The function below has at least one rational zero.
Use this fact to find all zeros of the function.

If there is more than one zero, separate them with commas. Write exact values, not decimal approximations.

Ask by Bird Carrillo. in the United States

Mar 31,2025

Question

The function below has at least one rational zero.
Use this fact to find all zeros of the function.

If there is more than one zero, separate them with commas. Write exact values, not decimal approximations.

Ask by Bird Carrillo. in the United States

Mar 31,2025

UpStudy AI · Accepted Answer

The first step is to use the Rational Root Theorem. The possible rational zeros of

$$
h(x)=5x^3+12x^2+12x+7
$$

are given by the factors of the constant term (7) divided by the factors of the leading coefficient (5). Thus the candidates are

$$
\pm 1, \pm 7, \pm \frac{1}{5}, \pm \frac{7}{5}.
$$

We test these candidates. Testing $$ x=-\frac{7}{5} $$, we have

$$
h\left(-\frac{7}{5}\right)=5\left(-\frac{7}{5}\right)^3+12\left(-\frac{7}{5}\right)^2+12\left(-\frac{7}{5}\right)+7.
$$

Calculate each term step by step:
 
1. $$ \left(-\frac{7}{5}\right)^3 = -\frac{343}{125} $$, so

$$
5\left(-\frac{343}{125}\right)=-\frac{1715}{125}.
$$

2. $$ \left(-\frac{7}{5}\right)^2 = \frac{49}{25} $$, so

$$
12\left(\frac{49}{25}\right)=\frac{588}{25}=\frac{588 \cdot 5}{125}=\frac{2940}{125}.
$$

3. $$ 12\left(-\frac{7}{5}\right)=-\frac{84}{5}=-\frac{84 \cdot 25}{125}=-\frac{2100}{125}. $$

4. $$ 7=\frac{875}{125}. $$

Now, summing all terms:

$$
-\frac{1715}{125}+\frac{2940}{125}-\frac{2100}{125}+\frac{875}{125} = \frac{-1715+2940-2100+875}{125} = \frac{0}{125}=0.
$$

Since $$ h\left(-\frac{7}{5}\right)=0 $$, we have found a rational zero, $$ x=-\frac{7}{5} $$.

Now, factor $$ h(x) $$ by dividing by the factor corresponding to this zero. Since $$ x=-\frac{7}{5} $$ is a zero, the factor is $$ 5x+7 $$. Dividing $$ h(x) $$ by $$ 5x+7 $$ (using synthetic or long division) yields:

$$
h(x)=(5x+7)(x^2+x+1).
$$

Next, find the zeros of the quadratic factor by solving

$$
x^2+x+1=0.
$$

The quadratic formula gives

$$
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-1\pm\sqrt{1-4}}{2}=\frac{-1\pm\sqrt{-3}}{2}=\frac{-1\pm i\sqrt{3}}{2}.
$$

Thus, all zeros of the function $$ h(x) $$ are

$$
x=-\frac{7}{5},\quad x=\frac{-1+i\sqrt{3}}{2},\quad x=\frac{-1-i\sqrt{3}}{2}.
$$
The zeros of the function $$ h(x) = 5x^3 + 12x^2 + 12x + 7 $$ are $$ x = -\frac{7}{5} $$, $$ x = \frac{-1 + i\sqrt{3}}{2} $$, and $$ x = \frac{-1 - i\sqrt{3}}{2} $$.

The function below has at least one rational zero.
Use this fact to find all zeros of the function.

If there is more than one zero, separate them with commas. Write exact values, not decimal approximations.

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The function below has at least one rational zero. Use this fact to find all zeros of the function. If there is more than one zero, separate them with commas. Write exact values, not decimal approximations.

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Solution

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The function below has at least one rational zero.
Use this fact to find all zeros of the function.

If there is more than one zero, separate them with commas. Write exact values, not decimal approximations.