SOLUTION: Given that P(x) = x^4 + ax^3 - x^2 + bx - 12 has factors x-2 and x+1,solve the equation P(x) = 0.

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Question 1183635: Given that P(x) = x^4 + ax^3 - x^2 + bx - 12 has factors x-2 and x+1,solve the equation P(x) = 0.
Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


Since x-2 is a factor of P(x), P(2) is 0:
16%2B8a-4%2B2b-12=0
8a%2B2b=0
4a%2Bb=0
b=-4a [1]

Since x+1 is a factor of P(x), P(-1)is 0:
1-a-1-b-12=0
a%2Bb=-12 [2]

Solving [1] and [2] gives a=4, from which b=-16

So the polynomial is

P%28x%29=x%5E4%2B4x%5E3-x%5E2-16x-12

Use synthetic division to factor out the two known linear factors:

  2 | 1  4  -1 -16 -12
    |    2  12  22  12
    ------------------
 -1 | 1  6  11   6   0
    |   -1  -5  -6
    --------------
      1  5   6   0

The reduced polynomial is

x%5E2%2B5x%2B6+=+%28x%2B2%29%28x%2B3%29

So the other two roots are -2 and -3.

ANSWER: P(x)=0 --> x = -3, -2, -1, and 2

A graph confirms those roots:

graph%28400%2C400%2C-5%2C5%2C-20%2C20%2Cx%5E4%2B4x%5E3-x%5E2-16x-12%29