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.

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) (Show Source): You can put this solution on YOUR website!

Since x-2 is a factor of P(x), P(2) is 0:

[1]

Since x+1 is a factor of P(x), P(-1)is 0:

[2]

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

So the polynomial is

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

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

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

A graph confirms those roots:

RELATED QUESTIONS
1)consider the polynomial p(x)=x^3+ax^2+bx-12 given that (x+3)and (x-4)are factors of... (answered by Edwin McCravy)