$$ \left\lvert\, \leftarrow \quad \begin{array}{l} ext { Sketch the graph of the polynomial function. Use synthetic division and the } \ ext { remainder theorem to find the zeros. } \ f(x)=-x^{3}+5 x^{2}+8 x-12 \ ext { What are the zeros of } f(x)=-x^{3}+5 x^{2}+8 x-12 ? \ x=\square \ ext { (Use a comma to separate answers as needed.) }\end{array} ight. $$

Question

$$ \left\lvert\, \leftarrow \quad \begin{array}{l}	ext { Sketch the graph of the polynomial function. Use synthetic division and the } \ 	ext { remainder theorem to find the zeros. } \ f(x)=-x^{3}+5 x^{2}+8 x-12 \ 	ext { What are the zeros of } f(x)=-x^{3}+5 x^{2}+8 x-12 ? \ x=\square \ 	ext { (Use a comma to separate answers as needed.) }\end{array}ight. $$

UpStudy AI · Accepted Answer

We begin with the polynomial
$$
f(x)=-x^3+5x^2+8x-12.
$$
It is often easier to work with a positive leading coefficient, so we factor out $$-1$$ from $$f(x)$$:
$$
f(x) = -\left(x^3-5x^2-8x+12\right).
$$
Thus, the zeros of $$f(x)$$ are the same as the zeros of
$$
x^3-5x^2-8x+12=0.
$$

**Step 1. Find a rational zero**

We use the Remainder Theorem. Testing $$x=1$$:
$$
1^3-5(1)^2-8(1)+12 = 1-5-8+12 = 0.
$$
Since the result is $$0$$, $$x=1$$ is a zero.

**Step 2. Use Synthetic Division**

We will divide the polynomial $$x^3-5x^2-8x+12$$ by $$(x-1)$$. The coefficients are $$1,\,-5,\,-8,$$ and $$12$$.

$$
\begin{array}{r|rrrr}
1 & 1 & -5 & -8 & 12 \ \hline
  &   & 1  & -4 & -12 \
  & 1 & -4 & -12 & 0 \
\end{array}
$$

The division shows the polynomial factors as:
$$
x^3-5x^2-8x+12 = (x-1)(x^2-4x-12).
$$

**Step 3. Factor the Quadratic**

Next, factor the quadratic $$x^2-4x-12$$. We look for two numbers that multiply to $$-12$$ and add to $$-4$$. These numbers are $$-6$$ and $$2$$:
$$
x^2-4x-12 = (x-6)(x+2).
$$

**Step 4. Write the Complete Factorization**

Substitute back into our factorization:
$$
x^3-5x^2-8x+12 = (x-1)(x-6)(x+2).
$$
Since
$$
f(x) = -\left[(x-1)(x-6)(x+2)\right],
$$
the zeros of $$f(x)$$ are the same as the zeros of $$(x-1)(x-6)(x+2)=0$$.

**Step 5. Identify the Zeros**

Set each factor equal to zero:
$$
x-1=0 \quad \Rightarrow \quad x=1,
$$
$$
x-6=0 \quad \Rightarrow \quad x=6,
$$
$$
x+2=0 \quad \Rightarrow \quad x=-2.
$$

Thus, the zeros of $$f(x)=-x^3+5x^2+8x-12$$ are:
$$
x=-2,\quad x=1,\quad x=6.
$$
The zeros of $$ f(x) = -x^3 + 5x^2 + 8x - 12 $$ are $$ x = -2, 1, 6 $$.

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