please help... factor completely: 8x^2(5x+7)-10x(5+7)-3(5+7)

Question

anonymous · Answer

is the last two terms in the parenthesis correct or should they both also be (5x+7) like that first one?

seattle12345 · Answer

oops sorry yes they should be (5x+7)... so I was thinking (5x+7)^3 but not sure where to go from there.

anonymous · Answer

(5x+7)^3 is one part, to get the other part, just group the other factors in each of the (5x+7), namely: 8x^2, -10x, -3

now the part with the above, can be farther factorize, but I'll let you try it out first.

anonymous · Answer

you may try taking out the common factor (5x+7). then factorise (8x^2-10x-3).
it would be like  (5x+7)*(8x^2-10x-3).

anonymous · Answer

u cannot take it as (5x+7)^3

anonymous · Answer

Hope u know to factorise.

anonymous · Answer

any problem............can i help?

phi · Answer

this problem
$$  8x^2(5x+7)-10x(5x+7)-3(5x+7)$$
is like this one
$$  8x^2y-10xy-3y$$
where y is short for (5x+7)
you can factor out the y
$$  (8x^2-10x-3)y$$
notice you can "undo" the factor out of y by "distributing" the y:  multiply y times each term inside the parenthesis to get the original expression.

now you have the job of factoring 
$$  (8x^2-10x-3)$$

try (4x -  ?)  (2x + ?)  with 1 and 3  or 
 (4x +  ?)  (2x - ?)