I need help!!!!!!!!!!!!!!! Factor the expression completely, 6(x+5)^2-5(x+5)+1

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I need help!!!!!!!!!!!!!!! Factor the expression completely, 6(x+5)^2-5(x+5)+1

Mathematics
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6x(x+5)-(5x(+5))+1=6x(x+5)
@dopek.i.d.s is incorrect
I figured... level 48 and 18 at helping people hmmm..:)

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i got to 6x^2 + 55x + 126
umm....its completely factored but i think you mean you want it simplified...? :)
x^2 = x*x (x+a)^2 = (x+a)(x+a) c(x+a)= c*x + c*a
after that i think it can be factored :)
Well I know the answer it is (2x+9)(3x+14)
yes so lets see how to do that
my teacher gave that to me but I do not know how to get it.
u guys go
So we need to simplify it, combine like terms, then re-factor it
i feel stupid now
6(x+5)^2-5(x+5)+1 6(x^2+10x+25)-5x-25+1 6x^2+55x+126
This is where it gets difficult to factor.
x=-b+-_/-b^2-4ac/2a?
_/-= The square root of.
Okay:)
\[6(x+5)^2-5(x+5)+1\]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
@Jeans123 hope that helps! :)
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
she* ;)
Sry just habbit of typing, didn't mean anything by it
Thank you both of you he and she;)
no harm :) ...Youre Welcome Austin! :) feel free to ask more anytime you need help (when im there of course :))
In about five minutes I will:)
i leave pretty soon so ask now if you can :)

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