When p(x)
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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