Question:
When $p(x)=x^{4}+2 x^{3}-3 x^{2}+x-1$ is divided by $(x-2)$, the remainder is
(a) 0
(b) $-1$
(c) $-15$
(d) 21
Solution:
(d) 21
$x-2=0 \Rightarrow x=2$
By the remainder theorem, we know that when $p(x)$ is divided by $(x-2)$, the remainder is $p(2)$.
Thus, we have:
$p(2)=2^{4}+2 \times 2^{3}-3 \times 2^{2}+2-1$
$=16+16-12+1$
$=21$
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