SOLUTION: Please help me factor this polynomial: {{{8x^3-12x^2+6x-1}}}. I am trying to figure out how to solve it using the factor theorem. My answer key uses 2 as the factor of the constant

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Please help me factor this polynomial: {{{8x^3-12x^2+6x-1}}}. I am trying to figure out how to solve it using the factor theorem. My answer key uses 2 as the factor of the constant      Log On


   

Question 363157: Please help me factor this polynomial: 8x%5E3-12x%5E2%2B6x-1. I am trying to figure out how to solve it using the factor theorem. My answer key uses 2 as the factor of the constant term. And then the synthetic division looks like this: 1-6 12-8
2- 8 8
________
1-4 4 R.0
I do not understand how you can use 2 as the factor of the constant term. Thank you for your assistance.
Found 2 solutions by ewatrrr, stanbon:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!

Hi, 
substituion verifies that 1/2 is a root (1 - 3 + 3 - 1 = 0)


          8x^2 -   8x +     2
        __________________________
(x -.5) | 8x^3 -  12x^2 +   6x - 1





applying quadratic equation

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

x+=+%288+%2B-+sqrt%28+8%5E2-64+%29%29%2F%2816%29+

x = 1/2

factoring 8x%5E3-12x%5E2%2B6x-1 would result as: 

(x-.5)(x -.5)(x -.5)

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Please help me factor this polynomial:8x^3-12x^2+6x-1 . I am trying to figure out how to solve it using the factor theorem. My answer key uses 2 as the factor of the constant term.
And then the synthetic division looks like this: 1-6 12-8
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I graphed the cubic and found a zero at x = 1/2.
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(1/2)....8....-12....6....-1
..........8......-8....2...| 0
Factors: (2x-1)(8x^2 - 8x + 2)
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You can factor further.
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Cheers,
Stan H.