SOLUTION: Find the value of p and q if: (x-3) and (x+1) are factors of x^3+px+q Where (x-3) I identified the answer as 3p+27+q (Equation 1) Where (x+1) I identified the answer as -p

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Find the value of p and q if: (x-3) and (x+1) are factors of x^3+px+q Where (x-3) I identified the answer as 3p+27+q (Equation 1) Where (x+1) I identified the answer as -p      Log On


   

Question 935077: Find the value of p and q if:
(x-3) and (x+1) are factors of x^3+px+q
Where (x-3) I identified the answer as 3p+27+q (Equation 1)
Where (x+1) I identified the answer as -p-1+q (Equation 2)

I thought that I would need to use synthetic division for both factors; then when I had the answer from each I'd use simultaneous equation to solve for one term, substitute it into the equation, then simplify for the second term.
So for p; I thought 4p=28; p=6
and for q; I got utterly lost tbh.
If I substitute the value for p where p=6 into Equation 1; I got q=45
But if I substitute the value for p where p=6 into Equation 2; I got -7.
The book says p=-6 and q=-7
So now I am bewildered where Ive gone wrong

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

you did this:
%28x-3%29 identified the answer as 3p%2B27%2Bq.... (Equation 1)
%28x%2B1%29 identified the answer as -p-1%2Bq .....(Equation 2)
just solve this system
3p%2B27%2Bq.... (Equation 1)
-p-1%2Bq .....(Equation 2)
__________________________subtract eq.2 from eq.1
3p%2B27%2Bq+-%28-p-1%2Bq+%29=0+
3p%2B27%2Bcross%28q%29+%2Bp%2B1-cross%28q%29+=0
3p%2B27+%2Bp%2B1+=0

4p%2B28+=0+
4p+=-28
highlight%28p+=-7%29
go to eq.2 plug in value for p
-p-1%2Bq .....(Equation 2)
-%28-7%29-1%2Bq+=0+
7-1%2Bq+=0+

6%2Bq+=0++
highlight%28q=-6%29