Question 1020827: Let P(x) = x^3-3x^2+6. If P(x)=q(x)(x+2)+r(x). What is the value of r(x)
Answer by ikleyn(52781)   (Show Source):
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Let P(x) = x^3-3x^2+6. If P(x)=q(x)(x+2)+r(x). What is the value of r(x)
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Substitute x = -2 into this equality: P(x)=q(x)(x+2)+r(x).
What will you get?
Right, P(-2) = r(-2).
Therefore, to find r(x) (which is simply and actually a constant), simply substitute x = -2 into P(x) and calculate the answer.
This all is around the Remainder Theorem.
See the lesson Divisibility of polynomial f(x) by binomial x-a in this site.
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