For more information, please see full course syllabus of Pre Calculus
For more information, please see full course syllabus of Pre Calculus
Roots (Zeros) of Polynomials
 The roots/zeros/xintercepts of a polynomial are the xvalues where the polynomial equals 0.
 If you have access to the graph of a function/equation, it is very easy to see where the roots are: where the graph cuts the horizontal axis (the xintercepts)! Why? Because that's where f(x) = 0 or y=0.
 While you can occasionally find the roots to a polynomial by trying to isolate the variable and directly solve for it, that method often fails or is misleading.
 To find the roots of a polynomial we need to factor the polynomial: break it into its multiplicative factors. Then we can set each factor to 0 and solve to find the roots.
 Factoring can be quite difficult if you're trying to factor a very large or complicated polynomial. There is no procedure that will work for factoring all polynomials.
 In general, if we have a quadratic trinomial (something in the form ax^{2} + bx + c), we can factor it into a pair of linear binomials as
ax^{2} + bx + c = ( x + ) ( x + ).  Whenever you're factoring polynomials, make sure you check your work! Even on an easy problem, there are ample opportunities to make a mistake. That means you should always try expanding the polynomial (it's fine to do it in your head) to make sure you factored it correctly.
 In general, factoring higher degree polynomials is similar to what we did above. Figure out how you can break down the polynomial into a structure like the above, then ask yourself how you can fill in the blanks.
 If you already know a root to a polynomial, it must be one of its factors. For example, if we know x=a is a root, then the polynomial must have a factor of (x−a). This makes factoring the polynomial that much easier.
 Not all polynomials can be factored. Sometimes it is impossible to reduce it to smaller factors. In such a case, we call the polynomial irreducible. [Later on, we'll discuss a a hidden type of number we haven't previously explored when we learn about the complex numbers. These will allow us to factor these supposedly irreducible polynomials. However, for the most part, we won't work with complex numbers so such polynomials will stay irreducible.]
 There is a limit to how many roots/factors a polynomial can have. A polynomial of degree n can have, at most, n roots/factors.
 We also get information about the possible shape of a polynomial's graph from its degree. A polynomial of degree n can have, at most, n−1 peaks and valleys (formally speaking, relative maximums and minimums).
Roots (Zeros) of Polynomials
 The zeros of f(x) are all those x where f(x)=0. That means we want to find all the x such that
0 = x^{2}+2x − 15.  We do this by factoring the polynomial. Break the polynomial into factors. Because we have a quadratic, we know it will probably break into factors of the form
( x + ) ( x + ) .  We know that the coefficients in the front blanks for each parenthetical must multiply to make 1, because that is the coefficient in front of x^{2}. We know the numbers in the other blanks must multiply to make −15. Finally, we know that when we add those numbers together, they must make 2 because we have 2x.
 Working through this, we find (x−3)(x+5) works, which we can check. If we expand those factors, they do indeed come out to be x^{2} + 2x −15, so we know the factoring is correct.
 Once we have 0=(x−3)(x+5), we can find the values for x. We set each of the factors equal to 0. [We can do this by the same logic that 0 = a·b means a and/or b must be 0.]
0 = x−3 0 = x+5
 The roots of g(t) mean the same thing as the zeros: all those t where g(t)=0. That means we want to find all the t such that
0 = t^{2} − 7t − 18.  We do this by factoring the polynomial. Break the polynomial into factors. Because we have a quadratic, we know it will probably break into factors of the form
( t + ) ( t + ) .  We find (t+2)(t−9) works, which we can check. If we expand those factors, they do indeed come out to be t^{2} − 7t − 18, so we know the factoring is correct.
 Once we have 0=(t+2)(t−9), we can find the values for t. We set each of the factors equal to 0. [We can do this by the same logic that 0 = a·b means a and/or b must be 0.]
0 = t+2 0 = t−9
 Because we're trying to solve something involving polynomials, we will need to use factoring. However, before factoring can be useful, we need to get one side equal to 0. [This is because we want to eventually have something of the form 0 = a·b and then say 0=a and 0=b.]
 Use algebra to get one side of the equation equal to 0. We obtain
x^{2} + 15x + 50 = 0.  Factor the left side to get
(x+10)(x+5) = 0.  Now that we have it in the form of 0 equals things multiplied together, we can set each of those things equal to 0:
0 = x+10 0 = x+5
 The zeros of h(x) are all those x where h(x)=0. That means we want to find all the x such that
0 = x^{3} − 9x^{2} + 23x − 15.  We do this by factoring the polynomial. Break the polynomial into factors. Start off easy by trying to break it into something of the form
( x + ) ( x^{2} + x + ).  While there are other possibilities (depending on which factor we pull out first), we might get
[With something difficult like this, it's even more important to check your work and expand the polynomial in your head to be sure you didn't make a mistake.](x−1)(x^{2}−8x+15).  Now factor the remaining x^{2}−8x+15 to obtain
(x−1)(x−3)(x−5).  We now have 0 = (x−1)(x−3)(x−5). Set each of the three factors to 0 and solve:
0 = x−1 0 = x−3 0 = x−5
 If you know a root, that automatically implies a factor. A root of x=a implies a factor of (x−a). Thus, since we know x=2 is a root of the polynomial, we know (x−2) is a factor, which will help us in our factoring.
 Pulling out (x−2), we know we'll have something of the form
now we just need to appropriately fill in the blanks.(x−2) ( x^{2} + x + ),  Begin by filling in the blanks that are easiest. You know the coefficient on x^{2} must be 1 because in the fully expanded polynomial it is x^{3}. Furthermore, you know the blank for the constant must be 5 because in the fully expanded polynomial you have −10. Then use that information to figure out the middle blank. Eventually you obtain
(x−2)(x^{2} + 6x +5).  Factor the right parenthetical to finish factoring the entire polynomial:
(x−2)(x+1)(x+5)  Thus, to find the roots, we have 0 = (x−2)(x+1)(x+5). Set each of these equal to 0, then solve:
0 = x−2 0 = x+1 0 = x+5
 If you know a root, you automatically know a factor. Remember, a root/zero is where the function equals 0. Thus, since we know f(3) = 0 and f(−2) = 0, we know p=3 and p=−2 are roots. A root of p=a implies a factor of (p−a). Therefore we already know two roots: (p−3) and (p+2).
 If we pull out those two factors, we'll have something of the form
(p−3)(p+2) ( p^{2} + p + ).  As it stands, it's a little difficult to fill in the blanks. To make it easier to see how the blanks should be filled in, expand the left side:
(p^{2} − p − 6) ( p^{2} + p+ )  Working it out, we obtain
(p^{2} − p − 6) (2p^{2}−6p−8).  Finish factoring the above:
[There are actually other ways to factor the right side. You could also obtain(p−3)(p+2)(2p−8)(p+1)
Still, whatever way you factor it, you will wind up getting the same answer for the roots.](p−3)(p+2)(p−4)(2p+2) or 2·(p−3)(p+2)(p−4)(p+1)  We now are looking to solve 0 = (p−3)(p+2)(2p−8)(p+1). Set each factor to 0:
0 = p−3 0 = p+2 0 = 2p−8 0 = p+1
 Each root implies a factor where the constant has the opposite sign:
x=−4, 3, 9 ⇒ (x+4), (x−3), (x−9)  If we multiply them all together, we have a polynomial with degree 3:
(x+4)(x−3)(x−9)  OPTIONAL: You could expand the polynomial if you wanted (or if a problem required it).
 OPTIONAL: If you wanted, you could multiply the entire polynomial by a constant number because it has no effect on the roots. Notice that a·(x+4)(x−3)(x−9) has the exact same roots, because the same values for x will cause the expression to become 0.
[Additionally, it would also be correct to have a constant multiple of the above, such as −2·(x+4)(x−3)(x−9) = −2x^{3}+16x^{2} +42x −216.]
 Each root implies a factor where the constant has the opposite sign:
x=2, 5 ⇒ (x−2), (x−5)  However, if we just multiply them all together, we only have a polynomial with degree 2:
(x−2)(x−5)  We need to increase the degree of the polynomial, but without adding any additional roots. We can do this by adding multiplicity to a single root. Have the same root show up multiple times in the expansion, such as
In such a way, we now have a polynomial of degree 4.(x−2)(x−5)(x−5)(x−5).  OPTIONAL: You could expand the polynomial if you wanted (or if a problem required it).
 OPTIONAL: If you wanted, you could multiply the entire polynomial by a constant number because it has no effect on the roots.
 OPTIONAL: Instead of just using multiplicity and having roots appear more than once, you could also use a polynomial factor that is irreducible, such as (x^{2} +1). Because we cannot solve 0=x^{2}+1 in the real numbers, it will not add any additional roots. For example, we could have a polynomial like
(x−2)(x−5)(x^{2}+1)


 The maximum number of roots/factors a polynomial can have is based on its degree. A polynomial of degree n can have, at most, n roots/factors. Notice that this is a maximum: there is no guarantee it will have that many, we can only be sure it will not have more than that.
 The degree of the polynomial in this problem is 5, so the maximum number of roots is 5.
 In actuality, if you factor this polynomial, it does not have 5 roots. It only has 3 roots because it factors to
While (x−3), (x−7), and (x+2) all produce roots [x=3, 7, −2], the factor (x^{2}+1) does not produce any. Why? Because x^{2}+1=0 cannot be solved with any real numbers, so it is irreducible. We will learn more about this idea in later lessons.(x^{2}+1)(x−3)(x−7)(x+2).
[If you're curious about the precise number of roots that it has, check out the final step for this problem.]
 Begin by setting up the varaibles:
w = width l = length  Create equations from the information given:
w ·l = 66 w+5 = l  Notice that we can plug in w+5 for l in the area equation to produce:
w (w+5) = 66  This is a quadratic equation, so we can solve it by factoring. Begin by getting 0 on one side of the equation, then factor it:
w (w+5) = 66 ⇒ w^{2} +5w − 66=0 ⇒ (w+11)(w−6) = 0  We have two possibilities for w: w = −11 and w=6. Notice that a negative value for length makes no sense, so we throw out that extraneous solution. This leaves us with w=6.
 Once we know the width, we can find the length. Don't forget to put units in your answer.
*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
Roots (Zeros) of Polynomials
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
 Roots in Graphs
 Naïve Attempts
 Factoring: How to Find Roots
 Factoring: How to Find Roots CAUTION
 Factoring is Not Easy
 Factoring Quadratics
 Factoring Quadratics, Check Your Work
 Factoring Higher Degree Polynomials
 Factoring: Roots Imply Factors
 Not all Polynomials Can be Factored
 Max Number of Roots/Factors
 Max Number of Peaks/Valleys
 Max, But Not Required
 Example 1
 Example 2
 Example 3
 Example 4
 Intro 0:00
 Introduction 0:05
 Roots in Graphs 1:17
 The xintercepts
 How to Remember What 'Roots' Are
 Naïve Attempts 2:31
 Isolating Variables
 Failures of Isolating Variables
 Missing Solutions
 Factoring: How to Find Roots 6:28
 How Factoring Works
 Why Factoring Works
 Steps to Finding Polynomial Roots
 Factoring: How to Find Roots CAUTION 10:08
 Factoring is Not Easy 11:32
 Factoring Quadratics 13:08
 Quadratic Trinomials
 Form of Factored Binomials
 Factoring Examples
 Factoring Quadratics, Check Your Work 16:58
 Factoring Higher Degree Polynomials 18:19
 Factoring a Cubic
 Factoring a Quadratic
 Factoring: Roots Imply Factors 19:54
 Where a Root is, A Factor Is
 How to Use Known Roots to Make Factoring Easier
 Not all Polynomials Can be Factored 22:30
 Irreducible Polynomials
 Complex Numbers Help
 Max Number of Roots/Factors 24:57
 Limit to Number of Roots Equal to the Degree
 Why there is a Limit
 Max Number of Peaks/Valleys 26:39
 Shape Information from Degree
 Example Graph
 Max, But Not Required 28:00
 Example 1 28:37
 Example 2 31:21
 Example 3 36:12
 Example 4 38:40
Precalculus with Limits Online Course
Transcription: Roots (Zeros) of Polynomials
Welcome back to Educator.com.0000
Today, we are going to talk about roots (also called zeroes) of polynomials.0002
We briefly went over what a root is when we talked about the properties of functions.0006
But let's remind ourselves: the zeroes of a function, the roots of an equation, and the xintercepts of a graph are all the same thing.0009
They are inputs (which are just xvalues) where the output is 0; it is where it comes out to be 0things that will make our expression give out 0.0017
The roots, zeroes, xintercepts of f(x) = x^{2}  1 and y = x^{2}  1 are x = 1 and x = +1,0026
because those two values, f(x) and y, are equal to 0.0036
If we plug in 1, (1)^{2} will become positive 1, minus 1 is 0.0040
If we plug in positive 1, 1^{2} is 1, minus 1 is 0.0045
Those two things cause it to give out 0, and the roots of the polynomial are the xvalues where the polynomial equals 0.0049
The roots of the polynomial x^{2}  1 are 1 and positive 1.0055
Being able to find roots in functions is important for many reasons; and it will come up very often when you are working with polynomials.0062
If you continue on to calculus, you will see how roots can be useful for finding lots of information about a function.0068
So, it is very important to have a grasp of what is going on there and be able to find roots.0073
Roots in graphs: if you have access to the graph of a function/equation, it is very easy to see where the roots are.0078
Of course, you might not see precisely, because it is a graph, after all, and it might be off by a little bit.0083
But you can get a very good sense of where they are: it is where the graph cuts the horizontal axisthe xintercepts.0088
Why? Because here we have f(x) = 0 or y = 0, depending on if it is a function or an equation.0094
Since that means our height is at 0 there, then every place where we cross the xaxis must be a rootit's as simple as that.0101
This also gives us a nice mnemonic to remember what the word "root" means0111
it can be a little hard to remember the word "root," since we aren't that used to using it.0115
But we could remember it as where the equation or the function grows out of the xaxiswhere it is 0.0118
It is like it is the ground; think of it as a plant rooted in the ground.0125
A function or equation has its roots in the xaxis; a tree has its roots in the earth, and a function has its roots in a height of 0.0129
So, a root is where it is growing up and down; it is where it is held in our plane, held in our axes.0140
And that is one way to remember what a root is.0150
How do we find the roots of a polynomial? Well, at first we might try a naive approach and attempt to solve the way we are used to.0153
Naive is just what you have done beforewhat seems to make sensewithout ever really having had a whole lot of experience about it.0160
So, the naive attempt would be probably to just isolate the variable on one side.0166
That is what we did with a bunch of other equations before, so let's do it again.0170
Now, in some cases, this will actually work, and we will find all of the real solutions.0173
For example, if we have f(x) = x  3, then we set it equal to 0, because we are looking for the roots; we are looking for when 0 = x  3.0176
We move that over, and we get x = 3; great.0184
Or if we had y = x^{3} + 1, we set that to 0, and we have 0 = x^{3} + 1,0187
because we are looking for what x's cause us to have 0 = x^{3} + 1.0192
So, we have x^{3} = 1; we take the cube root of both sides, and we get x = ^{3}√1, which is also just 1.0197
So, in both of these cases, the naive method of isolating for variables worked just fine.0204
But that is definitely not going to be the case for all situations.0208
The naive method of isolation will fail us quite quickly, even when used on simple quadratic polynomials.0212
Consider f(x) = x^{2}  x  2that is not a very difficult one.0217
But this method of trying to isolate will just fail us utterly if we use it here.0223
So, 0 = x^{2}  x  2...we might say, "Well, let's get the numbers off on one side."0227
We have 2 = x^{2}  x; but then, we don't have just x; so let's pull out an x; we get 2 = x(x  1).0232
And well, we are not really sure what to do now; so let's try another way.0240
0 = x^{2}  x  2; let's move x over, because we are used to trying to get just x alone.0243
So, we have x here; but then we also have x^{2} here; so let's divide by x.0248
We get 1 = (x  2)/x; once again, we are not really sure what to do.0252
Let's try again: 0 = x^{2}  x  2, so let's move over everything but the x^{2}.0256
Maybe x^{2} is the problem; so we will get x + 2 = x^{2}.0262
We take the square root of both sides; we remember to put in our ± signs: ±√(x + 2) = x.0265
I don't really know how to figure out what x's go in there to make that true.0271
So, in all three of these cases, it is really hard to figure out what is going on next.0274
If we are going to try to isolate, we are going to get these really weird things.0280
This method of isolation that we are used to isn't going to work here, because we can't get x alone; we can't get the variable alone on one side.0283
It is not going to let us find the roots of polynomials if we try to isolate.0290
But at least we could trust it in those previous examples; we saw that we can trust it when it does work.0293
So, we might as well try it firstno, it is even worse: the naive method of isolation0298
can make us miss answers entirely, even though we think we will know them all.0303
So, we will think we have found the answers; but in reality, we will only have found part of the answers.0307
Consider these two ones: if we have 0 = x^{2}  1, we move the 1 over; and then we take the square root of both sides.0311
The square root of 1 is 1; the square root of x^{2} is x; great.0317
For this one over here, we have 0 = p^{2} + 3p, so we realize that we can divide both sides by p.0320
And since 0/p is just 0, we have 0 = p^{2}/p (becomes p); 3p/p becomes 3; so we have 0 = p + 3.0327
We move the 3 over; we get p = 3; greatwe found the answers.0336
Not quite: those above things are solutions, but in each case, we have missed something.0340
We have been tricked into missing answers by trying to follow this naive method.0345
The other solutions for this would be x = 1 and p = 0, respectively.0349
The mistakes that we forgot were a ± symbol over on this one; we forgot to put a ± symbol when we took the square root;0353
and then, the other one was dividing by 0, because when we divided, we inherently forgot0361
about the possibility that, if we were actually dividing by 0, we couldn't divide by 0.0366
So, those are the two mistakes; but even if you personally wouldn't have made those same mistakes,0371
this example shows how it is easy to forget those things in the heat of actually trying to do the math.0375
You might forget about that; you might accidentally make one of these mistakes; so it is risky to try this method of isolation.0381
We need something that works better.0387
We find the roots of the polynomial by factoring; we break it into its multiplicative factors.0389
Let's look at how this works on the example that we couldn't solve with naive isolation.0395
We have f(x) = x^{2}  x  2; we have 0 = x^{2}  x  2, because we are looking for when f(x) is equal to 0.0399
And then, we say, "Let's factor it; let's break it into two things."0407
So, we have (x  2) and (x + 1); and if we check, that does become it: x times x becomes x^{2}; x times 1 becomes + x;0409
2 times x becomes  2x; 2 times 1 becomes 2; so yes, that checks out to be the same thing as x^{2}  x  2.0417
0 = x  2 and 0 = x + 1 is how we now set these two things equal to 0; we have 0 = x  2 and 0 = x + 1.0425
And we get x = 2 and x = 1, and we have found all of the solutions for this polynomial.0435
Why does this work, though? We haven't really thought about why it works.0441
And we don't want to just take things down and automatically say, "Well, my teacher told me that, so that must be the right thing."0444
You want to understand why it is the right thing.0449
Teachers can be wrong sometimes; so you want to be able to verify this stuff and say, "Yes, that makes sense,"0451
or at least have them explaining and saying, "Well, we don't understand quite enough yet;0456
but later on you will be able to see the proof for this"; you really want to be able to believe these things,0460
beyond just having someone tell you by word of mouth.0463
So, to figure out why this has to be the case, we will consider 0 = ab.0466
The equation is only true if a or b, or both of them, is equal to 0; if neither a nor b is equal to 0, then the equation cannot be true.0473
If a = 2 and b = 5, then we get 10, which is not equal to 0.0482
As long as a or b is 0, it will be true, because it will cancel out the other one.0489
But if both of them are not 0, then it fails, and it is not going to be 0.0494
It is the exact same thing happening with x^{2}  x  2; we have 0 = x^{2}  x  2,0499
which we then factor into (x  2) and (x + 1); so let's use two different colors, so we can see where this matches up.0505
We are pairing this to the same idea of the a times b equals 0; it is (x  2) (x + 1) = 0.0513
The only way that this equation can be true is if x  2 = 0 or x + 1 = 0.0520
Just as we showed up here, it has to be the case that a or b equals 0 for that to be true.0528
So, it must be the case that either (x  2) or (x + 1) equals 0, if this is going to be true.0534
So, our solutions are when either of the two possibilities is trueif the possibility is true, if one of them is true, then the whole thing comes out.0539
So, either case being true makes it acceptable; that gets us 2 (keeping with our color coding) and 1 as the two possibilities.0546
So, by breaking x^{2}  x  2 into its factors, we can find its roots.0557
So, this is how we find polynomial roots, in general: the first thing we do is set the whole thing as 0 = polynomial.0562
We have to have some polynomial, and then it is 0 equals that polynomial.0569
The next thing we do is factor it into the smallest possible factors; we break it down into multiplicative factors.0573
And then finally, we set each factor equal to 0, and we solve for each of them.0580
So, in step number 2, we are going to get things like 0 = (x + a)(x + b)(x + c)(x +d) and so on, and so on, and so on.0585
And then, in step 3, we set each factor to 0; so we get things like x + a = 0, at which point we can solve and say, "Oh, x = a."0595
That is one of our possible solutions; and from there, you can work out all of the roots of the polynomial.0604
Cautionthis is very important: notice that it is extremely, extremely important to begin by setting the equation as 0 = polynomial.0610
I have seen lots of mistakes where people forgot to set it as 0 = polynomial.0618
If it isn't, if it was something like 5 = (x  2)(x + 1), we can't solve for the solutions from those factors.0621
Those factors are now meaningless; they aren't going to help us.0628
We need the special property that 0 turns everything it multiplies into 0.0631
Without that special property, this method just won't work.0635
Consider if we had something like 5 = ab; there is no way that we could just figure out what the answers are here.0638
It is not just simply that a has to be 0 or b has to be 0, because a could be 5 and b could be 1.0644
Or b could be 5 and a could be 1, or a could be 25 and b could be 1/5; or a could be 100 and b could be 1/20.0649
We have lots of different possibilitiesa whole spectrum of things; there are way too many possible solutions.0658
We need that special property of 0 = ab to be able to really say, "That thing is 0, or that thing is 0"; that is what we know for sure.0663
That is how we get useful information out of it.0671
That is why it has to be 0 = polynomial; if you don't set it up as that before you try to factor it,0674
before you try to do the other steps, you are just not going to be able to get the answer,0679
because we need that special property that 0 has when it multiplies other things.0682
0 multiplying something automatically turns it to 0; if we don't have that special property, things just won't work.0686
Factoring is not necessarily easy: say we have something like0693
x^{5} + 6.5x^{4}  17x^{3}  41x^{2} + 24x, and we want to know what the zeroes are.0697
Well, if we knew its factors, we would be able to break it into (x + 8)(x + 2) times x times (x  0.5) times (x  3).0703
And it would be really easy to figure out what the polynomial's roots are.0710
At this point, we say, "Well, great; x + 8 becomes x = 8; x + 2 becomes x = 2; x + 0 becomes x = 0;0712
x  0.5 becomes x = positive .5; and x  3 becomes x = positive 3great; I found it; it is really easy to find its roots."0722
But how do you factor a monster like that?0732
Ah, there is the problem: factoring can be quite difficult.0734
Luckily, by this point, you have been certainly practicing how to factor for years in your algebra classes.0738
By now, you have done lots of factoring; you are used to this; you have played with polynomials a bunch in previous math classes.0743
And all that work has a use, and it is here, finding roots; we can break things down into their factors and find roots.0748
But there is no simple procedure for factoring polynomials.0754
Once again, remember: if you were confronted by something like this, you would probably have a really difficult time0757
figuring out what its factors werefiguring out how you can break that down into factors.0762
There is no simple procedure for factoring polynomials.0768
Highdegree polynomials can just be very difficult to factor; happily, we are not going to really see such polynomials.0770
Most courses on this sort of thing don't end up giving you very difficult things to factor at this stage.0777
So, we won't really have to worry about factoring really difficult polynomials.0782
We will be able to stick to the smaller things.0787
So, let's have a quick review of how you factor small things like quadratics.0790
We are going to see a bunch of quadratics; they are very importantthey pop up all the time in science.0795
So, let's look at a brief review of how to factor a quadratic polynomial.0799
Remember, quadratic is a degree 2, and a trinomial just means 3 terms.0801
So, if we have a quadratic trinomial in its normal form, then we have ax^{2} + bx + c.0806
If we want to factor that, we will turn it into a pair of linear binomials: degree 1 and 2 terms (linear and binomial).0811
We would want to break this into (_x + _) (_x + _).0819
Great; to be equivalent to the above, the coefficients of x must give a product of a.0824
So, a has to come out with the red dot here, times the red dot here.0830
And the constants have to give a product of c; so the blue dot here times the blue dot here has to come out to be c.0835
Also, b has to add up from the products of the outer blanks and the inner blanks.0843
So, it has to be that the red dot here times the blue dot here, plus the blue dot here times the red dot here, comes up to be this b here.0848
So, it is mixed out of the two of them.0863
Don't worry if that is a little bit confusing right now; you will be able to see it as we work through examples.0865
The b has to add up to the products of the outer blanks and the inner blanks.0868
The a has to come from the first blanks, and the c has to come from the last blanks.0872
Don't worry about memorizing this, thoughit is just a sense of what is going on in practice.0876
Let's look at an example: we want to factor 2x^{2}  5x  12.0880
So, we know, right away, that we want to break it into the form (_x + _)(_x + _).0884
The first thing we notice is that we have this 2 at the front; and 2 only factors into 2 times 1.0890
We can't break it up into anything else really easily; so let's put 2 times 1 down as 2x times x.0895
We have to put the 2 somewhere; so it is either going to be (2x + _)(1x + _), or it is going to be the 2 over here and the 1 here.0901
It doesn't really matter what order we put it in; so we will put the 2 at the front.0910
We have (2x + _)(x + _); now, what is going to go into those other blanks?0913
Now, we need to factor the 12; so let's factor 12 first: we notice that 12 can break into 1 times 12, 2 times 6, or 3 times 4.0917
And one of the factors has to be a negative, because we have a negative in front of the 12.0926
So, they have to be able to multiply to make a 12; so there is going to have to be a negative on either the 1 or the 12.0931
One of those two will have to have a negative on it; we don't know which one, but it is going to be one of them.0938
Or it is going to be 2, or 6; and then finally, for the last pair, it would be 3 or 4.0943
I am not going to use all of these at once; we will have to figure out which one is right.0948
But one of them will have to be negative, because of this negative sign up here.0950
We start working through this; and we know that there is this 2x here at the front.0955
We have this 2x at the front; so it is going to multiply this one, and it is going to effectively double whatever we put here.0960
So, the difference between one of the numbers doubled and its sibling (the one here times this one in front of it) must be 5, because we get 5x.0967
We notice that 3  2(4)...2 times 4 is 8; 3 is 3; so the difference between those is 5.0979
So, we can set it up as 3  2(4) = 5; and we get (2x + 3)(x  4); and we have factored this out; we have been able to work it out.0986
Now, there are various ways, various tricks that have been taught to you in previous classes.0998
But the important thing is just to set up and have an expectation of what form you are trying to get.1005
And then, plug things in and say, "Yes, that would work; that would get me what I am looking for" or1009
"No, if I plug that in, that won't work; that won't get me what I am looking for."1013
As long as you work through that sort of thing, you will be able to find the answer eventually.1016
It is always a good idea, though, to check your work; you will find the answer, but it is really easy to make mistakes.1020
So, even on the easiest of problems (like the one we were just working on), there are lots of chances to make mistakes; trust me.1026
I make mistakes; everybody makes mistakes; the important thing is to catch your mistakes before they end up causing problems.1032
So, it means that you should always try expanding the polynomial after you have factored it, to make sure you factored correctly.1038
And it is OK to do this in your head; once you get comfortable with doing this sort of thing1044
(and by now, honestly, you probably have had enough experience with this that you can just do this in your head reasonably quickly),1048
it is OK to do it in your head; the important part is that you want to have some step where you are checking back on what you are doing.1053
So, either do it on the paper (if it is a long one) or do it in your head (if it is something that is short enough, easy enough, for you to check).1058
But you want to make sure that you are checking your work.1063
For example, if we have 2x^{2}  5x  12, we figured out that that breaks into (2x + 3) and (x  4).1065
But we want to check and make sure that it is right.1071
So, we check and make sure: 2x + 3 times x  4...we get 2x^{2}, and then 2x(4), 3x...we get 8x + 3x...1073
3 times 4...we get 12; we combine our like terms of 8x + 3x, and we get 2x^{2}  5x  12.1083
Sure enough, it checks out; we have what we started with, so we know that our factoring was correct; we did a good job.1092
Factoring higherdegree polynomials: in general, factoring polynomials of any degree is going to be similar to what we just did on these previous few slides.1100
The only difference is that it will become more complex as they become longer, as we get to higher and higher degrees.1107
So, for example, if we had something like a cubicif we had ax^{3} + bx^{2} + cx + d1112
we would probably want to set it up as (_x + _)(_x^{2} + _x + _).1118
If we are going to be able to break it up and factor it, it is going to have to factor into these two things, (_x + _)(_x^{2} + _x + _).1123
Notice also that this is a degree 1, and this is a degree 2; and when you multiply these two together, you will get back to a degree 3 over here.1131
Adding up the degrees on the right side has to be what we had on the left side.1141
We could also work on quartic, a degree 4; and we might try one of these two templates.1145
If we have ax^{4} + bx^{3} + cx^{2} + dx + e, we might break it1149
into (_x + _)(_x^{3} + _x^{2} + _x + _); or we might break it into (_x^{2} + _x + _)(_x^{2} + _x + _).1153
And so on and so forth...clearly, this is going to become more difficult as we get to higher and higher degrees.1163
The higher the degree of a polynomial, the more complicated our template is going to have to be for where we are going to fit things in.1168
The more choices we are going to have, the more difficult it is going to be to do this.1173
Luckily, we are only occasionally going to need to factor cubics, things like this where it is degree 3.1177
And we are very, very seldom going to see anything of higher degree.1182
So, don't worry too much about having really difficult ones.1186
But just be aware that factoring really large, highdegree polynomials can actually be pretty difficult to do.1188
Roots imply factors: we have this useful trick if we want to break out these higher polynomials.1196
If we know one of a polynomial's roots, we automatically know one of its factors.1201
Remember: one use of finding the factors of a polynomial is to find its roots.1206
If you find a factor of the form (x  a), then you set that equal to 0, and you know that that is going to be x = a.1209
It turns out that the exact opposite is true; if we know a polynomial has a root at x = a, then it also means we have a factor of x  a.1216
So, if we have a root x = a, then that turns into a factor, x  a, just because of this equation here,1225
where we are setting that factor equal to 0, which gives us the root.1232
We won't prove this; it requires a little bit of difficult mathematics, and some things that we actually haven't covered in this course yet.1236
But we can see it as a theorem: Let p(x) be a polynomial of degree n; then if there is some number a,1242
such that p(a) = 0 (that is to say that a is a root of our polynomial, pif we plug a in, we get 0),1250
then there is some way to break it up so that it is p(x) = (x  a), our factor (x  a) that we know1257
from our root at x = a, times q(x), where q is some other polynomial of degree n  1,1264
because this here is degree 1; so when we multiply it by a degree n  1, we will be back up to our degree n polynomial that we originally had.1272
If we manage to find one or more roots, sometimes the problem will give them to use; other times, we will get lucky, and we might just guess one.1285
This theory means we automatically know that many factors of the polynomial.1291
For example, say we know that p(x) = x^{3}  2x^{2}  13x  10.1295
And we are told that p(5) = 0; we know that 5 is a root.1300
Then, we automatically know that at x = 5 we have a zero; so (x  5) is our root.1305
We plug that in; we know it is going to be (x  5)(_x^{2} + _x + _).1312
Now, we don't know what is in these blanks yet; we still don't know what is going to be in there.1317
But we are one step closer to figuring out what those factors are, for being able to figure out what has to go in those blanks.1325
Later on, in another lesson, we will use this fact to great advantage1334
this fact that knowing the root automatically means you know a factor1336
When we learn about the intermediate value theorem to help us find roots,1340
and the polynomial division using those roots to break down large polynomials into smaller, more manageable factors.1344
Not all polynomials can be factored, though; even with all of this talk of factoring polynomials, there are some that cannot be factors,1352
not because it is difficult or really hard to do, but because it is just simply impossible.1358
Consider this polynomial: f(x) = x^{2} + 1; if we try to find its roots, then we have 0 = x^{2} + 1.1363
So, we have x^{2} = 1; but there is no number that exists that can be squared to become a negative number.1370
No number can be squared to become a negative number; why?1377
Consider: if we have (2)^{2}, that becomes positive 4; if we square any negative number, it becomes a positive number.1380
If we square any positive number, it stays positive; if we square 0, it stays 0.1386
So, there is no number that we have, that we can square, and get a negative number out of it.1391
Thus, the polynomial has no roots; and since it has no roots from that theorem we just saw, it can't have any factors.1395
So, it has no roots; therefore, it cannot be reduced into smaller factors.1401
And something that cannot be reduced, we call irreducible; it is not reducible.1407
Now, I will be honest: what I just told you isn't really the whole story.1413
More accurately, we can't factor all polynomials yet.1418
The previous slide that we just saw is perfectly true, but only if you are working with just the real numbers (which have the symbol ℝ like that).1422
Now, that is what we are normally working with; so it is kind of reasonable to say this.1432
But it turns out that there is a hidden type of number that we haven't previously explored.1435
You might have seen this in previous math classes, even.1439
We will learn about the complex numbers later on; complex numbers can give us a way to factor these supposedly irreducible polynomials.1441
So, they are irreducible for real numbers; but they are not irreducible for complex numbers.1449
Now, we will learn about them in the lesson that is named after these numbersour lesson on complex numbers.1454
But for now, we are just working with real numbers; and in general, we will just be working with real numbers in this course.1460
Real numbers are really useful; you can do a lot of stuff with them.1464
So, it is enough for us to be working with real numbers, generally.1467
That means, for us right now, at least some polynomials are simply irreducible.1470
And we can't always find roots for everything in a polynomial, because we can't break it down,1474
because there are things that just don't have roots, based on how real numbers work.1480
Now, we will talk about complex numbers later on; but I just wanted to point this out1484
that I am not telling you the whole story right now, because we don't want to get confused with complex numbers.1488
But for our purposes right now, with real numbers, there are some things that are simply irreducible.1492
There is a limit to how many roots or factors a polynomial can have.1498
Now, roots and factors are basically two sides of the same thing.1501
Since x = a as a root is the same thing as knowing (x  a) as a factor, they are just two sides of the same thing.1505
So, we will consider them as roots/factors of polynomials.1516
A polynomial of degree n can have, at most, n roots/factors; so we can have a maximum of n of these.1518
Why is this the case? Well, consider this: every factor comes in the form (x + _), or even larger if the factor is irreducible.1527
It might be (_x^{2} + _x + _); but the very smallest has to be (x + _).1535
If we break a polynomial into its factors, we are going to get (x + _) (x + _)...so on and so on, up until (x + _).1540
Now, if we had more than n factors, then that would mean that we have (x + _) multiplying by itself more than n times.1547
So, if we have x multiplying more than n times, then it has to have a degree larger than n; its exponent is going to have to be greater than that.1555
If we wanted to max out at x^{2}, but we had (x + _)(x + _)(x + _)...well, that is going to become x^{3}1564
plus stuff after it, which is not going to be x^{2}...not going to be degree 2.1576
So, if we have a degree n polynomial, the most factors we can possibly have are n factors, n roots,1581
because otherwise, we would have too many factors, and we would blow out the degree of the polynomial.1589
Thus, the most roots/factors a polynomial can have is equal to its degree.1595
We also can get information about the possible shape of a polynomial's graph from its degree.1601
A polynomial of degree n can have, at most, n  1 peaks and valleys.1605
Formally speaking, that is relative maximums and minimums.1610
For example, if we have x^{4}, then that means we have n = 4 (our degree).1613
So, n  1 = 3; so we look over here, and we have one valley here, one peak here, one valley here.1618
This is also a relative minimum, a relative minimum (that is what we mean by "valley), and a relative maximum (that is what we mean by "peak").1630
So, n  1 tells us the most bottoms and tops we can have, before they either go off to positive infinity or go off to negative infinity.1642
Now, we can't prove this here, because it requires calculus; but it is connected with the maximum number of roots in a polynomial.1650
And if you go on to take calculus (which I heartily recommend), you will very clearly, very quickly see it.1656
It becomes very clear in calculus; it is one of the important points of what you do in calculus.1661
So, you will think, "Oh, that makes a lot of sense," because the possible peaks and valleys1665
are connected to a polynomial that has a degree that is 1 less; and that is why it is connected.1670
Don't worry about it too much right now; but it is very interesting, and very obvious, if you go on to take calculus.1675
Notice that, in both of the previous properties, it was described as "at most."1681
Just because a polynomial has degree n does not mean it will have n distinct roots or n  1 peaks and valleys.1687
We aren't necessarily going to have to have that many; it is just that we can have up to that many.1693
Consider f(x) = x^{5} + 1; this graphs like this, but from this graph, we can see clearly:1697
we only have one root, and we have no peaks or valleys.1704
The degree gives an upper limit on how many there can be, but it doesn't tell us how many there will be.1708
It just that the maximum is this; but you could definitely have fewer.1713
All right, we are ready for some examples.1719
The first one: we want to find the zeroes of f(x) = 3x^{2}  23x + 14.1721
So, this is a textbook exampleliterally a textbook example, since this is effectively a textbook.1726
So, you plug in 0, because we are looking for when f(x) is equal to 0.1731
0 = 3x^{2}  23x + 14; we know we are going to be looking for 0 =...something where it is going to be (_x + _)(_x + _).1735
What are we going to slot in there? Well, we notice that here is 3; the only way we can break up 3 is 3 times 1.1751
There are no other choices; so we either have to have 3 go for the first x or 3 go to the second x.1758
So, let's set it as 0 = (3x + _); and we will have 1x, so just x, plus _.1762
Great; now, at this point, we also say, "We have 14 over here; how can 14 break up?"1772
Well, we can have 1 times 14, or we can have 2 times 7; those are the only choices.1777
So, we are going to have to plug in either 1 times 14 or 2 times 7.1783
But now, we also have to take this 23 into consideration.1786
If we have 23, then we are going to have at least one negative over here.1789
And since it comes up as a positive, it is going to have to be that they are both negatives.1793
So, one of them is negative, so they are going to both be negatives; so it will be 1(14) or 2(7).1796
So, 1 times 14...we will notice that, either way we put that in, that won't work out.1802
But we can plug in 2 times 7, and we can amp up this 7; so + 7 here...put in the 2 here...and we get 0 = (3x  2)(x 7).1807
We have managed to factor it; let's really quickly check what we have heredoes this work out?1823
Check (3x  2)(x  7); we would get 3x^{2}  21x  2x + 14, so 3x^{2}  23x + 14; it checks out; sure enough, it is good.1828
So, at this point, we break this down into two different possibilities: either 3x  2 = 0, or x  7 = 0.1847
So, 3x  2 = 0 or x  7 = 0 are the two different worlds where this will be true,1855
where we will have found a root where the whole expression will be equal to 0.1861
So, 3x  2 = 0: we get 3x = 2; x = 2/3; over here, x  7 = 0; we have x = 7; so our answers are x = 2/3 and x = 7; those are the roots for this polynomial.1864
Great; if f(2) = 0, factor f(x) = x^{3}  7x + 6.1881
Remember: if we know that at x = 2 we have a zero (at x = 2 there is a root), then that means there is a factor in that polynomial of (x  2).1888
How do we figure that out? Well, we notice that x = 2; then it is x  2 = 0, so that implies that it has to be a factor of (x  2) in there.1904
We can use that piece of information; we know that f(x) is going to have to break down with an (x  2) in there.1912
So, let's set it up like normal: 0 = x^{3}  7x + 6; but what we just figured out here...we know that there is a factor of (x  2).1917
So, we can also write this as 0 = (x  2)(_x^{2} + _x + _); what are going to go into those blanks?1924
Well, at this point, we just use a little bit of logic and ingenuity, and we can figure this out.1935
Well, we know that...what is in front of this x^{3}? It is effectively a 1.1941
So, if there is a 1 in front, we have x times x^{2}; whatever goes into this blank is going to determine what coefficient is in front of it.1944
So, since we want a 1, it has to be that there is a 1 here, as well.1952
What about the very end? Well, the only thing that is going to create the ending constant is going to be the other constants.1957
So, the constants that we have here are 2 and whatever goes into that blank; so it must be that 2 times _ here becomes 6.1968
2 times 3 becomes 6; so we have a 3 here.1975
And finally, what is going to go into this blank here?1980
We think about this one, and we know that we want to have 0x^{2} come out of this; there are no x^{2}'s up here.1984
So, we have + 0x^{2}; so whatever we put into this blank must somehow get us a 0x^{2} to show up.1992
So, x times x^{2} is x^{3}; so we are not going to worry about that.2001
But x times _x is going to be x^{2}; let's do a little sidebar for this.2005
x times _x will become _x^{2}; and 2 times...we already filled in that blank...1x^{2} is going to be 2x^{2}.2010
Now, we want the 0x^{2} out of it; so it must be that, when we add these two things together, it comes out to be 0x^{2}.2023
What does this have to be? It has to be positive 2x^{2}.2031
We know that positive 2x^{2}, minus 2x^{2}, comes out to be 0x^{2}; so it must be that this is a positive 2x.2035
So, we write this whole thing out: 0 = (x  2)(x^{2} + 2x  3); and we have been able to figure out that that works.2042
We check this out and do a really quick check; so x^{3} + 2x^{2}  3x  2x^{2}  4x  6...2053
x^{3} checks out; 2x^{2}  2x^{2} cancel each other; that checks out.2067
3x  4x; that becomes 7x, so that checks out; 6...that checks out as well.2072
Great; we have a correct thing, so 0 = (x  2)(x^{2} + 2x  3) is correct.2078
We factored it properly; so at this point, the only thing that we have left to factor is the x^{2} + 2x 3.2087
0 = (x  2)(_x + _)(_x + _); what goes in those first blanks?2095
Well, we just have a 1 in front of that; so it is 1 and 1...we don't have to worry about it that much.2103
What else is going to go in there? Well, 3 is at the very end; we have +2x, so it must be that the negative amount is smaller than the positive amount.2108
So, it is going to be + 1 and + 3; 1 times 3 gets us 3, and everything else checks out.2117
x times x is x^{2}; plus 3x minus x...that gets + 2x; and minus 1 times 3 gets us 3; so that checks out.2128
We did another check in our heads really quickly.2136
So finally, we have 0 = (x  2)(x  1)(x + 3); we break this up into three different worlds;2138
set each world equal to 0: x  2 = 0; x  1 = 0; x + 3 = 0; so we have x = 2; x = 1; x = 3.2152
Those are all of the roots for this polynomial.2167
All right, the next example: give a polynomial with roots at the indicated locations and the given degree.2172
Now remember, a root can become a factor; so if we know that we have a root at 3, then that becomes a factor of...2177
if it was x = 3, then it becomes (x + 3); if we had x = 8, then that would become (x  47)...2186
I'm sorry; I accidentally read the wrong oneread forward one; that is (x  8).2196
And then finally, if we had x = 47, we would have (x  47) as our factor.2201
So, those three things together...we have (x + 3)(x  8)(x  47), and that right there is a polynomial.2207
We know it has degree 3, because we have x times x times x; that is going to be the largest possible exponent we can get on our variable.2217
That will come out to be x^{3}, so we have a degree 3; and we know it has roots in all of the appropriate places.2226
And we are donethat is it; we could expand this, and we could simplify, if we wanted.2230
We weren't absolutely required to by the problem, and this is a correct answer.2234
It is a polynomial; it is not in that general, standard form that we are used of _x to the exponent, _x to the exponent, _x to the exponent.2239
But it is still a polynomial, so it is a pretty good answer; we will leave it like that.2248
The next one: we have 2 and positive 2, so we have (x + 2) for the 2 and (x  2) for that.2252
Why? x = 2; we move that over; we get x + 2 = 0; x = +2...we move that over and we get x  2 = 0.2261
So, we get (x + 2)(x  2); but if we multiply those two together, we just have a degree of 2, and we want a degree of 4.2268
So, we need to somehow get the degree up on this thing, but have the same rootsnot have to accidentally introduce any more roots.2275
So, if we introduced multiplying just by x twice, we would have introduced a root at x = 0.2284
So, we can't just do that; but we do realize that if we just increase this to square it on both of them, they will still have the same roots.2289
It is just duplicate roots showing up; so (x + 2)^{2} + (x  2)^{2}...we have hit that degree of 4,2296
because each one of these will now have a degree of 2.2302
Alternately, we could have done this as (x + 2) to the 1, (x  2) cubed, or (x + 2)^{3}(x  2)^{1}.2305
Any one of these would have degree 4, and have our roots at the appropriate place.2315
Great; the final example: What is the maximum possible number of roots and peaks/valleys for each of the following polynomials?2320
So, for our first one, f(x), we notice that this has a degree of 3; so n = 3 means the maximum number of roots it can have is 3,2327
and the maximum peaks/valleys is one less; n  1 is 3  1, is 2.2339
So, the maximum number of roots is 3; the maximum peaks/valleys is 2.2350
We don't necessarily know it will have that many; all we know is that that is the maximum it could possibly have.2354
The next one: we notice that the degree for this one is 47; so if n = 47, then the maximum roots are going to be equal to that degree.2359
The maximum peaks/valleys are going to be one less than that degree, so we will get 46, one less than that.2369
The final one: for this one, we think, "Oh, 10^{3}, so it is 3!"no, we have to remember that this is not a variable.2379
This here is a variable; so it is x^{1}, so its degree is just 1.2388
For that one, degree 1...we will change over to the color green...n = 1 means the maximum roots are just 1;2394
and the maximum peaks/valleys are going to be one less than 1, so 1  1 = 0.2405
Now, why is that the case? Well, think about it: 10^{3}  5...10 cubed is just some constant;2415
it happens to be 1,000, but that is not really the point; so 1000x  5 is just going to be a very steep line.2420
x  5...it will intersect here; but does it ever go up and downdoes it ever undulate in weird ways?2429
No, it never does anything; we just have a nice, straight line, since it is a linear thing (linear like a line).2435
So, since it is a linear expression, it never undulatesnever has any peaks or valleys.2441
So, it never has any relative maximums, and no relative minimums; and that is why we have 0 thereit makes sense.2445
All right, I hope everything there made sense; I hope you got a really good understanding of roots,2451
because roots will come up in all sorts of places; they are really important to understand.2454
It is really important to understand this general idea, because you will see it in other things, being changed around.2457
But if you understand this general idea, you will be able to understand what is going on in later things and different courses.2461
All right, see you at Educator.com latergoodbye!2466
2 answers
Last reply by: Tiffany Warner
Wed May 25, 2016 7:17 PM
Post by Tiffany Warner on May 25 at 06:21:53 PM
Hi Professor!
I am really struggling with one of the practice questions given.
Find all the roots of f(p) = 2 p4 ? 8p3 ? 14p2 + 44p + 48 given that f(3) = 0 and f(?2) = 0.
The first part of the problem seems straight forward. We are given two roots, therefore two factors.
(p3)(p+2)
In example 2 of the lecture, you show us how to tackle factoring a cubic polynomial. However, I’m lost with this beast.
I figured expanding the two factors would make it more simple and the steps did reinforce that idea.
So we have (p3)(p+2)(_p^2+_p+_)
Which expanded becomes
(p^2p6)(_p^2+_p+_)
They show us in the steps what to fill in those blanks with. I understand how they got 8. (8)(6)=48
I also understand how they got the coefficient of 2 in front of p^2.
However I have no idea how they went about filling in that middle blank in front of p. I’m clearly missing something. If you could provide some guidance, I’d be very grateful!
Thank you!
0 answers
Post by Jamal Tischler on December 21, 2014
I've seen that theorem in Horner's factoring table.
3 answers
Last reply by: Professor SelhorstJones
Fri Dec 12, 2014 8:38 PM
Post by abendra naidoo on December 12, 2014
Hello,
These are good teaching modules.
I have 2 questions:
1)How is an equation different from a function.
2) I can correctly operate exponents and logs but still don't see why logs are necessary? Why not just use exponents  is it because log tables exist? How can I better understand this concept?
1 answer
Last reply by: Professor SelhorstJones
Mon Oct 20, 2014 11:59 AM
Post by Saadman Elman on October 19, 2014
In 21:49 21:52 you meant to say X5 is a factor. And X=5 is a zeros/roots. So X5 is a factor. But you said x5 is a root. Waiting for your reply.