Question

1. \\(\frac{2x^3 + 6x + 152}{x + 4}\\)

Explanation:

Step1: Use polynomial long division

We divide \(2x^3 + 6x + 152\) by \(x + 4\). First, divide the leading term \(2x^3\) by \(x\) to get \(2x^2\). Multiply \(x + 4\) by \(2x^2\) to get \(2x^3 + 8x^2\). Subtract this from the original polynomial:
\[

$$\begin{align*} (2x^3 + 6x + 152)-(2x^3 + 8x^2)&= -8x^2 + 6x + 152 \end{align*}$$

\]

Step2: Divide the new leading term

Divide \(-8x^2\) by \(x\) to get \(-8x\). Multiply \(x + 4\) by \(-8x\) to get \(-8x^2 - 32x\). Subtract this from the previous result:
\[

$$\begin{align*} (-8x^2 + 6x + 152)-(-8x^2 - 32x)&= 38x + 152 \end{align*}$$

\]

Step3: Divide the new leading term

Divide \(38x\) by \(x\) to get \(38\). Multiply \(x + 4\) by \(38\) to get \(38x + 152\). Subtract this from the previous result:
\[

$$\begin{align*} (38x + 152)-(38x + 152)&= 0 \end{align*}$$

\]
So, the division gives \(2x^2 - 8x + 38\) with a remainder of 0.

Answer:

\(2x^2 - 8x + 38\)