## anonymous 3 years ago I need help!!!!!!!!!!!!!!! Factor the expression completely, 6(x+5)^2-5(x+5)+1

1. anonymous

6x(x+5)-(5x(+5))+1=6x(x+5)

2. anonymous

@dopek.i.d.s is incorrect

3. anonymous

I figured... level 48 and 18 at helping people hmmm..:)

4. anonymous

i got to 6x^2 + 55x + 126

5. anonymous

umm....its completely factored but i think you mean you want it simplified...? :)

6. anonymous

x^2 = x*x (x+a)^2 = (x+a)(x+a) c(x+a)= c*x + c*a

7. anonymous

after that i think it can be factored :)

8. anonymous

Well I know the answer it is (2x+9)(3x+14)

9. anonymous

yes so lets see how to do that

10. anonymous

my teacher gave that to me but I do not know how to get it.

11. anonymous

u guys go

12. anonymous

So we need to simplify it, combine like terms, then re-factor it

13. anonymous

i feel stupid now

14. anonymous

6(x+5)^2-5(x+5)+1 6(x^2+10x+25)-5x-25+1 6x^2+55x+126

15. anonymous

This is where it gets difficult to factor.

16. anonymous

x=-b+-_/-b^2-4ac/2a?

17. anonymous

_/-= The square root of.

18. anonymous

Okay:)

19. anonymous

$6(x+5)^2-5(x+5)+1$lets simplify this: expand the $$(x+5)$$ and distribute the $$5$$ to the parenthesis$6(x^2+10x+25)-5x-25+1$distribute the $$6$$$6x^2+60x+150-5x-25+1$combine like terms:$6x^2+55x+126$Now , we can factor it...i would use Quadratic equation since i find ti to be the "safest":${-55\pm\sqrt{55^2-4(6)(126)}}\over2(6)$${-55\pm\sqrt{3025-3024}}\over12$${-55\pm\sqrt{1}}\over12$${-55\pm1}\over12$${-56\over12}~~~~or~~~~{-54\over12}$$x=-{14\over3}~~~~or~~~~x=-{9\over2}$ so that means that its...$(x+{14\over3})~~~and~~~(x+{9\over2})$so to make those 'un-fractioned we'll multiply by the denominators:$\large(3x+14)(2x+9)$ tada! :D

20. anonymous

@Jeans123 hope that helps! :)

21. anonymous

I have been proved wrong. @yummydum that was smart to use the quadratic. He found the solutions for x so he knew what the other terms had to be. Then he got rid of the fractions by multiplying by the denominator

22. anonymous

she* ;)

23. anonymous

Sry just habbit of typing, didn't mean anything by it

24. anonymous

Thank you both of you he and she;)

25. anonymous

no harm :) ...Youre Welcome Austin! :) feel free to ask more anytime you need help (when im there of course :))

26. anonymous

In about five minutes I will:)

27. anonymous

i leave pretty soon so ask now if you can :)