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If x^5 - x^4 + x^3 - px^2 + qx + 14 is divisible by both x+2 and x-1, then find p and q.

 Feb 26, 2022
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Let P(x) = x^5 - x^4 + x^3 - px^2 + qx + 14. From the remainder theorem, if P(x) is divisible by x+2, then P(-2) = 0. (The same holds true for x-1.)

 

So,

 

P(-2) = -32 - 16 - 8 + 4p - 2q + 14 = 0

P(1) = 1 - 1 + 1 - p + q + 14 = 0

 

4p - 2q = -42

-p + q = -15

 

Solving this system of equations will yield the answer.

 Feb 27, 2022

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