question 5 of 15, step 1 of 1
construct a polynomial function with the stated properties. reduce all fractions to lowest terms.
third - degree, with zeros of - 3, - 1, and 2, and passes through the point (1, 6).
answer
p(x) =
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# Explanation:
## Step1: Form the polynomial with zeros
If a polynomial has zeros $ r_1, r_2, r_3 $, then it can be written in factored form as $ p(x)=a(x - r_1)(x - r_2)(x - r_3) $, where $ a $ is a constant. Given zeros are $-3$, $-1$, and $2$, so the polynomial is $ p(x)=a(x + 3)(x + 1)(x - 2) $.

## Step2: Find the value of $ a $
The polynomial passes through the point $(1,6)$, so substitute $ x = 1 $ and $ p(1)=6 $ into the equation:
$$
\begin{align*}
6&=a(1 + 3)(1 + 1)(1 - 2)\
6&=a(4)(2)(-1)\
6&=a(-8)\
a&=\frac{6}{-8}\
a&=-\frac{3}{4}
\end{align*}
$$

## Step3: Substitute $ a $ back into the polynomial
Substitute $ a = -\frac{3}{4} $ into $ p(x)=a(x + 3)(x + 1)(x - 2) $:
$$
\begin{align*}
p(x)&=-\frac{3}{4}(x + 3)(x + 1)(x - 2)\
&=-\frac{3}{4}(x^2 + 4x + 3)(x - 2)\
&=-\frac{3}{4}(x^3 - 2x^2 + 4x^2 - 8x + 3x - 6)\
&=-\frac{3}{4}(x^3 + 2x^2 - 5x - 6)\
&=-\frac{3}{4}x^3 - \frac{3}{2}x^2 + \frac{15}{4}x + \frac{9}{2}
\end{align*}
$$

# Answer:
$ p(x)=-\frac{3}{4}x^3 - \frac{3}{2}x^2 + \frac{15}{4}x + \frac{9}{2} $ (or in factored form $ p(x)=-\frac{3}{4}(x + 3)(x + 1)(x - 2) $)

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Explanation:

Step1: Form the polynomial with zeros

Step2: Find the value of \( a \)

Step3: Substitute \( a \) back into the polynomial

Answer: