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High School Math : Finding Zeros of a Polynomial

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Example Question #1 : Finding Zeros Of A Polynomial

Find the zeros of the following polynomial:

$f(x) = x^{4} - 4x^{3} - 7x^{2} + 22x + 24$ $\displaystyle f(x) = x^{4} - 4x^{3} - 7x^{2} + 22x + 24$

Possible Answers:

$\displaystyle x = -3, -2, 1, 4$

$\displaystyle x = -4, -3, 1, 2$

$\displaystyle x = -4, -1, 2, 3$

$\displaystyle x = -4, -2, 1, 3$

$\displaystyle x = -2, -1, 3, 4$

Correct answer:

$\displaystyle x = -2, -1, 3, 4$

Explanation:

First, we need to find all the possible rational roots of the polynomial using the Rational Roots Theorem:

$\frac{\textup{factors of the constant}}{\textup{factors of the leading coefficient}}$

Since the leading coefficient is just 1, we have the following possible (rational) roots to try:

±1, ±2, ±3, ±4, ±6, ±12, ±24

When we substitute one of these numbers for $\displaystyle x$ , we're hoping that the equation ends up equaling zero. Let's see if $\displaystyle -1$ is a zero:

$f(-1)=(-1)^{4}-4(-1)^{3}-7(-1)^{2}+22(-1)+24$ $\displaystyle f(-1)=(-1)^{4}-4(-1)^{3}-7(-1)^{2}+22(-1)+24$

$\displaystyle f(-1)=1+4-7-22+24$

f(-1)=0 $\displaystyle f(-1)=0$

Since the function equals zero when $\displaystyle x$ is $\displaystyle -1$ , one of the factors of the polynomial is (x+1) $\displaystyle (x+1)$ . This doesn't help us find the other factors, however. We can use synthetic substitution as a shorter way than long division to factor the equation.

$\textup{-1 } | \textup{ 1 -4 -7 22 24}$

$\small \textup{ -1 }\textup{ 5 }\textup{ 2 }\textup{ -24}$

$\frac{ }{\textup{ 1 -5 -2 24 }\textup{ 0}}$

Now we can factor the function this way:

$f(x)=(x+1)(x^{3}-5x^{2}-2x+24)$ $\displaystyle f(x)=(x+1)(x^{3}-5x^{2}-2x+24)$

We repeat this process, using the Rational Roots Theorem with the second term to find a possible zero. Let's try $\displaystyle -2$ :

$f(-2)=[(-2)+1][(-2)^{3}-5(-2)^{2}-2(-2)+24]$ $\displaystyle f(-2)=[(-2)+1][(-2)^{3}-5(-2)^{2}-2(-2)+24]$

$\displaystyle f(-2)=(-1)(-8-20+4+24)=0$

When we factor using synthetic substitution for x=-2 $\displaystyle x=-2$ , we get the following result:

$f(x)=(x+1)(x+2)(x^{2}-7x+12)$ $\displaystyle f(x)=(x+1)(x+2)(x^{2}-7x+12)$

Using our quadratic factoring rules, we can factor completely:

$\displaystyle f(x)=(x+1)(x+2)(x-4)(x-3)$

Thus, the zeroes of f(x) $\displaystyle f(x)$ are $\displaystyle x=-2, -1, 3, 4.$

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High School Math : Finding Zeros of a Polynomial

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