Question 1020827
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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 <U>Remainder Theorem</U>. 


See the lesson <A HREF=http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Divisibility-of-polynomial-f%28x%29-by-binomial-x-a.lesson>Divisibility of polynomial f(x) by binomial x-a</A> in this site.