Question

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

Answer 1

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

Answer 2

@dopek.i.d.s is incorrect

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

after that i think it can be factored :)

Answer 8

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

Answer 9

yes so lets see how to do that

Answer 10

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

Answer 11

u guys go

Answer 12

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

Answer 13

i feel stupid now

Answer 14

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

Answer 15

This is where it gets difficult to factor.

Answer 16

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

Answer 17

_/-= The square root of.

Answer 18

Okay:)

Answer 19

\[6(x+5)^2-5(x+5)+1\]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

Answer 20

@Jeans123 hope that helps! :)

Answer 21

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

Answer 22

she* ;)

Answer 23

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

Answer 24

Thank you both of you he and she;)

Answer 25

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

Answer 26

In about five minutes I will:)

Answer 27

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