Factoring $10r^2 - 31r + 15$

Question

What is the answer for factoring:

$$10r^2 - 31r + 15$$

I have tried to solve it. This was my prior attempt:

$$10r^2 - 31r + 15\\ = (10r^2 - 25r) (-6r + 15)\\ = -5r(-2r+5) -3 (2r-5) $$

Answer 1

You lost a $+$ sign in the first line starting with $=$. Then note that in the next line, the expressions inside the parentheses are the same except for multiplying by $-1$. So change the sign inside one set of parentheses and the leading coefficient. $$-5r(-2r+5)-3(2r-5)=5r(2r-5)-3(2r-5)=(5r-3)(2r-5)$$

Answer 2

$10r^2 -31r + 15$

what are the factors of 10: 1, 2,5, and 10. Then what are the factors of 15: 1, 3, 5, and 15. From here, there two ways factoring or using the foil method.

The factoring method.

$10r^2 -31r + 15$

$(10r^2 -25r)(-6r + 15)$

$5r(2r - 5) - 3(2r - 5)$ since we have two of the same $(2r - 5)$, it counts as one.

at the end just combine,

$(5r - 3)(2r - 5)$ = $10r^2 -31 + 15$

I prefer the foil method. Then by knowing the factors of 15.

then, $(? - 3)(? - 5)$ and both negative because of (-31). Then by know what are the factors of 10.

$(5r - 3)(2r - 5)$ = $10r^2 - 31r + 15$