How do you use the remainder theorem to find P(c) $P (x) = x^{3} + 4 x^{2} - 8 x - 6$ $P(x)=x^3+4x^2-8x-6$ ,a=-2?

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Answer 1 · 2015-09-15T07:07:11

18

To find the remainder if P(x) is divided by x+a , plug in x= -a in P(x)

P(-a) = $- a^{3} + 4 a^{2} + 8 a - 6$ $-a^3 +4a^2 +8a -6$ . Here a is given to be -2, hence P(-2)= -8+16+16-6=18

How do you use the remainder theorem to find P(c) P(x)=x^3+4x^2-8x-6P(x)=x3+4x2−8x−6P(x)=x^3+4x^2-8x-6,a=-2?