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 Mejia Gardner. 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 Mejia Gardner. in the United States

Mar 31,2025

UpStudy AI · Accepted Answer

We are given
$$
f(x)=7x^3+2x^2-12x+5.
$$
We know there is at least one rational zero. By the Rational Root Theorem, any rational zero has the form
$$
\frac{p}{q},
$$
where $$p$$ divides the constant term $$5$$ and $$q$$ divides the leading coefficient $$7$$. Hence, the possible values for $$\frac{p}{q}$$ are
$$
\pm 1,\ \pm 5,\ \pm \frac{1}{7},\ \pm \frac{5}{7}.
$$

**Step 1. Test possible zeros**

Test $$x=\frac{5}{7}$$:
$$
\begin{aligned}
f\left(\frac{5}{7}\right) &= 7\left(\frac{5}{7}\right)^3 + 2\left(\frac{5}{7}\right)^2 - 12\left(\frac{5}{7}\right)+5\[1mm]
&= 7\left(\frac{125}{343}\right)+2\left(\frac{25}{49}\right) - \frac{60}{7}+5\[1mm]
&= \frac{875}{343}+\frac{50}{49}-\frac{60}{7}+5.
\end{aligned}
$$
Rewrite with denominator $$343$$:
$$
\begin{aligned}
\frac{875}{343}+\frac{50\cdot7}{343}-\frac{60\cdot49}{343}+\frac{5\cdot343}{343}
&=\frac{875+350-2940+1715}{343}\[1mm]
&=\frac{0}{343}=0.
\end{aligned}
$$
Thus, $$x=\frac{5}{7}$$ is a zero.

**Step 2. Factor the polynomial**

Since $$x=\frac{5}{7}$$ is a zero, the factor corresponding to this zero is
$$
x-\frac{5}{7}.
$$
It is often convenient to clear the fraction by writing the factor as
$$
7x-5.
$$
We can factor $$f(x)$$ as:
$$
f(x)=(7x-5)(\text{quadratic factor}).
$$
Assume the quadratic factor is
$$
x^2 + ax + b.
$$
Then,
$$
(7x-5)(x^2+ax+b)=7x^3+(7a-5)x^2+(7b-5a)x-5b.
$$
This product must equal the original polynomial:
$$
7x^3+2x^2-12x+5.
$$
Matching coefficients, we have:
- For $$x^2$$: $$7a-5=2$$ $$\Longrightarrow$$ $$7a=7$$ $$\Longrightarrow$$ $$a=1$$.
- For $$x$$: $$7b-5a=-12$$ $$\Longrightarrow$$ $$7b-5=-12$$ $$\Longrightarrow$$ $$7b=-7$$ $$\Longrightarrow$$ $$b=-1$$.
- For the constant: $$-5b=5 \Longrightarrow -5(-1)=5$$, which is correct.

Thus, the factorization is:
$$
f(x)=(7x-5)(x^2+x-1).
$$

**Step 3. Find the zeros of the quadratic factor**

Solve the quadratic equation:
$$
x^2+x-1=0.
$$
Using the quadratic formula:
$$
x=\frac{-1\pm\sqrt{1^2-4(1)(-1)}}{2(1)}
=\frac{-1\pm\sqrt{1+4}}{2}
=\frac{-1\pm\sqrt{5}}{2}.
$$

**Final Answer**

The zeros of $$ f(x) $$ are:
$$
x=\frac{5}{7},\quad x=\frac{-1+\sqrt{5}}{2},\quad x=\frac{-1-\sqrt{5}}{2}.
$$
The zeros of the function are $$ x = \frac{5}{7} $$, $$ x = \frac{-1 + \sqrt{5}}{2} $$, and $$ x = \frac{-1 - \sqrt{5}}{2} $$.

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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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.