If p(x) = x^3-5x^2+4x-3 and g(x)=x-2, show that p(x) is not a multiple of the factor g(x)
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Answered by
172
p(x)= x³-5x²+4x-3 g(x)=x-2 ⇒x=2
so,
⇒p(2)=2³-5(2)²+4(2)-3
⇒p(2)=8-20+8-3
⇒p(2)=16-23
⇒p(2)=-7
so,
⇒p(2)=2³-5(2)²+4(2)-3
⇒p(2)=8-20+8-3
⇒p(2)=16-23
⇒p(2)=-7
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Answered by
73
Answer:-7
Explanation:
g(x)=x-2
x-2=0
x=2
By the remainder theorem,we know that when p(x) is divided by (x-2)then the remainder is p(2)
p(x)=x^3-5x^2+4x-3
p(2)=2^3-5×2^2+4×2-3
=8-5×4+8-3
=8-20+5
=-7
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