Question

Factor completely.
\[10a ^{2}+25a-15\]

Answer 1

divide everything out by 5 first

Answer 2

2,5,-3

Answer 3

5(a+3)(2a-1)

Answer 4

oh no you're right! there needs to be a 2 there, haha

Answer 5

that was where this one got me confused. Because I kept getting what you got but it didn't add up. @Jhannybean

Answer 6

\[\large 5(2a^2+5a-3)\] in order to factor this, you can do what @lujanels1 has done,or you can try my method, works everytime. \[\large 5(\color{red}2a^2+5a-\color{red}3)\] multiply them both. \[\large 5(a^2+5a-6)\]find two numbers that multiply to 6 and add to 5 \[\large 5(a-1)(a+6)\]divide by the leading coefficient 2 \[\large (a-\frac{1}{\color{red}2})(a+\frac{6}{\color{red}2})\]simplify again \[\large (2a-1)(a+3)\]

Answer 7

oh i forgot to multiply the 5 with it too.

Answer 8

sorry.

Answer 9

\(\large 5(2a-1)(a+3)\)****

Answer 10

But i just did that?

Answer 11

It's fine. I just don't understand how to multiply it to check it with it in \[5(a+3)(2a-1)\] form.

Answer 12

I factorized it though :( It's alittle longer method but it works.

Answer 13

you use the FOIL method, @Jewels89 \[\large 5[a(2a-1)+3(2a-1)]\] you get \[\large 5[(a)(2a)-(a)(1)+(3)(2a)-(3)(1)]\]

Answer 14

ok. Now I get it. THANK YOU!!! :)

Answer 15

no problem :)