How do you use the remainder theorem to find P(-3) for $P (x) = x^{3} + 3 x^{2} - x - 8$ $P(x) = x^3 + 3x^2 - x - 8$ ?

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Answer 1 · 2015-09-17T15:52:52

It is $P (- 3) = - 5$ $P(-3)=-5$

In order to find the value of $P (- 3)$ $P(-3)$ you need to divide the P(x) with $x + 3$ $x+3$ hence

$P (x) = x^{3} + 3 x^{2} - x - 8 = (x + 3) \cdot k (x) + q (x)$ $P(x)=x^3+3x^2-x-8=(x+3)*k(x)+q(x)$

or $x^{3} + 3 x^{2} - x - 8 = (x^{2} - 1) \cdot (x + 3) - 5$ $x^3+3x^2-x-8=(x^2-1)*(x+3)-5$

Hence the $P (- 3) = - 5$ $P(-3)=-5$

How do you use the remainder theorem to find P(-3) for P(x)=x3+3x2−x−8P(x) = x^3 + 3x^2 - x - 8?