High School Math : Finding Zeros of a Polynomial

Study concepts, example questions & explanations for High School Math

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Example Questions

Example Question #1 : Finding Zeros Of A Polynomial

Find the zeros of the following polynomial:

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

Possible Answers:

\(\displaystyle x = -3, -2, 1, 4\)

\(\displaystyle x = -4, -2, 1, 3\)

\(\displaystyle x = -4, -3, 1, 2\)

\(\displaystyle x = -2, -1, 3, 4\)

\(\displaystyle x = -4, -1, 2, 3\)

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:

\(\displaystyle \frac{\textup{factors of the constant}}{\textup{factors of the leading coefficient}}\)

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:

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

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

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

Since the function equals zero when \(\displaystyle x\) is \(\displaystyle -1\), one of the factors of the polynomial is \(\displaystyle (x+1)\). 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.

\(\displaystyle \textup{-1 } | \textup{ 1 -4 -7 22 24}\)

             \(\displaystyle \small \textup{ -1 }\textup{ 5 }\textup{ 2 }\textup{ -24}\)

       \(\displaystyle \frac{ }{\textup{ 1 -5 -2 24 }\textup{ 0}}\)

Now we can factor the function this way:

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

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

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

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

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

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

Using our quadratic factoring rules, we can factor completely:

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

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

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