Question

Factor completely
6a^3 - 10a^2 + 3a - 5
I will give medal! Can you pls show me step by step how to do this? I want to know how to factor it next time by myself. Thx!

Answer 1

We can solve this one through grouping. First, group the terms.

(6a^3-10a^2) + (3a-5)

Then we will factor out anything we can from the two groups we have now. This looks like:
2a^2(3a-5) + (3a-5)

Make sense so far?

Answer 2

I don't get the second step

Answer 3

Okay we can look at that :)

Answer 4

btw I gtg I will be back in about 10 min. Sorry. but if you could show me the steps I will look over them. if you are still here in 10 min, I will be back

Answer 5

So after we group we have

(6a^3-10a^) + (3a-5)

The next step is to factor out anything in common.
A simpler example would be
(6x+3)
both of these terms can be divided by 3 so we can "take out" the three and put it on the outside of the parentheses.
3(2x+1)

In this case the common factor in 2a^2 because both 6a^3 and 10a^ can be divided by it.
So now we place it outside the parentheses and divide just like with our example which leaves us with.
2a^2(3a-5)

Does that help at all?

Answer 6

The second group can't be divide (or factored out) anymore so we leave it.
Now, we have 2a^2(3a-5) + (3a-5).
Now this part is a bit confusing so I'll give you this link ( http://www.mesacc.edu/~scotz47781/mat120/notes/factoring/grouping/grouping.html )which can probably explain it much better than me and you let me know if you have any other questions Okay? I will check back periodically if you don't see this soon so you can comment any questions (tag me) or PM me.

Answer 7

Basically at the end your answer will be (2a^2+1)(3a-5) You can check this by multiplying.

Answer 8

Thx I get it now! @alexishope47

Answer 9

Whooohooooo!!