Question

if $f(x) = 3x^6 + x^5 + 4$, then what is the remainder when $f(x)$ is divided by $x + 1$?

Explanation:

Step1: Recall the Remainder Theorem

The Remainder Theorem states that if a polynomial \( f(x) \) is divided by \( x - a \), the remainder is \( f(a) \). For \( x + 1 \), we can rewrite it as \( x - (-1) \), so \( a=-1 \).

Step2: Substitute \( x = -1 \) into \( f(x) \)

Given \( f(x)=3x^{6}+x^{5}+4 \), substitute \( x = -1 \):
\[

$$\begin{align*} f(-1)&=3(-1)^{6}+(-1)^{5}+4\\ &=3(1)+(-1)+4\\ &=3 - 1 + 4\\ &=2 + 4\\ &=6 \end{align*}$$

\]

Answer:

The remainder is \( 6 \).