select the correct answer.
one factor of the polynomial $2x^3 - 3x^2 - 3x + 2$ is $(x - 2)$. which expression represents the other factor, or factors, of the polynomial?
$\bigcirc$ a. $(2x + 1)(x - 1)$
$\bigcirc$ b. $(2x - 1)(x + 1)$
$\bigcirc$ c. $(2x^2 - x + 1)$
$\bigcirc$ d. $(2x^2 + 1)$

Question

Accepted Answer

# Explanation:
## Step1: Perform Polynomial Division or Use Synthetic Division
We can divide the polynomial $2x^3 - 3x^2 - 3x + 2$ by $(x - 2)$ using polynomial long division or synthetic division. Let's use synthetic division. The root for $(x - 2)$ is $x = 2$.

Set up synthetic division:
$$
\begin{array}{r|rrrr}
2 & 2 & -3 & -3 & 2 \
  &   & 4  & 2  & -2 \
\hline
  & 2 & 1  & -1 & 0 \
\end{array}
$$

The quotient polynomial is $2x^2 + x - 1$.

## Step2: Factor the Quotient Polynomial
Now, factor $2x^2 + x - 1$. We need two numbers that multiply to $2\times(-1)= -2$ and add to $1$. The numbers are $2$ and $-1$.

Rewrite the middle term: $2x^2 + 2x - x - 1$

Group and factor:
$$
\begin{align*}
(2x^2 + 2x) + (-x - 1) &= 2x(x + 1) - 1(x + 1) \
&= (2x - 1)(x + 1)
\end{align*}
$$
Wait, no, wait. Wait, when we did synthetic division, the coefficients were 2, 1, -1, so the quotient is $2x^2 + x - 1$. Wait, let's check the factoring again. Wait, maybe I made a mistake. Wait, let's use the quadratic formula on $2x^2 + x - 1$. The quadratic formula is $x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$ for $ax^2+bx+c$. Here, $a = 2$, $b = 1$, $c = -1$.

$$
x=\frac{-1\pm\sqrt{1 + 8}}{4}=\frac{-1\pm3}{4}
$$

So $x=\frac{-1 + 3}{4}=\frac{2}{4}=\frac{1}{2}$ and $x=\frac{-1 - 3}{4}=\frac{-4}{4}=-1$. So the factors are $(x - \frac{1}{2})(x + 1)$, but to eliminate the fraction, multiply by 2: $(2x - 1)(x + 1)$. Wait, but let's check by multiplying $(x - 2)(2x - 1)(x + 1)$.

First, multiply $(2x - 1)(x + 1)=2x^2 + 2x - x - 1 = 2x^2 + x - 1$. Then multiply by $(x - 2)$:

$$
\begin{align*}
(x - 2)(2x^2 + x - 1) &= x(2x^2 + x - 1) - 2(2x^2 + x - 1) \
&= 2x^3 + x^2 - x - 4x^2 - 2x + 2 \
&= 2x^3 - 3x^2 - 3x + 2
\end{align*}
$$

Which matches the original polynomial. Wait, but let's check the options. Option A is $(2x + 1)(x - 1)$, Option B is $(2x - 1)(x + 1)$, Option C is $2x^2 - x + 1$, Option D is $2x^2 + 1$.

Wait, maybe I made a mistake in synthetic division. Let's try polynomial long division. Divide $2x^3 - 3x^2 - 3x + 2$ by $x - 2$.

$$
\begin{array}{r|rrrr}
          & 2x^2 & + x  & -1 \
x - 2 \enclose{longdiv}{2x^3 - 3x^2 - 3x + 2} \
          & 2x^3 - 4x^2 \
\hline
          & 0 + x^2 - 3x \
          & x^2 - 2x \
\hline
          & 0 - x + 2 \
          & -x + 2 \
\hline
          & 0
\end{array}
$$

So the quotient is $2x^2 + x - 1$, which factors to $(2x - 1)(x + 1)$ as we did before. Wait, but let's check the options again. Option B is $(2x - 1)(x + 1)$, which is the correct factorization of the quotient. Wait, but let's check by multiplying $(x - 2)(2x - 1)(x + 1)$ as we did, which gives the original polynomial. So the other factor is $(2x - 1)(x + 1)$, which is option B.

Wait, but let's check the options again. Option A: $(2x + 1)(x - 1)=2x^2 - 2x + x - 1=2x^2 - x - 1$, which is different. Option B: $(2x - 1)(x + 1)=2x^2 + 2x - x - 1=2x^2 + x - 1$, which is the quotient. So the correct answer is B.

# Answer:
B. $(2x - 1)(x + 1)$

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QUESTION IMAGE

Question

Explanation:

Step1: Perform Polynomial Division or Use Synthetic Division

Step2: Factor the Quotient Polynomial

Answer: