SOLUTION: I have to solve this factoring problem : 6x^4-10x^3-6x^2 the # after the ^ is the exponents I found a GCF of 2x^2 and left off at: 2x^2(3x^2-5x-3) I did x+x=b which is -5 a

Algebra ->  Radicals -> SOLUTION: I have to solve this factoring problem : 6x^4-10x^3-6x^2 the # after the ^ is the exponents I found a GCF of 2x^2 and left off at: 2x^2(3x^2-5x-3) I did x+x=b which is -5 a      Log On


   

Question 1069316: I have to solve this factoring problem :
6x^4-10x^3-6x^2
the # after the ^ is the exponents
I found a GCF of 2x^2 and left off at:
2x^2(3x^2-5x-3)
I did x+x=b which is -5
and x*x=ac which is -9
so what two numbers would added give me -5 but multiplied give me -9?
Thats as far as i got. Please help me
Thanks
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Common factor, 2x%5E2;
2x%5E2%283x%5E2-5x-3%29

Discriminant?
%28-5%29%5E2%2B4%2A3%2A3=25%2B36=71, not a perfect square.

highlight%282x%5E2%283x%5E2-5x-3%29%29

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

I have to solve this factoring problem :
6x^4-10x^3-6x^2
the # after the ^ is the exponents
I found a GCF of 2x^2 and left off at:
2x^2(3x^2-5x-3)
I did x+x=b which is -5
and x*x=ac which is -9
so what two numbers would added give me -5 but multiplied give me -9?
Thats as far as i got. Please help me
Thanks 
Good job!! You're correct up to here =======> 2x%5E2%283x%5E2+-+5x+-+3%29

I see you tried to find factors with a product of - 9, and that sum to - 5 (ac method).

Those integers do not exist, so this is where the factoring ends, as it cannot be factored further.