Question 74704: I am having problems solving this equation 
This is what I have done:
x^3-6x^2-15x-8=0
x^3-6x^2-15x=8
x(x^2-6x-15)=8
i tried to factor x^2-6x-15 by completing square like this, since it is not a perfect squared trinomial.
x(x^2-6x+9)-15-9=8
x(x-3)^2-32=0
I am stucked here. How can I find the roots of the above polynomial?
Someone showed me the factored answer
(x+1)(x+1)(x-8)=0 and then the roots are x=-1, 8
If you multiply the above factors it will give you the above polynomial. And the roots satisfy the equation. But how can i get this answer?
I appreciate your help. Thank you very much.
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! Solve:
You could try using the "Factor theorem" (related to the "remainder theorem")which states that..."if (x-a) is a factor of the polynomial, there will be no remainder (when the polynomial is divided by (x-a)), therefore f(a) = 0.
How do you apply this to factoring your polynomial?
You will need to choose values that make sense as possible factors.
Using the constant term in the polynomial, (-8), what are the factors of -8?
1(-8) = -8
(-1)(8) = -8
2(-4) = -8
(-2)(4) = -8
So, the sensible choices are:
1, -1, 2, -2, 4, -4, 8, -8
Start with 1.
f(1) = 1^3-6(1)^2-15(1)-8
f(1) = 1-6-15-8
f(1) = -28 Since this is not = 0, then (x-1) is not a factor.
Try -1.
f(-1) = (-1)^3-6(-1)^2-15(-1)-8
f(-1) = -1-6+15-8
f(-1) = -15+15
f(-1) = 0, so (x+1) is a factor.
Now try 8 because you know that (-1)(8) = -8, the constant term.
It also makes sense that (x+1) is a repeated factor because of the fact that (-1)(8) = -8
f(8) = 8^3-6(8)^2-15(8)-8
f(8) = 512-384-120-8
f(8) = 512-512
f(8) = 0 so (x-8) is a factor.
So, now we have:
(x+1)(x+1)(x-8) = 0
x+1 = 0, so x = -1
x+1 = 0, so x = -1
x-8 = 0, so x = 8
Check:
 = 
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