SOLUTION: How do I factor 2x^3+10x^2+6x-18=0 . I know the answer is 2(x-1)(x+3)^2. How do I get there ? I know it is not a perfect square so I cannot factor by grouping. I have tried the ste

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: How do I factor 2x^3+10x^2+6x-18=0 . I know the answer is 2(x-1)(x+3)^2. How do I get there ? I know it is not a perfect square so I cannot factor by grouping. I have tried the ste      Log On


   

Question 1089948: How do I factor 2x^3+10x^2+6x-18=0 . I know the answer is 2(x-1)(x+3)^2. How do I get there ? I know it is not a perfect square so I cannot factor by grouping. I have tried the steps in a difference of cubes and end up having to square root 18 which gives me a decimal so that doesn't work either. Help
Found 2 solutions by josgarithmetic, Edwin McCravy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
2%28x%5E3%2B5x%5E2%2B3x-9%29=0
and then by Rational Roots Theorem some possible roots to try testing are -3, -1, 1, 3. 

1   |   1   5   3   -9
    |       1   6   9
    |_____________________
       1    6   9    0

     Meaning the next factor to break further if possible is  x%5E2%2B6x%2B9.

-
2%28x-1%29%28x%5E2%2B6x%2B9%29=0, and the quadratic factor is easily factored further:

2%28x-1%29%28x%2B3%29%28x%2B3%29=0

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!

2x³+10x²+6x-18=0 

2(x³+5x²+3x-9)=0

Observe that x=1 is a solution to that equation since 

2(1³+5∙1²+3∙1-9) = 2(1+6+3-9) = 0

Therefore x-1 must be a factor of x³+5x²+3x-9.
So we divide by long division:

        x²+6x+9
x-1)x³+5x²+3x-9
    x³- x²
       6x²+3x
       6x²-6x
           9x-9
           9x-9
              0

As we expect, the remainder is 0, thus the partial
factorization of is x³+5x²+3x-9 is this:   

(x-1)(x²+6x+9) 

and its complete factorization is

(x-1)(x+3)(x+3) or

(x-1)(x+3)²

Therefore the complete factorization of

2x³+10x²+6x-18=0

is

2(x-1)(x+3)² = 0

Edwin