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(
Now, we have:
$p(-2)=(-2)^{3}-3 \times(-2)^{2}+4 \times(-2)+32$
$=-8-12-8+32$
$=4$
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