5x + 7, notice that it's "missing" x^{3} and x^{2}. Before you can divide the polynomial, you have to put something in for those "missing" parts. We do that by filling it in with a 0 multiplying them:
x^{4} + 5x + 7 ⇒ x^{4} + 0x^{3} + 0 x^{2} + 5x + 7.
Intermediate Value Theorem and Polynomial Division

 By the intermediate value theorem, we know there must be a root (or more than one root) between x=0 and x=5. This is because f(0) is negative, while f(5) is positive. Since a polynomial can't "jump", the function must cross f(x) = 0 for some x between 0 and 5.
 Try guessing an x and plugging it in. Let's try x=2:
Now we know that x=2 is not a root. However, we do know that there must be a root between x=2 and x=5 because f(2) is negative and f(5) is positive.f(2) = 3(2)^{3}+9(2)^{2}−30(2)−72 = −72  Try guessing another x, using the information from our previous guess to help us decide where to guess. Let's try x=3:
Now we know that x=3 is a root! If we wanted to, we could convert that root into a factor of (x−3) and use it to help us factor the rest of the polynomial and find the rest of the roots.f(3) = 3(3)^{3}+9(3)^{2}−30(3)−72 = 0

 By the intermediate value theorem, we know there must be a root (or more than one root) between x=−10 and x=0. This is because f(−10) is negative, while f(0) is positive. Since a polynomial can't "jump", the function must cross f(x) = 0 for some x between −10 and 0.
 Try guessing an x and plugging it in. Let's try x=−5:
Now we know that x=−5 is not a root. However, we do know that there must be a root between x=−10 and x=−5 because f(−10) is negative and f(−5) is positive.f(−5) = (−5)^{5}+7(−5)^{4}−15(−5)^{3}−105(−5)^{2}+50(−5)+350 = 600  Try guessing another x, using the information from our previous guess to help us decide where to guess. Let's try x=−8:
Now we know that x=−8 is not a root. However, we do know that there must be a root between x=−8 and x=−5 because f(−8) is negative and f(−5) is positive.f(−8) = (−8)^{5}+7(−8)^{4}−15(−8)^{3}−105(−8)^{2}+50(−8)+350 = −3 186  Try guessing another x, using the information from our previous guess to help us decide where to guess. Let's try x=−7:
Now we know that x=3 is a root! If we wanted to, we could convert that root into a factor of (x+7) and use it to help us factor the rest of the polynomial and find the rest of the roots.f(−7) = (−7)^{5}+7(−7)^{4}−15(−7)^{3}−105(−7)^{2}+50(−7)+350 = 0

 We set up polynomial long division very similarly to how we set up long division many years ago in grade school:
x −3 ⎞
⎠x^{2} +5x −24  Begin by figuring out how many times the first term of the divisor (the thing doing the dividing: x−3) go into the first term of the dividend (the thing being divided: x^{2}+5x−24). Thus, our question is what do we get when we divide x^{2} by x? The result is [(x^{2})/x] = x, and we write that above the second term of the dividend because we have two terms in the divisor. [We can imagine this as if we lined up the dividend and the divisor, one over the other, then put our first result at the back of where they line up. This is very similar to how we did long division way back when we learned it in primary school.]
x x −3 ⎞
⎠x^{2} +5x −24  Next, we multiply our divisor by the result that we just obtained. In this case, that's (x−3) ·x = x^{2}−3x. Write this below the dividend, lining the first terms up with each other.
x x −3 ⎞
⎠x^{2} +5x −24 x^{2} −3x  Finally, subtract what we just wrote from the dividend above and write the result below. [It's critical to remember that you must subtract every term from the one above! It can be difficult to remember this, so you might want to write a big subtraction sign, then distribute it to each term to help you see what you need to do. The important thing is to remember to subtract what you got from the previous step.]
That completes the first round of long division, but we need to repeat the process until we reach the end of the dividend. Immediately after doing the subtraction, bring down the next term from the dividend so you can continue with the division:x x −3 ⎞
⎠x^{2} +5x −24 x^{2} −3x 8x x x −3 ⎞
⎠x^{2} +5x −24 x^{2} −3x 8x −24  We're now on to the next round of long division. This round is very similar to the previous round, except this time we're dividing what we just found previously. Once again, how many times does the first term of the divisor divide in? [8x/x] = 8, which we now write over the next open spot on the very top.
x +8 x −3 ⎞
⎠x^{2} +5x −24 x^{2} −3x 8x −24  Next, multiply the divisor by the result we just obtained: (x−3)·8 = 8x−24. Write this at the bottom, lining up the first terms.
x +8 x −3 ⎞
⎠x^{2} +5x −24 x^{2} −3x 8x −24 8x −24  Finally, subtract what we just found from the line above it.
x +8 x −3 ⎞
⎠x^{2} +5x −24 x^{2} −3x 8x −24 8x −24 0  The result of this last subtraction is the remainder. In this case, it is 0, so we know that (x−3) divides evenly into (x^{2}+5x−24). We can multiply the quotient (the result: x+8) by the divisor and, since the remainder was 0, we'll get back the original dividend. It's always a good idea to check long division problems because it's easy to make a mistake somewhere. Check simply by multiplying your answer (the quotient) with the divisor (what you've been dividing by):
Because we get back our dividend (the thing being divided), we know that we did the division correctly.Check: (x−3)(x+8) = x^{2} + 5x − 24

 To use synthetic division, our divisor (thing doing the dividing) must be in the form (x−k). In this case, our divisor is (x−3), so we can use synthetic division. The first step is to figure out what the value of k is. Here we have k=3.
 Next, figure out the value of the coefficients on each term in the polynomial being divided (the dividend). If the polynomial is in the form ax^{2} + bx + c, where a, b, and c are all coefficients, for our polynomial, we would have a = 1, b=5, and c=−24.
 Once you have found k and the coefficients to the polynomial, set them up in the following structure:
So, plugging in the values we found for this problem, we havek ⎢
⎢a b c ⎢
⎢3 ⎢
⎢1 5 −24 ⎢
⎢  Once you've set up the structure, begin by bringing down the first number from the right hand side.
3 ⎢
⎢1 5 −24 ⎢
⎢1  Next, multiply that number by whatever value k is (the number on the left hand side, in this case, 3). Write that number under the next term on the right side.
3 ⎢
⎢1 5 −24 ⎢
⎢3 1  Add the two numbers that are above one another and write that number directly below them.
3 ⎢
⎢1 5 −24 ⎢
⎢3 1 8  Continue this process of multiplying on the diagonal, then adding on the vertical. Next, we have 8·3.
3 ⎢
⎢1 5 −24 ⎢
⎢3 24 1 8  And then we add the two numbers together.
3 ⎢
⎢1 5 −24 ⎢
⎢3 24 1 8 0  The process is now completed. The final number that we get out (the number in the bottom right, bolded below) is the remainder.
Because the remainder is 0, we know the divisor (x−3) divides evenly into the dividend (x^{2} + 5x − 24). We can find the quotient (the result of the division) by looking at the nonremainder numbers. In this case, we found 1 8 and these make up the coefficients to the quotient:3 ⎢
⎢1 5 −24 ⎢
⎢3 24 1 8 0
[If you're a little unsure how to turn the numbers into a polynomial, work from the right and move left. The rightmost number is the constant, then one to the left is the coefficient for the variable to the power of 1, then another to the left is the coefficient for the variable to the power of 2, and just keep going as such until you run out of numbers. In this case we only had two numbers, so we only made it up to x^{1}.]x+8  It's always a good idea to check your answer, and we can do so by multiplying the divisor by the quotient (our answer):
Because we get back our dividend (the thing being divided), we know that we did the division correctly.Check: (x−3)(x+8) = x^{2} + 5x − 24

 (Notice: A detailed explanation of how to do long division will not be given in these steps. The steps will be shown, but no careful explanation of why each step occurs. If you already understand the mechanics of polynomial long division, you'll be able to understand what follows. If polynomial long division does not currently make sense, first go try to understand how it works. Watch the lesson carefully, see how it is used, and check out the steps to the first problem about long division in this question set. That problem has very detailed steps that explain how it works. Come back to this problem once you understand the mechanics of polynomial long division.)
 Set up the long division. However, there is an issue: (x^{3}−18x+8) does not have an x^{2} term. This will cause problems later on, so give it an x^{2} term. Since it has 0 x^{2}, we have
Now we're ready to write out the long division.(x^{3}−18x+8) = (x^{3}+0x^{2}−18x+8) x −4 ⎞
⎠x^{3} +0x^{2} −18x +8  Divide the first term by the first term of the divisor.
x^{2} x −4 ⎞
⎠x^{3} +0x^{2} −18x +8  Multiply the result by the divisor, then subtract and bring down the next term from the dividend.
x^{2} x −4 ⎞
⎠x^{3} +0x^{2} −18x +8 x^{3} −4x^{2} 4x^{2} −18x  Divide the first term by the first term of the divisor.
x^{2} +4x x −4 ⎞
⎠x^{3} +0x^{2} −18x +8 x^{3} −4x^{2} 4x^{2} −18x  Multiply the result by the divisor, then subtract and bring down the next term from the dividend.
x^{2} +4x x −4 ⎞
⎠x^{3} +0x^{2} −18x +8 x^{3} −4x^{2} 4x^{2} −18x 4x^{2} −16x −2x +8  Divide the first term by the first term of the divisor.
x^{2} +4x −2 x −4 ⎞
⎠x^{3} +0x^{2} −18x +8 x^{3} −4x^{2} 4x^{2} −18x 4x^{2} −16x −2x +8  Multiply the result by the divisor, then subtract and bring down the next term from the dividend.
x^{2} +4x −2 x −4 ⎞
⎠x^{3} +0x^{2} −18x +8 x^{3} −4x^{2} 4x^{2} −18x 4x^{2} −16x −2x +8 −2x +8 0  The result of the last subtraction is the remainder. We got 0, so the divisor divides evenly into the dividend. Our answer is x^{2}+4x−2.
 It's really easy to make a mistake on long division problems, so always check by expanding it.
Check: (x−4)(x^{2}+4x−2) = x^{3} −18x +8

 (Notice: A detailed explanation of how to do synthetic division will not be given in these steps. The steps will be shown, but no careful explanation of why each step occurs. If you already understand the mechanics of synthetic division, you'll be able to understand what follows. If synthetic division does not currently make sense, first go try to understand how it works. Watch the lesson carefully, see how it is used, and check out the steps to the first problem about synthetic division in this question set. That problem has very detailed steps that explain how it works. Come back to this problem once you understand the mechanics of synthetic division.)
 To use synthetic division, our divisor must be in the form (x−k). In this case, our divisor is (x−4), so we can use synthetic division.
 Set up the synthetic division structure. However, there is an issue: (x^{3}−18x+8) does not have an x^{2} term. This will cause problems later on, so give it an x^{2} term. Since it has 0 x^{2}, we have
Now we're ready to write out the structure.(x^{3}−18x+8) = (x^{3}+0x^{2}−18x+8) 4 ⎢
⎢1 0 −18 8 ⎢
⎢  Start from the left, add on verticals, multiply (by k=4) on diagonals.
4 ⎢
⎢1 0 −18 8 ⎢
⎢4 1 4 ⎢
⎢1 0 −18 8 ⎢
⎢4 16 1 4 4 ⎢
⎢1 0 −18 8 ⎢
⎢4 16 −8 1 4 −2
This is the last step, and the final number in the bottom right is the remainder. It's 0 in this case, so we know that the divisor divides in evenly. We can find the quotient from the nonremainder numbers on the bottom:4 ⎢
⎢1 0 −18 8 ⎢
⎢4 16 −8 1 4 −2 0 x^{2} + 4x −2  It's always a good idea to check your work, which we can do by expanding the divisor and quotient together
Check: (x−4)(x^{2}+4x−2) = x^{3} −18x +8

 (Notice: A detailed explanation of how to do long division will not be given in these steps. The steps will be shown, but no careful explanation of why each step occurs. If you already understand the mechanics of polynomial long division, you'll be able to understand what follows. If polynomial long division does not currently make sense, first go try to understand how it works. Watch the lesson carefully, see how it is used, and check out the steps to the first problem about long division in this question set. That problem has very detailed steps that explain how it works. Come back to this problem once you understand the mechanics of polynomial long division.)
 Set up the long division.
x +3 ⎞
⎠2x^{2} +3x −8  Divide first terms.
2x x +3 ⎞
⎠2x^{2} +3x −8  Multiply and subtract/bring down next term.
2x x +3 ⎞
⎠2x^{2} +3x −8 2x^{2} +6x −3x −8  Divide first terms.
2x −3 x +3 ⎞
⎠2x^{2} +3x −8 2x^{2} +6x −3x −8  Multiply and subtract. We're at the end of the dividend, so whatever results from the subtraction is our remainder.
2x −3 x +3 ⎞
⎠2x^{2} +3x −8 2x^{2} +6x −3x −8 −3x −9 1  Our quotient is 2x−3 but we also have a remainder of 1. Thus, (x+3) "cleanly" divides out a (2x−3) factor if we also have it divide the remainder. Thus our final answer is
2x −3 + 1 x+3.  As always, it's a good idea to check and make sure that our answer is correct. Furthermore, it will help us see how that strange part with the fraction needs to be a part of the answer:
Check: (x+3) ⎛
⎝2x−3+ 1 x+3⎞
⎠= (x+3)(2x−3) + (x+3) ⎛
⎝1 x+3⎞
⎠= (2x^{2} + 3x − 9) + 1 = 2x^{2} + 3x − 8

 (Notice: A detailed explanation of how to do synthetic division will not be given in these steps. The steps will be shown, but no careful explanation of why each step occurs. If you already understand the mechanics of synthetic division, you'll be able to understand what follows. If synthetic division does not currently make sense, first go try to understand how it works. Watch the lesson carefully, see how it is used, and check out the steps to the first problem about synthetic division in this question set. That problem has very detailed steps that explain how it works. Come back to this problem once you understand the mechanics of synthetic division.)
 To use synthetic division, our divisor must be in the form (x−k). In this case, our divisor is (x+5), so we can use synthetic division. However, notice that k=−5 because the format is (x−k) and so we must have a negative value for k to get the +5 to show up.
 Set up the synthetic division structure. However, there is an issue: (−4x^{3}−16x^{2}−35) does not have an x term. This will cause problems later on, so give it an x term. Since it has 0 x, we have
Now we're ready to write out the structure.(−4x^{3}−16x^{2}−35) = (−4x^{3}−16x^{2}+0x−35) −5 ⎢
⎢−4 −16 0 −35 ⎢
⎢  Start from the left, add on verticals, multiply (by k=−5) on diagonals.
−5 ⎢
⎢−4 −16 0 −35 ⎢
⎢20 −4 −5 ⎢
⎢−4 −16 0 −35 ⎢
⎢20 −20 −4 4 −5 ⎢
⎢−4 −16 0 −35 ⎢
⎢20 −20 100 −4 4 −20
This is the last step, and the final number in the bottom right is the remainder. In this case, our remainder is 65. The numbers before it [−4 4 −20] are the coefficients of the quotient, but we also need the remainder to appear and be divided to find the answer to our division:−5 ⎢
⎢−4 −16 0 −35 ⎢
⎢20 −20 100 −4 4 −20 65 −4x^{2} + 4x −20 + 65 x+5 As always, it's a good idea to check and make sure that our answer is correct. Furthermore, it will help us see how that strange part with the fraction needs to be a part of the answer:
Check: (x+5) ⎛
⎝−4x^{2} + 4x −20 + 65 x+5⎞
⎠= (x+5)(−4x^{2} + 4x −20) + (x+5) ⎛
⎝65 x+5⎞
⎠= (−4x^{3} − 16x^{2} − 100) + 65 = −4x^{3} − 16x^{2} − 35
 Begin by noticing that synthetic division can not be used for this problem. This is because synthetic division can only be used if the divisor is in the form (x−k), but (x^{2}+4) is not in that format. Therefore we must use long division, so set up the long division structure. However, there is a slight issue: the divisor (x^{2}+4) does not have an x term. While we can do the problem without one, it will make it a bit easier to see how things line up if it has one. Write it instead as (x^{2}+0x+4).
x^{2} +0x +4 ⎞
⎠x^{4} +8x^{3} −3x^{2} +32x −28  Divide the first term by the first term of the divisor.
x^{2} x^{2} +0x +4 ⎞
⎠x^{4} +8x^{3} −3x^{2} +32x −28  Multiply and subtract/bring down next term.
x^{2} x^{2} +0x +4 ⎞
⎠x^{4} +8x^{3} −3x^{2} +32x −28 x^{4} +0x^{3} + 4x^{2} 8x^{3} −7x^{2} + 32x  Repeat process until finished.
x^{2} +8x x^{2} +0x +4 ⎞
⎠x^{4} +8x^{3} −3x^{2} +32x −28 x^{4} +0x^{3} + 4x^{2} 8x^{3} −7x^{2} + 32x x^{2} +8x x^{2} +0x +4 ⎞
⎠x^{4} +8x^{3} −3x^{2} +32x −28 x^{4} +0x^{3} + 4x^{2} 8x^{3} −7x^{2} + 32x 8x^{3} +0x^{2} + 32x −7x^{2} +0x −28 x^{2} +8x −7 x^{2} +0x +4 ⎞
⎠x^{4} +8x^{3} −3x^{2} +32x −28 x^{4} +0x^{3} + 4x^{2} 8x^{3} −7x^{2} + 32x 8x^{3} +0x^{2} + 32x −7x^{2} +0x −28
The process is now complete with a remainder of 0, so we see that (x^{2}+4) divides in evenly. The quotient of x^{2}+8x−7 is our answer.x^{2} +8x −7 x^{2} +0x +4 ⎞
⎠x^{4} +8x^{3} −3x^{2} +32x −28 x^{4} +0x^{3} + 4x^{2} 8x^{3} −7x^{2} + 32x 8x^{3} +0x^{2} + 32x −7x^{2} +0x −28 −7x^{2} +0x −28 0  It's always a good idea to check your work, so multiply the divisor and quotient to make sure we have the correct answer.
Check: (x^{2}+4)(x^{2}+8x−7) = x^{4} + 8x^{3} − 3x^{2} + 32x − 28
 No matter what we do, we need to factor the polynomial to find its roots. There are multiple ways to approach factoring it. First, since (x^{2}−9) is a factor, we could put it in the form
Using logic, we could eventually figure out what has to go in those blanks, but depending on the problem it might be somewhat difficult. Next, we could attempt to use synthetic division, but we can't because (x^{2}−9) is not in the form (x−k). We could break it into two different factors of (x−3) and (x+3), then divide one after the other, but that takes multiple steps. Instead, for this problem, we'll do it with polynomial long division. [However, the other two methods above would work just fine if you wanted to do them instead.](x^{2}−9)( x^{2} + x + ) = x^{4}−6x^{3}−5x^{2}+54x−36 .  Set up the long division:
x^{2} +0x −9 ⎞
⎠x^{4} −6x^{3} −5x^{2} +54x −36 x^{2} x^{2} +0x −9 ⎞
⎠x^{4} −6x^{3} −5x^{2} +54x −36 x^{4} +0x^{3} −9x^{2} −6x^{3} +4x^{2} + 54x x^{2} −6x x^{2} +0x −9 ⎞
⎠x^{4} −6x^{3} −5x^{2} +54x −36 x^{4} +0x^{3} −9x^{2} −6x^{3} +4x^{2} + 54x −6x^{3} +0x^{2} + 54x 4x^{2} +0x −36
Our quotient is (x^{2} − 6x + 4) when we divide out (x^{2}−9). [Notice that the remainder came out to be 0. This is an automatic check on our work: since the problem said (x^{2}−9) was a factor of x^{4}−6x^{3}−5x^{2}+54x−36, we know it must divide in evenly. If it didn't come out with a remainder of 0, we would have made a mistake somewhere.]x^{2} −6x +4 x^{2} +0x −9 ⎞
⎠x^{4} −6x^{3} −5x^{2} +54x −36 x^{4} +0x^{3} −9x^{2} −6x^{3} +4x^{2} + 54x −6x^{3} +0x^{2} + 54x 4x^{2} +0x −36 4x^{2} +0x −36 0  Using long division, we have factored the polynomial as
We can easily factor (x^{2}−9) further as (x+3)(x−3), sox^{4}−6x^{3}−5x^{2}+54x−36 = (x^{2}−9)(x^{2} − 6x + 4).
which means we have roots at x+3 = 0 and x−3=0, so x=−3, 3. However, we still need to figure out if (x^{2}−6x+4) has any roots. Looking at it, we realize that it would be very difficult to factor because it will involve noninteger numbers in the factors.(x+3)(x−3)(x^{2} − 6x + 4),  Luckily, we have an easy way to find the roots of a quadratic polynomial: the quadratic formula! [See the lesson on the quadratic formula if you don't know how to use it yet. Set it up using (x^{2}−6x+4) to find its roots:
x = −(−6) ±
√(−6)^{2} − 4 ·1 ·42·1  Working through it, we get
Therefore, the roots of (x^{2} − 6x+4) are x = 3 − √5, 3 + √5.x = 6 ±
√202= 6 ±2√5 2= 3±√5.  Assemble these roots with the roots we found earlier to have all the roots of the polynomial.
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Answer
Intermediate Value Theorem and Polynomial Division
Lecture Slides are screencaptured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
 Intro
 Introduction
 Reminder: Roots Imply Factors
 The Intermediate Value Theorem
 Intermediate Value Theorem, Proof Sketch
 Finding Roots with the Intermediate Value Theorem
 Dividing a Polynomial
 Long Division Refresher
 The Division Algorithm
 Polynomial Long Division
 Synthetic Division
 Synthetic Division, Example
 Which Method Should We Use
 Example 1
 Example 2
 Example 3
 Example 4
 Intro 0:00
 Introduction 0:05
 Reminder: Roots Imply Factors 1:32
 The Intermediate Value Theorem 3:41
 The Basis: U between a and b
 U is on the Function
 Intermediate Value Theorem, Proof Sketch 5:51
 If Not True, the Graph Would Have to Jump
 But Graph is Defined as Continuous
 Finding Roots with the Intermediate Value Theorem 7:01
 Picking a and b to be of Different Signs
 Must Be at Least One Root
 Dividing a Polynomial 8:16
 Using Roots and Division to Factor
 Long Division Refresher 9:08
 The Division Algorithm 12:18
 How It Works to Divide Polynomials
 The Parts of the Equation
 Rewriting the Equation
 Polynomial Long Division 16:20
 Polynomial Long Division In Action
 One Step at a Time
 Synthetic Division 22:46
 Setup
 Synthetic Division, Example 24:44
 Which Method Should We Use 26:39
 Advantages of Synthetic Method
 Advantages of Long Division
 Example 1 29:24
 Example 2 31:27
 Example 3 36:22
 Example 4 40:55
Precalculus with Limits Online Course
Transcription: Intermediate Value Theorem and Polynomial Division
Hiwelcome back to Educator.com.0000
Today, we are going to talk about the intermediate value theorem and polynomial division.0002
Previously, we talked about how we need to factor a polynomial to find its roots.0006
But we recently saw the quadratic formula, which gave us the root of any quadratic without even having to factor at all.0009
So, maybe we don't need factoring; maybe there are formulas that allow us to find the roots for any polynomial; wouldn't that be great?0015
We would just be able to plug things in, put out some arithmetic, and we would have answers, no matter what polynomial we were dealing with.0021
Not reallywhile there are root formulas for polynomials of degree 3 and 4, they are so long and complicated, we are not even going to look at them.0028
The formula for finding the roots of a cubic, of a degree 3 polynomial, is really long and really complex.0036
And it is just something that we don't really want to look at right now.0044
And a degree 4 would be even worse; so we are just not going to worry about them.0049
And then, not only thatthey just simply don't exist for degree 5 or higher.0054
So, if you are looking at a degree 5 or higher, there is no such formula for any degree 5 or higher thing.0060
It was proven in 1824 that no such formula can exist, that would be able to do that.0067
Thus, it looks like we are stuck factoring, if you want to find the precise roots of a higherdegree polynomial.0073
If we are working with a higherdegree polynomial, and we need to know its roots for some reason, we have to figure out a way to factor it.0077
In this lesson, we are going to learn some methods to help factor these complicated polynomials.0082
We will first learn a theorem to help us guess where roots are located, and then a technique for helping us break apart big polynomials.0086
All right, let's go: In the lesson Roots of Polynomials, we mentioned the theorem that every root implies a factor.0092
That is, if f(x) is a polynomial, then if we have f(k) = 0 (k is a root), then that means we know x  k is a factor of f(x),0102
because if k = 0, then that means that x = k causes a root; so x  k = 0; and thus, we have a factor from our normal factor breakdown.0112
For example, if we have g(x) = x^{3}  3x^{2}  4x + 12,0123
and we happen to realize that when we plug in a 2, that all turns into a 00128
(and it does), then we would know that g(x) is equal to (x  2), this thing becoming (x  2), times (_x^{2} + _x + _).0131
We know that there is some way to factor that polynomial where something is going to go in those blanks.0147
2 is a root; this theorem tells us that (x  2) must be a factor.0152
It doesn't tell us what will be left; but it does make the polynomial one step easier.0156
We know we can pull up (x  2), so then we can use some logic, play some games, and figure out what has to go in those blanks.0162
But notice: it doesn't directly tell us what is going to be there.0168
If we are lucky, we can sometimes find a root or two purely by guessing.0173
We might think, "Well, I don't know where the roots are; but let's try 5; let's try √2; let's try π."0176
We might just try something, and surprisingly, it ends up working; that is great.0185
But is this always going to end up being the case?0189
If we manage to pick something where we figure out the root, and then we figure out something that gets us a root;0193
we plug in a number, and we get 0, and we know we have a root.0201
And if we have a root, that means we have a factor.0204
Knowing a factor makes it that much easier to factor the whole polynomial.0207
But it is hard to guess correctly every time; guessing is guessingyou can't guess every single time.0211
Luckily, there is a theorem that will give us a better idea of where the roots are located.0218
The intermediate value theorem will help us find roots; it goes like this:0223
If, first, f(x) is a polynomial (for example, we have this nice red curve here; that is our f(x));0227
then a and b are real numbers, such that a < b.0234
What that means is just that a and b are going from left to right; a is on the left side, and then we make it up to b.0238
That is what this a < b isjust that we know an order that we are going in.0247
And then, u is a real number, such that f(a) < u < f(b).0251
So, we look and figure out that part; we see that, at a, we are at some height f(a); and at b, we are at some height f(b).0256
That is how we get that graph in the first place.0267
Then, u is just something between those two heights; so u is just some height level,0269
where we put an imaginary horizontal line that ends up saying,0276
"Here is an intermediate value between f(a) and f(b), some intermediate height between those two."0280
The intermediate value theorem tells us that there exists some c contained in ab, such that f(c) = u.0289
So, we are guaranteed the existence of some c that is going to end up giving us this height, u.0298
Basically, if we have some height u, and that height u crosses between two different heights0305
that go...we have two points going from left to right, so we are going from left to right, and we cross over some height during that thing;0316
we start lower, and then we end above; we are guaranteed that we had to actually cross it.0332
We had to go across it; and since we had to go across it, there must be some c where we do that crossingwhere we end up landing on that height.0338
There is something that will give us that intermediate value.0348
Why does this have to be true? Because polynomials are continuous; there are no breaks in their graphs.0352
The only way f could possibly manage to not end up being on this height uthe only way f could dodge the height u is by jumping this intermediate height.0358
The only way that we have this...the graph is going; the graph is going; the graph is going; the graph is going; the graph is going.0369
And then, all of a sudden, it would have to jump over that height to be able to manage to not end up touching it.0375
If our graph touches the height at any point, then we have whatever point is directly below where it touched that height;0381
that is the location that intersects that intermediate value.0388
That is the thing that is going to fulfill our intermediate value theorem.0391
So, on any polynomial or any continuous function (in fact, the intermediate value theorem is true for any continuous function,0395
but we are just focusing on polynomials), they can't jump; polynomialscontinuous functionsthey are not allowed to jump.0400
There are no breaks in their graphs; so since there are no breaks, they have to end up crossing over this height.0406
Since they cross over this height, there is some place on the graph...we just look directly below that,0412
and that guarantees us our c, where f(c) is going to be equal to u; there we go.0416
This means we can use the intermediate value theorem to help us find roots.0423
If we know that f(a) and f(b) are opposite signsso, for example, if we know that at a, f(a) is positive0427
we have a positive for f(a)and then, at b, we know that we are negativewe have a negative for f(b)0434
then we know that there has to be a root.0440
Why? Well, we have to have some f that is going to make it from here somehow to here.0442
It has to manage to get both of those things.0448
So, the only way it can do it is by crossing over at some location.0451
It might cross over multiple times, but it has to cross over somewhere.0455
Otherwise, it is not going to be able to make it to that point that we know is below y = 0.0462
Since f must cross over y = 0, we are guaranteed the existence of this c at some point where it ends up crossing over.0467
Now, this does not mean that there is only one root; like we just saw in that second thing I drew, it could cross over multiple times.0478
This theorem guarantees the existence of at least one root, but there could be multiple roots.0485
if it bounces back and forth over that y = 0 on the way to making it to the second point.0491
OK, so say we find a root of a polynomial by a combination of luck and the intermediate value theorem.0498
We somehow manage to figure them out, or the problem just tells us a root directly from the beginning.0503
In either case, with the root, we now know a factor; a root tells us a factor.0508
But how can we actually break up the polynomial if we know a factor?0513
How do we divide a polynomial by a factor?0516
For example, say we know x = 3 is a root of this polynomial.0519
Then we know that there is some way to divide out x  3 so that we have x  3 pulled out,0523
and some _x^{3} + _x^{2} + _x + _0527
there is some other polynomial that is going to go with it, that the two will multiply.0531
Otherwise, it would not have divided out cleanly.0534
It couldn't be a factor unless there was going to be this other polynomial where it does divide out cleanly.0536
So, how do we actually find out that thing that happens after we divide out this polynomial?0541
How do we do polynomial division?0545
To explore this idea, let's refresh ourselves on long division from when we were young.0548
So, long ago, in grade school and primary school, we were used to doing problems like 1456 divided by 3.0553
Let's break out long division: we have 1456, so the first thing we do is see how many times 3 goes into 1.0559
3 goes into 1 0 times; so it is 0 times 3; that gets us 0 down here; we subtract by 0; nothing interesting happens yet.0570
1; and then we bring down the 4; so we have 14 now.0577
How many times does 3 go into 14? It goes in 4 times.0583
3 times 4 gets us 12; we subtract by 12: minus 12; so minus 12...we get 2.0586
Then, we bring down the next one in the running; let's keep the colors consistent.0596
We have 5 coming down, so we now have 25; how many times does 3 go into 25?0602
It goes in 8 times...16, 24...so now we subtract by 24; 25  24 gets us 1.0607
We bring down the 6; we have 16; how many times does 3 go into 16? It goes in 5; we get 15.0617
So, minus 15...we get 1; now, we don't have any more numbers to go here.0625
There is nothing else, so that means we are left with a remainder of 1.0633
We have whatever our very last thing was, once we ran out of stuff; that becomes our remainder: 485 with a remainder of 1.0640
If we wanted to express this, we could say 1456 divided by 3; another way of thinking of that is 1456 is equal to 3(485) + 1.0647
Or, alternately, if we wanted to, we could say 1456/3 is equal to 485 plus the remainder, also divided,0660
because we know that we get the 485 cleanly, but the 1 is a remainder.0670
So, it doesn't come out cleanly; so it comes out as 1/3.0675
You could also see the connection between these two things, because we simply divide both sides by 3;0677
and that is how we are getting from one place to the other place.0683
That is what we are getting by going through long division.0686
Really quickly, let's also look at this 1456 = 3(485) + 1; we call this right here the dividend; the thing being divided is the dividend.0691
This one here is the thing doing the dividing; we call the thing doing the dividing the divisor.0705
Then, what we get out of it is our quotient; what comes out of division is the quotient.0715
And finally, what we have left at the very end is the remainder.0724
All right, these are some special terms; you might not remember those from grade school.0732
But these are the terms that we use to talk about it.0736
Why does that matter? because we are now going to want to be able to express it in a more abstract, interesting way,0738
where we are talking, not just about real numbers, but being able to talk about polynomials.0743
We found that 1456 divided by 3 became 3 times 485 plus 1.0747
Clearly, we can do this method for any two numbers; and it turns out that we can do a very similar idea for polynomials.0752
We call this the division algorithmthis idea of being able to do this.0757
And it says that if f(x) and d(x) are both polynomials, and the degree of f is greater than or equal to d0761
that is to say, f is a bigger polynomial than dand d(x) is not equal to 0 (why does d(x) not equal 0?0770
because we are not allowed to divide by 0, so if d(x) is just simply 0 all the time, forever,0776
then we can't divide by it, because we are not allowed to divide by 0)given these things (f(x) and d(x), both polynomials;0781
f(x) is a bigger polynomialthat is to say, higher degree than dand d(x) is not simply 0 everywhere),0786
then there exist polynomials q(x) and r(x) such that f(x) = d(x) times q(x) plus r(x).0792
So, how is this parallel? f(x) is the thing being divided.0800
The thing being divided is our dividend, once again.0804
The thing doing the dividing is the divisor.0808
What results after we have done that division is the quotient.0818
And finally, what is left at the end is our remainder.0826
So, the remainder is right here, and our quotient is right here.0836
So, we have parallels in this idea of 1456/3; we have this same thing coming up here.0844
1456/3 is not equal to 3(485); it becomes this idea; so it is not actually equal: 1456/3 is 485 plus 1/3; but it becomes this idea.0850
So, the dividend here, the thing that we are breaking up, in this idea, is 1456; let's just knock this out, so we don't get confused by it.0864
Our divisor, the thing doing the dividing, is 3; what we get in the end, our quotient, is 485; and the remainder is that 1; 1 is left out of it.0874
Now, we could also have an alternative form where we write this as f(x)/d(x) = [q(x) + r(x)]/d(x),0884
where we just turn this into dividing; we get between these two by dividing by...not 3...but dividing by d(x), dividing by our polynomial.0892
So, basically the same thing is happening over here, so this would not be 3 times 45; it should be 1/3.0906
1456/3 is equal to 485 + 1/3, because that is 1456/3.0913
We have a real connection between these two things.0920
The division algorithm is giving us this idea that if we have some polynomial f(x),0924
we can break it into the divisor, times the quotient, plus some remainder, which, alternatively, we can express as0928
the polynomial that we are dividing, divided by its divisor, is equal to the quotient, plus the remainder, also divided by the divisor.0934
This is effectively a way of looking at f(x) dividing d(x)seeing what is happening here.0941
Now, notice: r(x) is the remainder, so in the case when r(x) = 0, then that means we have no remainder,0945
which we describe as d(x) dividing evenly; so when it divides evenly into f(x), then that means d(x) is simply a factor.0954
5 divides evenly into 15, so that means 5 is a factor of 15.0964
How do we actually use the division algorithm to break apart a polynomial?0970
Let's look at two methods: we will first look at long division, and then we will look at synthetic division.0974
First, long division: we will just take a quick run at how we actually use polynomial long division.0978
And it works a lot like long division that we are already used to.0983
So, let's see it in action first; and then we will talk about how it just worked.0986
We have x^{4}  5x^{3}  7x^{2} + 29x + 30, divided by x  3.0989
So, we are dividing x  3; it is dividing that polynomial: x^{4}  5x^{3}  7x^{2} + 29x + 30.0996
OK, the first thing we do is ask, "All right, how many times does x  3 go into x^{4}  5x^{3}?"1013
Well, really, we are just concerned with the front part; so just look at the first term, x.1021
How many times does x go into x^{4}?1025
Well, x^{4} divided by x would be x^{3}; so the x^{3} goes here.1028
Now, that part might be a little confusing: why didn't we end up having it go at the front?1033
Well, think of it like this: if we have 12 divide into 24, does 2 show up at the front?1037
No, 2 doesn't show up at the front; 2 shows up on the side, because 12 is 2 digits long; so we end up being at the secondplace digit, as well.1043
It is 2 digits long, so we go with the secondplace digit.1052
So, the same thing is going on over here: x  3 is two terms long, so we end up being at the second term, as well.1054
All right, so all of the ideas...we are going to knock them out really quickly, now that we have them explained.1061
So, that is why we are not starting at the very first placebecause we have to start out at where they line up appropriately.1066
We check first term to first term, but then we go as far wide as that thing dividing is.1074
So, x^{3} is what we get out of x^{4} divided by x.1078
So, now we take x^{3}, and we multiply (x  3), just as we did in long division.1083
x^{3} times (x  3) becomes x^{4}  3x^{3}.1087
Now, we also, in long division, subtracted now; so subtract.1092
Let's put that subtraction over it; minus, minus, 2 minus's become a plus...so we have x^{4} attacking x^{4}, so we have 0 here.1097
And 3x^{3} + 5x^{3} becomes 2x^{3}.1105
And then, the next thing we do is bring down the 7x^{2}.1115
So, 7x^{2}: now we ask ourselves, "How many times does x go into 2x^{3}?"1119
Well, that is going to go in 2x^{2}, so we get 2x^{2}.1127
2x^{2} times x  3 becomes 2x^{3}; 2x^{2} times 3 becomes + 6x^{2}.1133
Now, we subtract by all this stuff; we distribute that; that becomes positive; this becomes negative.1142
We now add these things together, so 2x^{3} + 2x^{3} becomes 0 once again.1149
7x^{2}  6x^{2} becomes 13x^{2}.1154
The next thing we do is bring down the 29x, so + 29x.1159
How many times does x go into 13x^{2}? That goes in 13x.1164
The 13x times x  3 gets us 13x^{2} + 39x; that is this whole quantity; subtracting that whole thing,1171
we distribute that, and it becomes addition there, and subtraction there.1182
So, we have 13x^{2} + 13x^{2}; that becomes 0 once again; 29x  39x becomes 10x.1185
Once again, we bring down the 30; so we have + 30 here.1193
And now, how many times does x go into 10x? x goes in 10 times.1199
So, 10 times x  3 is 10 + 30; we subtract this whole thing; we distribute that; and we get 0 and 0.1205
So, we end up having a remainder of 0, which is to say it goes in evenly.1215
So, if that is the case, we now have x^{3}  2x^{2}  13x  10 as what is left over after we divide out x  3.1219
So, we know our original x^{4}  5x^{3}  7x^{2} + 29x + 30 factors as...1230
let's write this in blue, just so we don't get it confused...(x  3)(x^{3}  2x^{2}  13x  10).1236
That is what we have gotten out of it; cool.1248
To help us understand how that worked, let's look at the steps one at a time.1252
You begin by dividing the first term in the dividend by the first term in the divisor.1255
So, our dividend is this thing right here; its first term is x^{4}; the first term of our divisor,1260
the thing doing the dividing, is x; so how many times does x go into x^{4}?1266
Well, x^{4} divided by x...if we are confused by the exponents, we have x times x times x times x, over x;1269
so, one pair of them knock each other out; so we have x^{3} now, x times x times x; great.1279
That is why the x^{3} goes here; it is the very first thing that happens.1287
The next thing: we take x^{3}, and we multiply it onto (x  3).1291
So, x^{3}(x  3) becomes x^{4}  3x^{3}.1298
You multiply the entire divisor (this right here) by the result, our x^{3}.1304
And then, we subtract what we just had from the dividend.1310
x^{4}  3x^{3}: we subtract that from x^{4}  5x^{3}.1315
This gets distributed, so we get a negative here, a plus here; and so that becomes 2x^{3}.1319
The next thing we do is bring down the next term; so our next term to deal with is this 7x^{2}.1325
It gets brought down, and we have 2x^{3}  7x^{2}.1331
And then, once again, we do the same thing: how many times does x go into 2x^{3}?1337
It goes in 2x^{2}; so then, it is 2x^{2}(x  3); and we get 2x^{3} + 6x^{2}.1343
We subtract that, and we keep doing this process until we are finally at the end.1354
We might have a remainder if it doesn't come out to be 0 after the very last step.1358
Or if it comes out to be 0, we are good; we don't have a remainder.1362
All right, there is also a shortcut method that goes a bit faster if the divisor is in the form x  k.1365
And notice: it has to be in the form x  k; if it is in a different form, like x^{2} + something, we can't do it.1372
Now, notice that you could deal with x + 3; it would just mean that k is equal to 3, so that is OK.1378
It just needs to be x, and then a constant; so that is the important thing if we are going to use synthetic division.1386
So, it goes like this: we let a, b, c, d, e be the coefficients of the polynomial being divided.1391
For example, if we have ax^{4} + bx^{3} + cx^{2} + dx + e,1396
then we set it up as follows; this k right here is on the outside of our little bracket thing.1401
And then, we set them up: a, b, c, d, e.1407
Now, the very first step: every vertical arrowyou bring whatever is above down below the line.1410
So, a, since there is nothing underneath it...we add terms on vertical arrows, so they come down adding together.1417
So, a comes down; there is nothing below it, so it becomes just a.1423
Then, you multiply by k on the diagonal arrow; so we have a; it comes up; we multiply by k, and so we get k times a.1426
Then, once again, we are doing another vertical arrow where we are adding.1435
We go down: k times a...b + ka becomes ka + b.1439
The next thing that will happen (you will probably want to simplify it, just to make it easier, but) we multiply that whole thing by k once again.1446
And we keep up the process until we get to the very last thing.1452
And the very last thing is our remainder.1456
All the terms preceding that, all of the terms in these green circles, are the coefficients of the quotient.1459
So, if this is one, then we will have a constant here, starting from the right; and this will be x's coefficient;1468
this will be x^{2}'s coefficient; this will be x^{3}'s coefficient.1473
And that makes sense: since we started with that to the fourth,1476
and we were dividing by something in degree 1, we should be left with something of degree 3.1479
All right, let's see it in action now.1483
Once again, dividing the same thing, we have x^{4}  5x^{3}  7x^{2} + 29x + 30.1485
So, we have x  3; remember, it is x  k, so that means our k is equal to 3, because it is already doing the subtraction.1491
We have 3; and we set this up; our first coefficient here is just a 1; 1 goes here.1500
What is our next coefficient? 5; 5 goes here.1511
What is our next coefficient? 7; 7 goes here.1516
Our next coefficient is 29; what is our next coefficient? 30, and that is our last one, because we just hit the constant.1519
All right, so on the vertical parts, we add; so 1 + _ (underneath it) becomes 1.1526
Then, 3 times 1 becomes 3; 5 plus 3 becomes 2; 2 times 3 is 6; 6 plus 7 becomes 13.1533
3 times 13 becomes 39; 29 plus 39 becomes 10; 3 times 10 becomes 30; 30 + 30 becomes 0.1546
Now, remember: this very last one is our remainder; so our remainder is 0, so it went in evenly, which is great,1558
because since we just did this with polynomial long division, and we saw it went in evenly, it had better go in evenly here, as well.1564
So, this is our constant right here (working from the right); this is our x; this our x^{2}; this is our x^{3}.1571
So, we get x^{3}  2x^{2}  13x  10; that is what is remaining.1578
So, we could multiply that by x  3; and then this whole expression here would be exactly what we started with in here, before we did the division.1587
Great; all right, so which of these two methods should we use?1598
At this point, we have seen both polynomial long division and synthetic division.1602
And so, which is the better methodwhich one should we use when we have to divide polynomials?1606
Now, synthetic division, as you just saw, has the advantage of being fastit goes pretty quickly.1610
But it can only be used when you are dividing by (x  k); remember, it has to be in this form1614
of linear things dividing only: x and plus a constant or minus a constant.1618
Ultimately, it is just a trick for one very specific kind of problem, where you have some long polynomial,1623
and you are dividing by a linear factorby something x ± constant.1628
Long division, on the other hand, while slower, is useful in dividing any polynomial.1633
We can use it for dividing any polynomial at all.1638
I think it is easier to remember, because it goes just like the long division that we are used to from long, long ago.1641
There is a slight change in the way we are doing it, but it is pretty much the exact same format.1646
How many times does it fit it? Multiply how many times it fits in by what you started with, and then subtract that.1651
And just repeat endlessly until you get to the end.1655
And then, lastly, it is connected to some deep ideas in mathematics.1660
Now, you probably won't end up seeing those deep ideas in mathematics until you get to some pretty heavy college courses.1663
But I think it is really cool how something you are learning at this stage can be connected to some really, really amazing ideas in later parts of mathematics.1668
So, personally, I would recommend using long division.1676
I think long division is the clear winner for the better one of these to use,1679
unless you are doing a lot of the (x  k) type divisions, or the problem specifically says to do it in synthetic.1682
If your teacher or the book says you have to do this problem in synthetic, then you have to do it in synthetic, because you are being told to do that.1689
But I think long division is easier to remember; it is more useful in more situations;1694
and it is connected to some really deep ideas that help you actually understand1699
what is going on in mathematics, as opposed to just being a trick.1701
Honestly, the only reason we are learning synthetic division in this lessonin this course1704
is because so many other teachers and books teach it.1709
I personally don't think it is that great.1712
It is a useful trick; it is really useful in the specific case of linear division.1715
If you had to do a lot of division by linear factors, it would be really great.1718
But we are just sort of seeing that we can break up polynomials, so I think the better thing is long division.1721
It is easier to remember; you can actually pull it out on an exam after you haven't done it for two months,1727
and you will remember, "Oh, yes, it is just like long division," which by now is burned into your memory from learning it so long ago.1731
And so, synthetic division is really just watereddown long division; I would recommend keeping long division in your memory.1737
It is interesting; it is not that hard to remember; it is useful in any situation; and it is connected to some deep stuff.1743
And synthetic division is really only useful for this one specific situation.1748
So, it is really just a trick; I am not a big fan of tricks, because it is easy to forget them and easy to make mistakes with them.1751
But long division is connected to deep ideas, and it is already in your memory; you just have to figure out,1756
"How do I apply that same idea to a new format?"1761
All right, let's see some examples: Let f(x) = 2x^{3} + 4x^{2}  50x  100.1764
Use the fact that f(3) = 32, and f(1) = 48, to help you guess a root.1771
This sounds a lot like the intermediate value theorem: notice, 32 starts positive; this one is negative.1777
So, that means that between these two things, at 3, we are somewhere really positive.1782
At 1, we are somewhere really negative; so we know that somewhere on the way, it manages to cross; so we know we have a root there.1788
So, how are we going to guess it? Well, we might as well try the first thing that is in the middle of them.1796
So, let's give a try to f(2); now notice, there is no guarantee that f(2) is going to be the answer.1800
It could be f(2.7); it could be f(1.005); it could be something that is actually going to require square roots to truly express.1807
But we can get a better sense of where it is; and we are studentsthey are probably going to make it not too hard on us.1817
So, let's try 2; let's guess it; let's see what happens.1823
We plug in 2; we have 2 being plugged in, so (2)^{3} + 4(2)^{2}  50(2)  100.1826
OK, f(2) is going to be equal to 2 times...what is 2 cubed?1837
2 times 2 is 4, times 2 is 8; keep that negative sign; plus 4 times 2 squared (is positive 4);1842
minus 50 times 2; these will cancel out to plus signs; we will get 50 times 2, which is 100; minus 100.1853
So, those cancel out (100 + 100).1861
2 times 8 is 16, plus 4 times 4 is 16; they end up being not too difficult on us.1865
And sure enough, we get = 0; so we just found a root: f(2) = 0, so we have a root, or a zero, however you want to say it, at x = 2.1873
Great; there is our answer.1886
All right, Example 2: f(x) is an evendegree polynomial, and there exists some a and b,1887
such that f(a) and f(b) have opposite signs (one positive, the other negative).1893
Why is it impossible for f, our polynomial, to have just one root?1897
OK, so to do this, we need to figure out how we are going to do it.1902
Well, we first think, "Oh, one positive; the other negative; that sounds a lot like the intermediate value theorem that they just introduced to us."1905
So, it is likely that we are going to end up using that.1912
Let's think in terms of that: f(a) and f(b)...we have two possibilities: f(a) could be positive, while f(b) is negative;1914
or it could end up being the case that f(a) is the negative one, while f(b) is the positive one.1924
They didn't tell us which one; so we have to think about all of the cases.1932
Now, how can we see what this is?1935
We have an evendegree polynomial...let's start doing this by drawing.1937
We could have a world where we have a positive a (there is some a here, and then some b here,1941
where it is negative); and then we could also have another world where we have f(a) start as being negative somewhere,1951
and then f(b) is positive somewhere; they don't necessarily have to be on opposite sides of the yaxis.1961
But we are just trying to get an idea of what it is going to look like.1966
We know that we are somewhere on the left, and then we go up to the right.1968
So, how is this going to work?1972
And then, we could draw in a polynomial now; we could try to draw in a picture.1973
And we might say, "OK, we have a polynomial coming like this."1978
And we remember, "Oh, the polynomial could also come from the bottom."1981
So, now we have another two possibilities.1984
The polynomial is coming from up, or the polynomial is coming from down.1987
There is poly_{up}, if it is coming from above and then going down, or the polynomial that starts below, so poly_{down}.1992
It is coming from the bottom part, and it is going up.2001
And maybe that was a little bit confusing as a way to phrase it; but we have one of two possibilities.2004
The polynomial is coming in from either the top, or it is coming in from the bottom; polynomial at the top/polynomial at the bottom.2007
Those are our two possibilities; OK.2016
So, we could be coming from the top; and let's put in our points, as well, againthe same points, positive to negative.2020
Or we could be coming from the bottom, like this.2026
Then, on our negative to positive, we could have negative down here and positive here.2030
And once again, we could be coming from the top or...oops, I put that on the higher one...we could be coming from the bottom.2037
So now, let's see how it goes.2047
Well, we know for sure that the polynomial has to end up cutting through here, because we are told that it has that value.2048
And then, it has to also cut through here.2054
In this one, it cuts through here; and then, it has to get somehow to here, so it cuts through here.2057
And that is basically the idea of the theorem: we are coming from the bottom, and we go up and come down.2063
So, we have already hit more than one root; we have two roots minimum here already.2068
What about this one? Well, we go down, and now we have to go up to this one, as well.2075
So, we go up, and we are done already; we have two roots for this one.2081
In this one, we come up; we go through this one, and then we come up; and we go through this one.2084
So, these are the two where we still are unsure what has to happen next.2089
Well, remember: what do we know about evendegree polynomials?2093
They mentioned specifically that it is an even degree; even degree always means that the two ends,2096
if we have it going down this way...then it means that on the right side, it goes down here, as well.2104
On the other hand, if it goes up on the left side, then it has to go up on the right side.2109
Hardtosee yellow...we will cover that with a little bit of black, so we can make sure we can see it.2112
So, if we are an even degree, they have to be the same direction on both the left and the right side.2118
They could both go down, or they could both go up; but it has to, in the end, eventually go off in that way.2123
So, who knows what happens for a while here?2128
It could do various stuff; but eventually, at some point later on, it has to come back down,2131
which means that it has to end up crossing the xaxis a second time.2136
So, this one has to be true, as well; the same basic idea is going on over here.2140
Who knows what it is going to do for a while.2147
But because it is an evendegree polynomial, we know it eventually has to do the same thing.2148
So, it is going to have to come back up; so it is going to cross here and here; so it checks out.2153
All of our four possible cases, +/ or /+, combined with coming from the top or coming from the bottom2158
the four possible casesno matter what, by drawing out these pictures, we see that it is impossible,2164
because it is either going to have to hit the two just to make it there;2168
or because it has the even degree, it is going to be forced to come back up and reverse what it has done previously.2172
And we are getting that from the intermediate value theorem.2178
Great; Example 3: Let f(x) = x^{5}  3x^{4} + x^{3}  20x + 60, and d(x) = x  3.2182
And we want to use synthetic division to find f(x)/d(x).2191
All right, the first thing to notice: x^{5}, x^{4}, x^{3}...there is no x^{2}!2195
So, we need to figure out what x^{2} is, so we can effectively put it in as 0x^{2}.2202
Remember: the coefficient that must be on the x^{2} to keep it from appearing, to make it disappear, is a 0.2207
So, what we really have is the secret 0x^{2} + 60, because we have to have coefficients2213
for every single thing, from the highest degree on down, to use synthetic division.2218
So, what is our k? Well, it is x  k for synthetic division; so our k equals 3.2223
We have 3 here; and now we just need to place in all of our various coefficients.2229
Our various coefficients: we have a 1 at the front; we have a 3 in front of x^{4};2235
we have a hidden 1 in front of the x^{3}; we have a 0 on our completelyhidden x^{2}.2241
We have a 20 on our x, and we have a 60 on the very end for our constant.2247
That is all of them; we have made it all the way out to the constant.2254
So remember, on vertical arrows, when we go down, we add.2257
So, it is adding on vertical arrows: 1 + _ becomes just 1; and then, on these, it is multiplying by whatever our k is.2260
So, 1 times 3...we get 3; we add 3 and 3; we get 0; 0 times 3...we get 0 still;2271
0 + 1 is 1; 1 times 3 is 3; 0 + 3 is 3; 3 times 3 is 9; 20 + 9 is 11; 11 times 3 is 33; positive 27.2279
Now, remember: the very last spot is always the remainder; so this right here is our remainder of 27.2291
From there on, 11 is our constant; we work from the right to the left.2299
Then come our x coefficient, our x^{2} coefficient, our x^{3} coefficient, and our x^{4} coefficient.2304
And it makes sense that it is going to be one degree lower on the thing that eventually comes out of it, the quotient.2311
So, we write this thing out now; we have x^{4} + 0x^{3}, so we will just omit that;2316
plus 1x^{2} + 3x  11; but we can't forget that remainder of 27.2323
So, we have a remainder of + 27; but the remainder has to be divided, because that is the one part where it didn't divide out evenly.2331
So, 27/(x  3)that is what we originally divided by; this is f(x)/d(x).2338
Great; and that is our answer, that thing in parentheses right there.2348
And now, if we wanted to do a check, we could come by, and we could multiply by x  3.2351
If you divide out the number, and then you multiply it back in, you should be exactly where you started.2358
So, we multiply by x  3; for x^{4}, we get x^{5}; minus 3x^{4}...x^{2}...2362
+ x^{3}  3x^{2} + 3x(x)...+ 3x^{2}...+ 3x(3), so  9x; 11(x), so 11x...2371
minus 3 times...11 times 3 becomes positive 33; and then finally, all of 27/(x  3) times (x  3)...2384
As opposed to distributing it to the two pieces, we say, "x  3 and x  3; they cancel out," and we are just left with 27.2394
Now, we work through it, and we check out that this all works.2400
So, we have x^{5}; there are no other x^{5}'s, so we have just x^{5} that comes down.2404
3x^{4}...do we have any other x^{4}'s?...no, no other x^{4}'s, so it is  3x^{4}.2408
x^{3}: do we have any other x^{3}'s?...no, no other x^{3}'s, so it is + x^{3}.2414
 3x^{2}: are they any others?...yes, they cancel each other out, so it is 0x^{2}.2420
 9x  11x; that becomes  20x; let's just knock them out, so we can see what we are doing; + 33 + 27 is + 60.2425
Great; we end up getting what we originally started with; it checks out, so our answer in red is definitely correct.2433
So, remember: that remainder is the one thing where it didn't come out evenly, so it has to be this "divide by,"2442
whatever your remainder is here, divided by the thing you are dividing by,2449
because it is the one thing that didn't come out evenly.2453
All right, the final example: Find all roots of x^{4}  2x^{3}  11x^{2}  8x  602456
by using the fact that x^{2} + 4 is one of its factors.2462
The first thing that is going to make this easier for us: if we want to keep breaking this down into factors,2466
if we want to find the answers, if we want to find what it iswe have to factor it, so that we can get to the roots.2469
So, we want to factor this larger thing: we know that we can pull out x^{2} + 4.2476
If we are going to pull it out, can we use synthetic division?...no, because it is not in the form x  k.2481
We have this x^{2}, so we have to use polynomial long division.2486
So, x^{2} + 4: we plug in x^{4}  2x^{3}  11x^{2}  8x  60.2490
Great; x^{2} + 4 goes into x^{4}...oh, but notice: do we have an x in here?2502
We don't, so it is once again + 0x; so let's rewrite this; it is not just x^{2} + 4.2509
We can see this as a threetermed thing, where one has actually disappeared; + 0x + 4.2514
Notice that they are the same thing; but it will help us see what we are doing.2520
How many times did x^{2} go into x^{4}? It goes in x^{2}.2523
But we don't put it here; we put it here, where it would line up for three different terms: 1 term, 2 terms, 3 terms.2525
It lines up on the third term over here, 11x, so it will be x^{2} here.2534
x^{2} times (x^{2} + 0x + 4): we get x^{4}; this is just blank still; + 4x^{2}.2540
Now, we subtract by that; we put our subtraction onto both of our pieces, so now we are adding.2549
x^{4}  x^{4} becomes 0; 11x^{2}  4x^{2} becomes 15x^{2}.2554
We bring down our 2x^{3}; we bring down our 8x; so we have everything.2560
2x^{3}  15x^{2}  8x: once again, we just ask,2566
"How many times does the first term go into the first term here?"2572
2x^{3} divided by x^{2} gets us just 2x^{2}.2575
Oops, I'm sorry: we are dividing by x^{2}, because 2 is just x^{1}.2582
So, 2 times x^{2} gets us 2x^{3}; and then, 2x times 4 becomes 8x.2586
We subtract by this; we distribute to start our subtraction, so that becomes positive; that becomes positive.2594
Now we are adding: 2x^{3} + 2x^{3} becomes 0.2599
8x + 8x becomes nothing; and now we bring down the thing that didn't get touched, the 15x^{2} and the 60.2603
And we have 15x^{2}  60; and hopefully, it will line up perfectly.2612
In fact, we know it has to line up perfectly, because we were told explicitly that it is one of the factors.2617
So, there should be no remainder; otherwise, something went wrong.2621
x^{2} + 0x + 4; how many times does that fit into 15x^{2}  60?2624
Once again, we just look at the first part: 15x^{2} divided by x^{2} becomes just 15.2628
15x^{2}...multiplying it out...4 times 15, minus 60; we now subtract by all that.2634
Subtraction distributes and cancels those into plus signs.2640
We add 0 and 0; we have a remainder of 0, which is good; that should just be the case, because we were told it was a factor.2643
So, we get x^{2}  2x  15.2650
What is our polynomialwhat is another way of stating this polynomial?2654
We could also say this as (x^{2} + 4)(x^{2}  2x  15).2658
Let's keep breaking this up: x^{2}  2x  15...how can we factor that?2666
x^{2} + 4...can we factor that any more?...no, we can't; it is irreducible.2670
If we were to try to set that to 0, we would have to have x^{2}, when squared, become a negative number.2674
There are no real numbers that do that, so that one is irreducible; we are not going to get any roots out of that; no real roots were there.2679
x^{2}  2x  09 how can we factor this?2685
Just 1 is in front of the x^{2}, so that part is easy; it is going to be x and x, and then... what about the next part, 15?2688
We could factor that into...one of them is going to have to be negative;2695
we factor it into 5 and 3; 5 and 3 have a difference of 2, so let's make it 5 and +3.2697
We check that: x^{2} + 3x  5x...2x; 5 times 3 is 15; great.2704
So, at this point, we set everything to 0; x^{2} + 4 will provide no answers.2710
x^{2} + 4 = 0...nothing there; there are no answers there.2715
x  5 = 0; turn this one inthat gets us x = 5; x + 3 = 0 (this gives us all of the roots): x =  3.2721
Our answers, all of the roots for this, are x = 3 and 5.2732
And we were able to figure this by being able to break down a much more complicated polynomial2737
into something that was manageable, something we can totally factor; and we had to do that through polynomial long division.2741
All right, coolI hope you got a good idea of how this all works.2746
Just remember: polynomial long division is probably your best choice.2748
Just think of it the same way that you approach just doing normal long division with plain numbers that you did many, many years ago.2751
It is basically the same thing, just with a slightly different format.2758
How many times does it fit in? Multiply; subtract; repeat; repeat; repeat; get to a remainder.2761
All right, we will see you at Educator.com latergoodbye!2766
1 answer
Last reply by: Professor SelhorstJones
Sat Oct 24, 2015 1:06 AM
Post by Fadumo Kediye on October 18, 2015
P(x) = 2x^3 + ax^2 +bx + 6 is divided by x + 2, the remainder is 12. If x  1 is a factor of the polynomial, find the values of a and b.
1 answer
Last reply by: Professor SelhorstJones
Mon Aug 26, 2013 11:12 PM
Post by Humberto Coello on August 23, 2013
How are there no roots for X2 + 4 = 0
isn't it equal to (x+2)(x2)=0
and don't we get the roots x=2, x=2 from it?