SOLUTION: {{{x^2(3x^2+13x-10)+x(3x^2+13x-10)-2(3x^2+13x-10)=0}}} can anyone help me factor this? with step by step explanation? thanks!

Algebra ->  Equations -> SOLUTION: {{{x^2(3x^2+13x-10)+x(3x^2+13x-10)-2(3x^2+13x-10)=0}}} can anyone help me factor this? with step by step explanation? thanks!      Log On


   

Question 622204: x%5E2%283x%5E2%2B13x-10%29%2Bx%283x%5E2%2B13x-10%29-2%283x%5E2%2B13x-10%29=0
can anyone help me factor this? with step by step explanation? thanks!
Found 2 solutions by solver91311, matineesuxxx:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Do a little substitution trick to make this look simpler. Let . Then your problem looks like:



Factor out the and then substitute back.

John

My calculator said it, I believe it, that settles it
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Answer by matineesuxxx(27) About Me  (Show Source):
You can put this solution on YOUR website!
now first thing you need to notice with this function,x%5E2%283x%5E2%2B13x-10%29%2Bx%283x%5E2%2B13x-10%29-2%283x%5E2%2B13x-10%29=0 is that 3x%5E2%2B13x-10 is a common factor so you get %283x%5E2%2B13x-10%29%28x%5E2%2Bx-2%29 now to factor the first quadratic equation, multiply a by c so it is 3%2A-10 now what two numbers multiply together to give you -30 yet add together to give you a sum of 13. in this case we use +15 and -2 and we get %283x%5E2%2B15x%29%2B%28-2x-10%29=0 you can factor out a 3x from the first bracket and a -2 from the second giving you 3x%5E2%28x%2B5%29-2%28x%2B5%29 and now %28x%2B5%29 is a common factor giving us a result of %283x-2%29%28x%2B5%29
now for the second part.
we have %28x%5E2%2Bx-2%29 this factors easily enough into %28x%2B2%29%28x-1%29

so our factored form is %283x-2%29%28x%2B5%29%28x%2B2%29%28x-1%29