SOLUTION: The instruction tell me to factor completely given that the binomial following is the factor of the binomial. Please help i thought i had the hang of it it but guess not!!! x3 + 2

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Question 289099: The instruction tell me to factor completely given that the binomial following is the factor of the binomial. Please help i thought i had the hang of it it but guess not!!!
x3 + 2x2 - 5x - 6, x + 3
Found 3 solutions by stanbon, CharlesG2, richwmiller:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website!
The instruction tell me to factor completely given that the binomial following is the factor of the binomial. Please help i thought i had the hang of it it but guess not!!!
x3 + 2x2 - 5x - 6, x + 3
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Using synthetic division:
-3)....1....2....-5....-6
.......1...-1....-2...|..0

Quotient: x^2-x-2
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Solve x^2-x-2 = 0
(x-2)(x+1) = 0
x = 2 or x = -1
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Cheers,
Stan H.

Answer by CharlesG2(834)   (Show Source):
You can put this solution on YOUR website!
The instruction tell me to factor completely given that the binomial following is the factor of the binomial. Please help i thought i had the hang of it it but guess not!!!
x3 + 2x2 - 5x - 6, x + 3
x + 3 --> x^3 + 2x^2 - 5x - 6 (x^2)
x^3 + 3x^2
-x^2 - 5x - 6 (-x)
-x^2 - 3x
- 2x - 6 (-2)
- 2x - 6
0
(x + 3)(x^2 - x - 2)
(x + 3)(x - 2)(x + 1) (x^2 + x - 2x - 2 by FOIL last 2 terms)

Answer by richwmiller(17219)   (Show Source):
You can put this solution on YOUR website!
Use long division or synthetic division to get the quotient of the equation.
Once you have it that far you can use normal quadratic formula etc to solve.
BTW 2x2 should be 2x2
and x3 should be x^3