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zeekarn zeekarn
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11 years ago
Find the FACTORS of 6x3 - 13 x 2 - 41x - 12
Enter the FACTORS  in the form (ax+b) where 'a' and 'b' are integers. Begin with the factor that has the smallest "a" value and end with the factor that has the largest 'a' value.  I'm lost. Thanks
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dbassandbassan
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11 years ago
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11 years ago
(x-4)(2x+3)(3x+1)
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#3
11 years ago
The rational root theorem says the possible rational roots of 6x^3 - 13x - 41x - 12 = 0 are factors of 12 divided by factors of 6, positive and negative.

+ or  -
1, 1/2, 1/3, 1/6, 2, 2/3, 3, 3/2, 1/2, 6

That is 20 possible rational roots to substitute for x and see if the polynomial equals 0.

A faster method is to use a graphing calculator to find the zeros or roots of the polynomial equation.  We found x = 4, x = -1/3, and x = -3/2.

Each root has a related factor of the polynomial equation:

(x - 4)(x - (-1/3))(x - (-3/2)) = 0
(x - 4)(x + 1/3)(x + 3/2) = 0
(x - 4)3(x + 1/3)2(x + 3/2) = 0
(x - 4)(3x + 1)(2x + 3) = 0

Write the left side of the equation to meet the requirements of the problem:

ANSWER:   (x - 4)(2x + 3)(3x + 1)

Check:
(x - 4)(2x + 3)(3x + 1)
= (2x^2 - 5x - 12)(3x + 1)
= 6x^3 - 15x^2 - 36x + 2x^2 - 5x - 12
= 6x^3 - 13x^2 - 41x - 12
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