Question

factor the trinomial by grouping (10x)^3-(26x)^2-12x

Answer 1

i cant figure out the comon factor

Answer 2

from here do you distribute the \[2x\]

Answer 3

can is now be solved in the form of \[ax ^{2}+bx+c?\]

Answer 4

if your expression out in expanded form you have

\[1000x^3 - 676x^2 - 12x\]

so the most obvious common factor is 4x

\[4x(250x^2 - 169x - 3)\]

Answer 5

what do you get for an answer because im not coming up with anytihng close to where your getting your numbers :/

Answer 6

well you had \[(10x)^3 - (26x)^2 - 12x = 10^3x^3 - 26^2x^2 - 12x\]

hence

\[1000x^3 - 676x^2 - 12x\]

so my information can from what you typed

Answer 7

i might have typed it wrong let me re type it....

Answer 8

\[10x ^{2}-26x ^{2}-12x\]

Answer 9

ok... so if its

\[10x^3 -26x^2 - 12x \]

each term has 2x as a factor so taking that out as the common factor you have

\[2x(5x^2 - 13x - 6)\]

Answer 10

yeah i have that and now im confused on where to go from here

Answer 11

not asking for the answer , im looking for some help in getting the answer

Answer 12

well the quadratic can be factored as well...

Answer 13

okay let me see what i come up with...

Answer 14

\[2x(5x ^{2}-15x+2x-6)\] is what i got from factoring out the quadratic

Answer 15

That's correct so far, now group out the common factor from \(5x^2 - 15x\) and the common factor from \(2x - 6\)

Answer 16

\[5x(x-3)+2(x-3)\]

Answer 17

for my answer i get \[(x-3), (5x+2)\]

Answer 18

yes, but don't forget you also need the 2x in the factorisation :p

Answer 19

So the complete factorisation of \(10x^3 - 26x^2 - 12x\) would be: :)?

Answer 20

so would it be \[2x(x-3)(5x+2)\]

Answer 21

yep! :)

Answer 22

thanks meepi for the help, im sure ill be back on with more trinomials, not my strongest thing

Answer 23

Alright, good luck with them

Answer 24

thanks