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# Finding Zeros of Polynomials

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Practice Finding Zeros of Polynomials
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The Factor Theorem

A rectangular shaped container is built in such a way that its volume can be represented by the polynomial $V(w)=w^3+7w^2+16w+12$ , where $w$ is the width of the container.

a) Factor the polynomial.

b) If $w = 2 \ ft$ , what are the dimensions of the container?

### Guidance

You know techniques for factoring quadratics and special cases of cubics, but what about other cubics or higher degree polynomials? With the factor theorem, you can attempt to factor these types of polynomials. The factor theorem states that if $(x-a)$ is a factor of $p(x)$ , then $p(a)=0$ . To use the factor theorem:

1. Guess factors of the given polynomial $p(x)$ . Factors should be of the form $(x-a)$ where $a$ is a factor of the constant term of the polynomial divided by a factor of the first coefficient of the polynomial.
2. Test potential factors by checking $p(a)$ . If $p(a)=0$ , then $x-a$ is a factor of the polynomial.
3. Divide the polynomial by one of its factors.
4. Repeat Steps 2 and 3 until the result is a quadratic expression that you can factor using other methods.

#### Example A

Use the factor theorem to determine if $x + 1$ is a factor of $p(x) =2x^3+3x^2-5x-6$ . If so, find the other factors.

Solution: If $x+1$ is a factor, then $p(-1)=0$ . Test this:

$p(x)&= 2x^3+3x^2-5x-6 \\x = -1: p(-1) & = 2(-1)^3+3(-1)^2-5(-1)-6 \\p(-1)&= -2+3+5-6 \\p(-1)&= 0 \quad (\text{IS a factor})$

Now that you have one of the factors, use division to find the others.

Step 1:

Step 2:

Step 3:

Step 4:

Step 5:

Step 6:

So:

$p(x)&=2x^3+3x^2-5x-6 \\p(x) &=(x+1)(2x^2+x-6)\\p(x)&=(x+1)(2x-3)(x+2)$

#### Example B

Use the factor theorem to determine if $x + 3$ is a factor of $s(x)=5x^2-13x-84$ . If so, find the other factors.

Solution: If $x+3$ is a factor, then $p(-3)=0$ . Test this:

$s(x) &= 5x^2-13x-84 \\ \\x = -3: s(-3)& =5(-3)^2-13(-3)-84 \\s(-3)&= 45+39-84 \\s(-3)&= 0 \quad (\text{IS a factor})$

Now that you have one of the factors, use division to find the other factor.

Step 1 : Divide the first term in the numerator by the first term in the denominator; put this in your answer. Therefore $\frac{5x^2}{x}=5x$ .

$(x+3) \overset{{\color{red}5x}}{|\overline{5x^2-13x-84}}$

Step 2 : Multiply the denominator by this number (variable) and put it below your numerator, subtract and get your new polynomial.

$&(x+3) \overset{{\color{red}5x}}{|\overline{5x^2-13x-84}}\\&\qquad \quad \underline{\;\;\; 5x^2+15x \;\;\;\;\;\;\;}\\& \qquad \qquad \quad \ -28x$

Step 3: Repeat the process until you cannot repeat it anymore.

$&(x+3)\overset{5x{\color{red}-28}}{|\overline{5x^2-13x-84}}\\&\qquad \quad \underline{\;\;\; 5x^2+13x \;\;\;\;\downarrow \;\;\;\;}\\& \qquad \qquad \qquad \ -28x-84\\&\qquad \quad \underline{\;\;\;\;\;\;\;\;\;\;\;\;\; -28x-84 \;}\\& \qquad \qquad \qquad \qquad \quad \ 0 \\$

Therefore: $\frac{(5x^2-13x-84)}{(x+3)}=(5x-28)$

So:

$s(x)&=5x^3-13x-84\\s(x)&=(x+3)(5x-28)$

#### Example C

Factor $f(t)=t^3-8t^2+17t-10$ .

Solution: In order to begin to find the factors, look at the number –10 and find the factors of this number. The factors of –10 are –1, 1, –2, 2, –5, 5, –10, 10. Next, start testing the factors to see if you get a remainder of zero.

$f(t)&=t^3-8t^2+17t-10\\t=-1:f(-1)&=(-1)^3-8(-1)^2+17(-1)-10\\f(-1)&=-36 \quad (\text{NOT a factor})\\\\t=1:f(1)&=(1)^3-8(1)^2+17(1)-10\\f(1)&=0 \quad (\text{IS a factor})$

Now that you have one of the factors, use division to find the others.

Step 1:

Step 2:

Step 3:

Step 4:

Step 5:

Step 6:

So :

$f(t)&=t^3-8t^2+17t-10\\f(t)&=(t-1)(t^2-7t+10)\\f(t)&=(t-1)(t-5)(t-2)$

Therefore: $f(t)=(t-1)(t-5)(t-2)$

#### Concept Problem Revisited

A rectangular shaped container is built in such a way that its volume can be represented by the polynomial $V(w)=w^3+7w^2+16w+12$ , where $w$ is the width of the container.

a) In order to begin to find the factors, look at the number 12 and find the factors of this number. The factors of 12 are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6$ , and $\pm 12$ . Next, start testing the factors to see if you get a remainder of zero.

$V(w)&=w^3+7w^2+16w+12\\w=-1: V(-1)&=(-1)^3+7(-1)^2+16(-1)+12\\V(-1) &= 2 \quad (\text{NOT a factor})\\\\w=-2: V(-2)&=(-2)^3+7(-2)^2+16(-2)+12\\V(-2) &= 0 \quad (\text{IS a factor})$

Now that you have one of the factors, use division to find the others.

Step 1:

Step 2:

Step 3:

Step 4:

Step 5:

Step 6:

So:

$V(w)&=w^3+7w^2+16w+12\\V(w)&=(w+2)(w^2+5w+6)\\V(w)&=(w+2)(w+2)(w+3)$

b) If $w = 2$ , what are the dimensions of the container?

$(w + 2) &= 2 + 2 = 4 \\(w + 2) &= 2 + 2 = 4 \\(w + 3) &= 2 + 3 = 5$

Therefore. the dimensions of the container are $4 \ ft \times 4 \ ft \times 5 \ ft$ .

### Vocabulary

The Factor theorem
The factor theorem states that if $p(a)=0$ , then $x-a$ is a factor of $p(x)$ .

### Guided Practice

1. Determine if $e + 3$ is a factor of $2e^3-e^2+e-1$ .

2. Factor: $x^3+4x^2+x-6$ .

3. A tennis court is being built where the volume is represented by the polynomial $p(L)=3L^3+8L^2+3L-2$ , where $L$ represents the length of the court. Determine if $L + 1$ is a factor and if so, find the other factors. If $L = 5ft$ , what are the dimensions of the court.

1. $e=-3:2(-3)^3-(-3)^2+(-3)-1=-67$ . Therefore $(e+3)$ is not a factor of $2e^3-e^2+e-1$ .

2. In order to begin to find the factors, look at the number –6 and find the factors of this number. The factors of –6 are $\pm 1, \pm 2, \pm 3,$ and $\pm 6$ . Next, start testing the factors to see if you get a remainder of zero.

$& x^3+4x^2+x-6\\& x=-1:(-1)^3+4(-1)^2+(-1)-6=-4 \quad (\text{NOT a factor})\\& x=1:(1)^3+4(1)^2+(1)-6=0 \quad (\text{IS a factor})$
Now that you have one of the factors, use division to find the others.
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
Step 6:
Therefore: $x^3+4x^2+x-6=(x-1)(x^2+5x+6)=(x-1)(x+2)(x+3)$

3. Start by testing the factor $L + 1$ to see if you get a remainder of zero.

$p(L)&=3L^3+8L^2+3L-2L=-1:\\p(L)&=3(-1)^3+8(-1)^2+3(-1)-2\\p(1)&=0 \quad (\text{IS a factor})$
Now that you have one of the factors, use division to find the others.
Step 1:
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
So:
$p(L)&=3L^3+8L^2+3L-2\\p(L)&=(L+1)(3L^2+5L-2)\\p(L)&=(L+1)(3L-1)(L+2)$
If $L = 5 \ ft$ , what are the dimensions of the container?
$(L+1)&=5+1=6\\(3L-1)&=3(5)-1=14\\(L+2)&=5+2=7$
Therefore the dimensions of the container are $6 \ ft \times 14 \ ft \times 7 \ ft$ .

### Practice

Determine if $a - 4$ is a factor of each of the following.

1. $a^3-5a^2+3a+4$
2. $3a^2-7a-20$
3. $-a^4+3a^3+5a^2-16$
4. $a^4-2a^3-8a^2+3a-4$
5. $2a^4-5a^3-7a^2-21a+4$

Factor each of the following:

1. $x^3+2x^2+2x+1$
2. $x^3+x^2-x-1$
3. $2x^3-5x^2+2x+1$
4. $2b^3+4b^2-3b-6$
5. $3c^3-4c^2-c+2$
6. $2x^3-13x^2+17x+12$
7. $x^3+2x^2-x-2$
8. $3x^3+2x^2-53x+60$
9. $x^3-7x^2+7x+15$
10. $x^4+4x^3-7x^2-34x-24$