use the following information to answer questions 10 - 11. the polynomial function $g(x)=2x^{3}+5x^{2}-28x - 15$ has a known factor of $(x - 3)$. 10. use polynomial division to show that $(x - 3)$ is a factor of $g(x)$. explain how you know. 11. rewrite $g(x)$ as the product of linear factors. $g(x)=$ (answer) 12. let $p$ be a polynomial function, and $p(x)=x^{4}-dx^{3}+8x^{2}-14x + 16$. if $(x - 2)$ is a factor of the polynomial, what is the value of $d$? explain or show how you know. 13. identify the number of real and non - real zeros for the polynomial $p(x)=\frac{1}{3}(x^{3}-x^{2}-x - 15)$. complete each blank in the statement below by circling the correct response. $p(x)$ has no or one or two or three real zero(s) (circle one) and no or one or two or three non - real zero(s) (circle one) 14. factor completely: $g(x)=3x^{3}-192$ $g(x)=$ (answer)

Question

Accepted Answer

### 10.
# Explanation:
## Step1: Perform polynomial long - division
Divide $g(x)=2x^{3}+5x^{2}-28x - 15$ by $(x - 3)$.
We know that if $g(x)=(x - 3)q(x)+r$, where $q(x)$ is the quotient polynomial and $r$ is the remainder.
Using polynomial long - division:
$$
\begin{align*}
2x^{3}+5x^{2}-28x - 15&=(x - 3)(2x^{2}+11x + 5)\
\end{align*}
$$
The remainder $r = 0$.
## Step2: Determine if it's a factor
Since the remainder of the division of $g(x)$ by $(x - 3)$ is $0$, by the factor theorem, $(x - 3)$ is a factor of $g(x)$.
# Answer:
We performed polynomial long - division of $2x^{3}+5x^{2}-28x - 15$ by $(x - 3)$ and got a remainder of $0$, so $(x - 3)$ is a factor of $g(x)$.

### 11.
# Explanation:
## Step1: Factor the quotient
We already found that $g(x)=(x - 3)(2x^{2}+11x + 5)$.
Factor $2x^{2}+11x + 5$:
$$
\begin{align*}
2x^{2}+11x + 5&=2x^{2}+10x+x + 5\
&=2x(x + 5)+1(x + 5)\
&=(2x + 1)(x+5)
\end{align*}
$$
## Step2: Write $g(x)$ as a product of linear factors
So $g(x)=(x - 3)(2x + 1)(x + 5)$.
# Answer:
$g(x)=(x - 3)(2x + 1)(x + 5)$

### 12.
# Explanation:
## Step1: Use the factor theorem
If $(x - 2)$ is a factor of $P(x)=x^{4}-dx^{3}+8x^{2}-14x + 16$, then $P(2)=0$ by the factor theorem.
## Step2: Substitute $x = 2$ into $P(x)$
$$
\begin{align*}
P(2)&=2^{4}-d\times2^{3}+8\times2^{2}-14\times2 + 16\
&=16-8d + 32-28 + 16\
&=(16 + 32+16-28)-8d\
&=36-8d
\end{align*}
$$
## Step3: Solve for $d$
Set $P(2)=0$, so $36-8d=0$.
$$
\begin{align*}
8d&=36\
d&=\frac{36}{8}=\frac{9}{2}
\end{align*}
$$
# Answer:
$d=\frac{9}{2}$

### 13.
# Explanation:
## Step1: Consider the degree of the polynomial
The polynomial $P(x)=\frac{1}{3}(x^{3}-x^{2}-x - 15)$ has degree $n = 3$.
By the fundamental theorem of algebra, the number of roots (counting multiplicities) is $3$.
## Step2: Use Descartes' rule of signs
For $P(x)=\frac{1}{3}(x^{3}-x^{2}-x - 15)$, the number of sign - changes in $P(x)$ is $1$, so there is exactly $1$ positive real root.
For $P(-x)=\frac{1}{3}(-x^{3}-x^{2}+x - 15)$, the number of sign - changes is $2$, so there are either $2$ or $0$ negative real roots.
We can also use the rational root theorem to test possible rational roots $\pm1,\pm3,\pm 5,\pm15$. By testing, we find that there is one real root.
Since the total number of roots is $3$, and we have $1$ real root, the number of non - real roots is $2$.
# Answer:
$P(x)$ has one real zero(s) and two non - real zero(s)

### 14.
# Explanation:
## Step1: Factor out the greatest common factor
First, factor out the GCF from $g(x)=3x^{3}-192$. The GCF of $3x^{3}$ and $192$ is $3$.
$g(x)=3(x^{3}-64)$
## Step2: Use the difference of cubes formula
Since $a^{3}-b^{3}=(a - b)(a^{2}+ab + b^{2})$ and $x^{3}-64=x^{3}-4^{3}$.
$$
\begin{align*}
g(x)&=3(x - 4)(x^{2}+4x + 16)
\end{align*}
$$
# Answer:
$g(x)=3(x - 4)(x^{2}+4x + 16)$

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QUESTION IMAGE

Question

Explanation:

10.

Step1: Perform polynomial long - division

Step2: Determine if it's a factor

Step1: Factor the quotient

Step2: Write $g(x)$ as a product of linear factors

Step1: Use the factor theorem

Step2: Substitute $x = 2$ into $P(x)$

Step3: Solve for $d$

Answer:

11.