# Solve this

Question:

When $p(x)=x^{3}-3 x^{2}+4 x+32$ is divided by $(x+2)$, the remainder is

(a) 0

(b) 32

(c) 36

(d) 4

Solution:

(d) 4

$x+2=0 \Rightarrow x=-2$

By the remainder theorem, we know that when p(x) is divided by (x + 2), the remainder is p(-">-2).
Now, we have:

$p(-2)=(-2)^{3}-3 \times(-2)^{2}+4 \times(-2)+32$

$=-8-12-8+32$

$=4$

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