SOLUTION: x^3-13x+12=0

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Question 891074: x^3-13x+12=0
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Can you factorize? Try checking possible rational roots. Choose from plus or minus 1,2,3,4,6, maybe 12.

Using synthetic division to check rational roots:
1 is a root. The binomial factor is x-1.
This means, the cubic expression in the equation is %28x-1%29%28x%5E2-12x-12%29.

Now check the quadratic factor:

Discriminant is 144-4%2A%28-12%29=144%2B48=192
192=4%2A48=4%2A12%2A4=4%2A4%2A4%2A3

x=%2812%2B-+sqrt%284%2A4%2A4%2A3%29%29%2F2
x=%2812%2B-+sqrt%282%5E6%2A3%29%29%2F2
x=%2812%2B-+8sqrt%283%29%29%2F2
highlight%28x=6%2B-+4sqrt%283%29%29----the other two roots.

Answer by MathTherapy(10551) About Me  (Show Source):
You can put this solution on YOUR website!

x^3 - 13x + 12 = 0


Is it that difficult to state what you want?

Assuming you want this trinomial solved, one of its roots is - 4, so x + 4 is a factor.

%28x%5E3+-+13x+%2B+12%29%2F%28x+%2B+4%29 = x%5E2+-+4x+%2B+3, and factors to: (x - 3)(x - 1)

Therefore, x%5E3+-+13x+%2B+12+=+0 becomes: %28x+%2B+4%29%28x+-+3%29%28x+-+1%29+=+0, so solutions or roots are: highlight_green%28highlight_green%28x+=+-+4_3_and_1%29%29