3. $(x^3 - 13x + 12) div (x + 4)$

Question

Accepted Answer

# Explanation:
## Step1: Use polynomial long division
Divide $x^3 - 13x + 12$ by $x + 4$. First, divide the leading term $x^3$ by $x$ to get $x^2$. Multiply $x + 4$ by $x^2$ to get $x^3 + 4x^2$. Subtract this from the original polynomial:
$$
\begin{align*}
&(x^3 - 13x + 12) - (x^3 + 4x^2)\
=&x^3 - 13x + 12 - x^3 - 4x^2\
=& -4x^2 - 13x + 12
\end{align*}
$$
## Step2: Divide the new leading term $-4x^2$ by $x$ to get $-4x$. Multiply $x + 4$ by $-4x$ to get $-4x^2 - 16x$. Subtract this from the previous result:
$$
\begin{align*}
&(-4x^2 - 13x + 12) - (-4x^2 - 16x)\
=& -4x^2 - 13x + 12 + 4x^2 + 16x\
=& 3x + 12
\end{align*}
$$
## Step3: Divide the new leading term $3x$ by $x$ to get $3$. Multiply $x + 4$ by $3$ to get $3x + 12$. Subtract this from the previous result:
$$
\begin{align*}
&(3x + 12) - (3x + 12)\
=& 3x + 12 - 3x - 12\
=& 0
\end{align*}
$$
So, putting it all together, the quotient is $x^2 - 4x + 3$ and the remainder is $0$. We can also factor the quotient: $x^2 - 4x + 3=(x - 1)(x - 3)$.
# Answer:
The result of $(x^3 - 13x + 12)\div(x + 4)$ is $x^2 - 4x + 3$ (or $(x - 1)(x - 3)$) with a remainder of $0$.

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QUESTION IMAGE

Question

Explanation:

Step1: Use polynomial long division

Step2: Divide the new leading term \(-4x^2\) by \(x\) to get \(-4x\). Multiply \(x + 4\) by \(-4x\) to get \(-4x^2 - 16x\). Subtract this from the previous result:

Step3: Divide the new leading term \(3x\) by \(x\) to get \(3\). Multiply \(x + 4\) by \(3\) to get \(3x + 12\). Subtract this from the previous result:

Answer: