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Jeans123
I need help!!!!!!!!!!!!!!! Factor the expression completely, 6(x+5)^2-5(x+5)+1
6x(x+5)-(5x(+5))+1=6x(x+5)
I figured... level 48 and 18 at helping people hmmm..:)
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.
So we need to simplify it, combine like terms, then re-factor it
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.
_/-= The square root of.
\[6(x+5)^2-5(x+5)+1\]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
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 :)