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INSTRUCTORSCarleen EatonGrant Fraser
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Lecture Comments (3)

0 answers

Post by julius mogyorossy on December 16, 2013

I was wondering how do you know when you should continue to factor to find the remaining 0's, you say the way you did it in Ex. 4 is simpler, not to me, yet, but I realized, I think, that since the degree is 3, you know if you can only find one positive or real root, that you can, must, factor, to find the remaining roots, the complex roots, is this correct?

1 answer

Last reply by: Dr Carleen Eaton
Thu Jun 13, 2013 11:59 PM

Post by Kavita Agrawal on June 13, 2013

Isn't the constant in the first example, 4x^2 + x - 3, -3 and not 3? I don't think this should make difference, though...

Rational Zero Theorem

  • Understand the Rational Zero Theorem and the special case where the leading coefficient is 1. Use it to list all possible rational roots of a polynomial.
  • Use synthetic substitution to test each possible rational root in your list.
  • After you find the first root, try to factor the quotient to find the remaining roots.

Rational Zero Theorem

List all possible rational zeros x3 + 8 = 0
  • Find [p/q] where p are the factors of the constant and q are the factors of the leading coefficient.
  • p = 8
  • Factors of p: 1,2,4,8
  • q = 1
  • Factors of q: 1
  • [Factors of p/Factors of q] = [1,2,4,8/1]
  • Now Find all the possibilities
[p/q] = ±[1/1]; ±[2/1]; ±[4/1]; ±[8/1] = ±1; ±2; ±4; ±8
List all possible rational zeros 3x4 − 40x3 + 10x2 + 40x − 13 = 0
  • Find [p/q] where p are the factors of the constant and q are the factors of the leading coefficient.
  • p = 13
  • Factors of p: 1,13
  • q = 3
  • Factors of q: 1,3
  • [Factors of p/Factors of q] = [1,13/1,3]
  • Now Find all the possibilities
[p/q] = ±[1/1]; ±[1/3]; ±[13/1]; ±[13/3] = ±1; ±[1/3]; ±13; ±[1/3]
List all possible rational zeros 5x2 + 10x + 2 = 0
  • Find [p/q] where p are the factors of the constant and q are the factors of the leading coefficient.
  • p = 2
  • Factors of p: 1,2
  • q = 5
  • Factors of q: 1,5
  • [Factors of p/Factors of q] = [1,2/1,5]
  • Now Find all the possibilities
[p/q] = ±[1/1]; ±[1/5]; ±[2/1]; ±[2/5] = ±1; ±[1/5]; ±2; ±[2/5]
List all possible rational zeros 5x2 + 4x − 1 = 0
  • Find [p/q] where p are the factors of the constant and q are the factors of the leading coefficient.
  • p = 1
  • Factors of p: 1
  • q = 5
  • Factors of q: 1,5
  • [Factors of p/Factors of q] = [1/1,5]
  • Now Find all the possibilities
[p/q] = ±[1/1]; ±[1/5]; = ±1; ±[1/5]
List all possible rational zeros 9x4 − 4 = 0
  • Find [p/q] where p are the factors of the constant and q are the factors of the leading coefficient.
  • p = 4
  • Factors of p: 1,2,4
  • q = 9
  • Factors of q: 1,3,9
  • [Factors of p/Factors of q] = [1,2,4/1,3,9]
  • Now Find all the possibilities
[p/q] = ±[1/1]; ±[1/3]; ±[1/9]; ±[2/1]; ±[2/3]; ±[2/9]; ±[4/1]; ±[4/3]; ±[4/9] = ±1; ±[1/3]; ±[1/9]; ±2; ±[2/3]; ±[2/9]; ±4; ±[4/3]; ±[4/9]
List all possible rational zeros − 15x4 + 37x3 + 38x2 − 19x − 5 = 0
  • Find [p/q] where p are the factors of the constant and q are the factors of the leading coefficient.
  • p = 5
  • Factors of p: 1,5
  • q = 15
  • Factors of q: 1,3,5,15
  • [Factors of p/Factors of q] = [1,5/1,3,5,15]
  • Now Find all the possibilities
[p/q] = ±[1/1]; ±[1/3]; ±[1/5]; ±[1/15]; ±[5/1]; ±[5/3]; ±[5/5]; ±[5/15] = ±1; ±[1/3]; ±[1/5]; ±[1/15]; ±5; ±[5/3];
Find all the zeros of x3 − 9x2 − 17x + 10 = 0
Find the possible rational zeros
  • [p/q] = [1,2,5,10/1]
  • Possible rational zeros then are
  • ±1; ±2; ±5; ±10
  • Test Using Synthetic Division. Begin by testing ,1, − 1, 2, − 2. If the remainder is Zero, then that number is a rational root.
  • 1 1 -9 -17 10
        1 -8 -25
      1 -8 -25 −15
  • -1 1 -9 -17 10
        -1 10 7
      1 -10 -7 17
  • 2 1 -9 -17 10
        2 -14 -62
      1 -7 -31 −52
  • -2 1 -9 -17 10
        -2 22 -10
      1 -11 5 0
  • So far x = 2 is a zero, leaving behind x2 − 11x + 5 = 0 to factor out.
  • By observation, you will see that the remaining polynomial is not factorable, the last thing to do is to use the Quadratic Formula to find the missing roots.
Roots = { − 2,[(11 + √{101} )/2],[(11 − √{101} )/2]}
Find all the zeros of 3x3 + x2 − 3x − 1 = 0
  • Find the possible rational zeros
  • [p/q] = [1/1,3]
  • Possible rational zeros then are
  • ±1; ±[1/3]
  • Test Using Synthetic Division. Begin by testing ,1, − 1, [1/3], − [1/3]. If the remainder is Zero, then that number is a rational root.
  • 1 3 1 -3 -1
        3 4 1
      3 4 1 0
  • 1 3 1 -3 -1
        -3 2 1
      3 -2 -1 0
  • [1/3] 3 1 -3 -1
        1 [2/3] −[7/9]
      3 2 −[7/3]
    −[16/9]
  • −[1/3] 3 1 -3 -1
        1 0 1
      3 0 −3
    0
  • Using synthetic division, you found all possible zeros.
Roots/Zeros = { 1, − 1, − [1/3]}
Find all the zeros of 5x3 − 2x2 − 5x + 2 = 0
  • Find the possible rational zeros
  • [p/q] = [1,2/1,5]
  • Possible rational zeros then are
  • ±1; ±[2/5] ±[1/5]; ±2;
  • Test Using Synthetic Division. Begin by testing ,1, − 1, [2/5], − [2/5]. If the remainder is Zero, then that number is a rational root.
  • 1 5 -2 -5 2
        5 3 -2
      5 3 -2 0
  • -1 5 -2 -5 2
        -5 7 -2
      5 -7 2 0
  • [2/5] 5 -2 -5 2
        [10/5] 0 −[10/5]
      5 0 -5 0
  • −[2/5] 5 -2 -5 2
        −[10/5] [8/5] [34/25]
      5 -4 −[17/5]
    [84/25]
  • Using synthetic division, you found all possible zeros.
Roots = { 1, − 1, − [2/5]}
Find all the zeros of 2x3 + 5x2 + x − 2 = 0
  • Find the possible rational zeros
  • [p/q] = [1,2/1,2]
  • Possible rational zeros then are
  • ±1; ±2; ±[1/2];
  • Test Using Synthetic Division. Begin by testing , − 1, − 2, − [1/2]. If the remainder is Zero, then that number is a rational root.
  • 1 2 5 1 -2
        -2 -3 2
      2 3 -2 0
  • -2 2 5 1 -2
        -4 -2 2
      2 1 -1 0
  • −[1/2] 2 5 1 -2
        -1 -2 [1/2]
      2 4 -1
    −[3/2]
  • Using synthetic division, you found two zeros, x = − 1;x = − 2. There is one more missing.
  • If you looked carefuly when we tested x = − [1/2], it looked like it was going to work. Test x = [1/2]
  • to see if that will work.
  • [1/2] 2 5 1 -2
        1 3 2
      2 6 4
    0
  • It worked! You have now found all three zeros/roots.
Roots = { − 1, − 2,[1/2]}

*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

Rational Zero Theorem

Lecture Slides are screen-captured 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 0:00
  • Equation 0:08
    • List of Possibilities
    • Equation with Constant and Leading Coefficient
    • Example: Rational Zero
  • Leading Coefficient Equal to One 7:19
    • Equation with Leading Coefficient of One
    • Example: Coefficient Equal to 1
  • Finding Rational Zeros 12:58
    • Division with Remainder Zero
  • Example 1: Possible Rational Zeros 14:20
  • Example 2: Possible Rational Zeros 16:02
  • Example 3: Possible Rational Zeros 19:58
  • Example 4: Find All Zeros 22:06

Transcription: Rational Zero Theorem

Welcome to Educator.com.0000

In today's lesson, we will be discussing the rational zero theorem.0002

What this theorem tells you is the possible rational zeroes of a polynomial.0009

And again, it just helps you generate a list of possibilities.0017

So, if f(x) is some polynomial function with integer coefficients, then you can generate a list of possible rational zeroes0020

by looking at two things: the leading coefficient and the constant.0032

If p/q is a fraction in simplest form that is a zero of f(x), then p is a factor of a0,0042

which is the constant; and q is a factor of an, which is the leading coefficient.0053

So, let's think about what this is saying.0061

p/q: here, p is a factor of the constant; so the numerators of all the elements of the lists that we are going to generate consist of factors of the constant.0069

The denominators consist of factors of the leading coefficient.0091

Using this equation, using the rational zero theorem, what I can do is make a list of the factors of the constant,0105

make a list of the factors of the leading coefficient, and then find all possible combinations,0116

using those factors as the numerator (the factors of the constant) and the factors of the leading coefficient as the denominator.0124

Find all the possible combinations; and then, that will give me a list of possible rational zeroes.0133

Now, this is only a list; and we are going to talk about how you check the list and determine which ones,0138

if any, are actually zeroes; because it may turn out that none of these are zeroes.0145

It may turn out that the zeroes are irrational or complex numbers.0150

But if there are rational zeroes, they will be members of this list.0159

OK, using an example: f(x) = 4x2 + x - 3.0165

Here, my leading coefficient is 4, and the constant is 3.0175

Now, I am going to first just think about the factors: the factors of 4 are going to be 1, 2, and 4.0190

And these could be positive or negative; and I am going to worry about the signs in a minute.0208

Right now, I am just going to look for the factors.0212

The factors of 3 are 1 and 3; now, these factors of the leading coefficient are going to comprise the denominator.0219

So, these are possibilities for the denominator.0230

Here, these factors of the constant are going to be possibilities for the numerator.0237

So, I am going to take each one; I am going to go in an organized manner;0246

I am looking for different possibilities for p/q, for the factors of 3 over the factors of 4.0250

p/q could be...I could have 1 in the numerator and 1 in the denominator, and it could be positive or negative,0257

because the factors here could include -1 times -4 (that would give me 4), and so on.0267

So, I could have all different combinations of positive and negative.0274

OK, I could have 1 for the numerator and 2 for the denominator, plus or minus.0277

I could have 1/4; I could have 3/1, plus or minus; I could have 3/2, plus or minus; and plus or minus 3/4.0286

And then, I am just writing this in simpler form: this is just ±1, ±1/2, ±1/4; this is just 3 (3/1 is just 3); ± 3/2; ±3/4.0306

So, this gives me the list of possible rational zeroes.0328

Now, let's factor this out and see what happens, just to check and make sure that what we come up with actually is on this list.0341

Let's take the corresponding equation and find the zeroes.0348

4x2 + x - 3 = 0: you can actually factor this out, and this is going to come out to (4x - 3) (x + 1).0352

So, this would be 4x2, and this would be 4x - 3x (would give me x); -3 times 1 is -3.0367

Using the zero product property, I am going to get 4x - 3 = 0 and x + 1 = 0.0377

So, this is going to give me 4x = 3, or x = 3/4; this will give me x = -1.0385

I look, and I have 3/4 on the list, and I have -1 on the list; both are on the list.0395

So, what this gave me is a list of 1, 2, 3, 4, 5, 6 possibilities, positive or negative: that is 12 possibilities.0402

And I found two of them; I was able to find this by factoring.0410

But what it is showing is that these were on the list; but again, this is just a list of possibilities.0415

And you need to be able to figure out which possibilities are correct.0422

So, what we are going to do is generate lists of possibilities, and then talk about0426

what you can do with that information--how you can find which possibilities are correct.0433

OK, if the leading coefficient of the polynomial is equal to 1, then the rational zero theorem tells us that any rational zero of f(x) is a factor of the constant.0438

Let's think about what this means: we said that the list of rational zeros equal p/q,0454

where p is factors of the constant--let's just say the constant--and q are factors of the leading coefficient.0465

Well, if the leading coefficient is 1, then this becomes p/1, so it is just going to be factors of the constant.0487

So, in the situation where the leading coefficient is 1, you just have to look at the factors of the constant.0515

And those will give you your list of possible rational zeroes.0520

For example, if I have f(x) = x4 - 3x3 + 2x2 + x - 8.0526

I have a leading coefficient of 1, so I don't need to worry about that: p/q = p/1 = p.0543

Now, here p is going to equal the factors of the constant, which will be factors of 8, equal ±1, ±2, ±4, ±8.0553

Now, I mentioned that these are just possibilities; so I need to check these.0576

And if you will recall, we can check to see if something is a factor by using synthetic division.0581

We also talked about how, if a number is a 0, then if you take the value of the function,0592

you are going to find that if f(a) = 0, then x - a is a factor.0605

So, for example, if x - 2 is a factor of f(x), then f(2) will equal 0.0615

And recall that using synthetic division, we can find the value of a function by looking at the remainder.0635

So, what we can do is go back to synthetic division (or synthetic substitution, as we call it in this case)0641

and divide the polynomial by x - a, where a is one of the factors we are checking.0646

And if the remainder is 0, then we know we have a factor or a rational zero, in this case, if the remainder is 0.0653

So, these factors are also zeroes, because f(a) is 0.0664

So, if I go ahead and divide, for example, f(x) by x - 2, that will allow me to check to see if 2 is a factor.0669

If 2 is a factor, the remainder will equal 0.0683

Let's go ahead and do that: here we are dividing by 2, and I do not have any missing terms.0690

I have a coefficient of 1, a coefficient of -3, a coefficient of 2, another coefficient of 1, and a coefficient of -8.0697

Bring down 1; multiply 1 by 2; that gives me 2.0708

And now, I take 2 plus -3 to get -1; multiply -1 times 2 to get -2.0714

Combining 2 and -2, I get 0; 0 times 2 is 0; 1 + 0 is 1; 1 times 2 is 2, to give me a final value of -6; this is the remainder.0722

2 is not a zero of f(x); if the remainder here was 0, then this would tell me that 2 is a zero of f(x),0745

because it is telling me that f(2) equals 0; in other words, when x = 2, the function equals 0.0760

And by definition, that is a zero; it is crossing the x-intercept.0767

OK, to sum up: if you are looking for rational zeroes, list all possible rational zeroes using the theorem.0772

And the theorem says that p/q equals the factors of the constant, over the factors of the leading coefficient.0783

Once you have done that, use synthetic division to determine which possibilities are actually zeroes.0803

Division will yield a remainder equal to 0, if the value is a zero.0812

So, you get your values for p and q; you take the function, and you divide it by each of the values that you are trying, p/q.0832

And if the remainder equals 0, p/q is a zero.0844

If the remainder is anything other than 0, then it is not.0854

So, list the rational zeroes, and then use synthetic division to check.0856

OK, list all possible rational zeroes of this function.0862

I need to find values of p/q, and I am going to first look at the constant; this is a0 = 9, and the factors are 1, 3, and 9.0866

So, these are values for p.0887

Here, the leading coefficient equals 2; factors are just 1 and 2.0893

So now, I am going to go through all of these different combinations of numerator (1, 3, 9) and denominator (1 and 2).0903

So, possible rational zeroes are ±1/1 (I am just going to simplify that to 1 right away), or ±1/2.0911

Next is ±3/1 (which is just 3); ±3/2; next, ±9/1 (which gives me 9); ±9/2.0922

So, I have 1, 2, 3 times 2; that is 6 possibilities; and then plus or minus for each gives me 12 possible rational zeroes.0944

I don't know which, if any, of these are actually zeroes; I would have to check using synthetic division.0956

In this next example, we are asked to actually find the rational zeroes, not just to list the possibilities.0963

So, I am going to need to use synthetic division.0969

Since the leading coefficient is 1, then possible rational zeroes equal the factors of 3.0971

I don't need to worry about this; the denominator is just going to be 1.0986

OK, factors of 3 are 1 and 3; so my possibilities are ±1 and ±3.0990

Now, I need to check these using synthetic division.1003

So, I am starting out with 1; and I am looking, and there are no missing terms: x3, x2, x, and a constant.1006

So, I can just go ahead and put these coefficients: 1, -2, -2, -2, -3.1015

This is 1 times 1 is 1; combining that with -2 gets -1; 1 times -1 is -1, plus -2 is -3; -3 times 1 is -3; this is -6.1022

The remainder equals -6, so 1 is not a zero.1043

OK, now let's try -1 right here: again, my coefficients are the same: 1, -2, -2, -3.1050

OK, this is going to give me 1 times -1 (is -1); -2 and -1 is -3; -3 times -1 is 3.1065

I am combining that with -2 to get 1; times -1 gives me -1 and -3; this is -4; so this is not a zero.1080

OK, the next possibility: let's try 3; the coefficients are 1, -2, -2, -3.1094

This is going to give me 1; times 3 is 3; 3 and -2 is going to give me 1.1105

1 times 3 is going to give me 3 again; and -2 is going to give me 1.1115

1 times 3 is 3; 0; 3 is a zero, because the remainder equals 0; 3 is a zero.1125

OK, I found one rational zero: I have one more possibility left--I checked 1; I checked -1; I checked 3; now I need to check -3.1138

1, -2, -2, -3: bring down the 1 and multiply by -3 to get -3.1150

Combine with -2 to give me -5; -5 times -3 is 15; plus -2 leaves me with 13; times -3 is -39.1158

And that is -42, so this is not a zero.1174

The rational zero for this function is 3; 3 is a rational zero.1181

I had this list of possibilities, and the only one that turned out to be correct, an actual zero, is 3.1188

List all possible rational zeroes of this function.1199

So, I am going to be looking for p/q, which equals the factors of the constant over factors of the leading coefficient.1202

What this is going to give me is 09 1, 3, 5, and 15, over factors of 4: 1, 2, and 4.1228

So, I need to figure out all of these possibilities and use positive and negative for each.1251

So, 1/1 is ±1; and then, 1/2 is ±1/2; 1/4 is ±1/4.1255

OK, now going on to 3: 3/1 is just ±3; 3/2 is ±3/2; 3/4 is ±3/4.1269

OK, next, 5: 5/1 is just 5; 5/2; 5/4...1285

Then, 09 15/1--just 15; 15/2; and then 15/4, plus or minus.1300

So, there are a lot of possibilities here; and I listed them all out.1316

But if you were actually asked to find the rational zeroes, you would need to check these through synthetic division.1320

OK, here we are asked to find all zeroes of this function; and notice, it doesn't just say "rational zeroes"; it says "all zeroes."1325

But I am going to start out by looking at the rational zeroes.1331

And since the leading coefficient is 1, the possible rational zeroes will equal factors of the constant.1334

And that is because, if I am taking p/q, and the denominator is the factors of the leading coefficient...1349

here the only factor is 1; so I am just going to end up with p.1357

So, factors of 4 are ±1, ±2, and ±4.1362

Now, I have a list of possible rational zeroes, and I am going to use synthetic division to determine which of these are actual zeroes (not just possibilities).1374

The coefficient here is 1; the coefficient here is 2; -2; and -4.1387

Now, there are no missing terms: I have cubed, squared, x, and constant.1394

So, I am going to go ahead and use synthetic division; this is 1 times 1, is 1.1399

So, this gives me 2 and 1, which is 3; 3 times 1 gives me 3; that combined with -2 is 1.1406

1 times 1 is 1, combined with -4 is -3.1418

This is not a zero: 1 is not a zero, because the remainder is something other than 0.1423

Let's try -1: the coefficients are, again, 1, 2, -2, and -4.1428

Here, I have 1 times -1 to yield -1; combined with 2 is going to give me 1; this times -1 is -1.1439

Combined with -2, it is -3; -3 times -1 is 3; plus -4 is -1; again, this is not a zero.1453

So, not a zero, not a zero...let's try 2.1467

OK, again, same coefficients, same process: bring down the 1; multiply by 2; combine to get 4.1472

4 times 2 is 8; plus -2 leaves me with 6; times 2 is 12; plus -4 is 8--again, not a zero.1488

Now, I am going to try -2: bring down the 1; times -2; combining 2 and -2 leaves me 0, times -2...1502

Combining, you get -2 times -2, is positive 4; -4 and positive 4 is 0.1520

So, looking at this, since the remainder equals 0, -2 is a zero of this function.1530

Now, I could continue on to check these two; but there is an easier way to proceed.1538

because what this is telling me is that (remember, this is the opposite sign) x + 2 (because I took the opposite sign) is a factor of this.1545

So, that leaves me with x + 2, times...this is x3, so times 1x2, plus 0x (that drops out), minus 2.1563

This is what I am actually left with: (x + 2) (x2 - 2).1575

So, instead of just checking these two (and they might be right; they might not be right),1581

what I am going to need to do, if these are not correct (because there are three zeroes, because the degree is 3)1588

is: then I am going to need to find irrational or complex zeroes.1595

So, instead of bothering to even check these, I just use the fact that I now have (x2 - 2) (x + 2) to go ahead and use factoring to solve.1600

So, I know that I have x + 2, and I look, and I can't factor any more.1613

So, I use the zero product property; and I know already that x + 2 = 0, and x = -2; I figured that out.1621

But I can go on and do it with this, as well: x - 2 = 0.1630

And this is going to give me x2 = 2; I am then going to take the square root of both sides to get x = ±√2.1634

I have two zeroes here, and I have one here: the zeroes are -2, √2, and -√2.1645

And this is an irrational number, so it wasn't on my list of possibilities.1656

It is easier to proceed by finding one zero and then using the information that you have here to write the factored-out1660

(or at least partially-factored-out, in some cases) form of the equation,1668

and then go on and use the zero product property to find the remaining zeroes.1672

Thanks for visiting Educator.com, and I will see you next lesson!1678