Question

how do I factor the trinomial by grouping?
12m^2-11m-5

Answer 1

By any chance are you completing the square in this unit?

Answer 2

it doesn't specify

Answer 3

Do you know how to factor using completing the square?

Answer 4

no

Answer 5

Ok so that's not going to work. Let me look at this from a less-complicated viewpoint, ok?

Answer 6

yes

Answer 7

Ok, got it. Look at it like this: the goal here is to either combine like variables that can factor somehow, or numbers that are common so those can factor.

Answer 8

Here, the numbers have nothing in common: there is no one number that can be divided evenly into our constants, so we have to look at the variable m.

Answer 9

Group together the terms that have something in common. Here, like I said, it's your terms with the m in them.

Answer 10

\[(12m ^{2}-11m)-5\] What can you factor out of the parenthesis?

Answer 11

so do we multiply the 12^2 and add the answer with the -11

Answer 12

What can you factor out of \[(12m ^{2}-11m)\]?

Answer 13

133m

Answer 14

m(12m-11)

Answer 15

Very good! So what you are left with is this:

Answer 16

-5

Answer 17

\[m(12m - 11) -5\]The commutative property of multiplication tells us that this is an equivalent way of writing the above expression:\[(12m - 11)(m-5)\]Those are your 2 factors now. See that?

Answer 18

yes, thank you

Answer 19

oops, I made an error. If you come back I will correct it and show you the right way (it's been a while and this, being a more complicated expression, threw me off). Please reply to this and let me correct the error of my ways! So sorry!

Answer 20

The answer is actually (3m + 1)(4m - 5). I'll show you how to get it if you look back to this post. Again, so sorry! I over looked the fact that the coefficient in front of the m^2 was something other than a 1. Big mistake.

Answer 21

Ok just in case you come back to this and I am not available, I'm going to tell you how to do this. This case is with a trinomial (3 terms). It's actually quite easier when you have 4 terms or 6...but nonetheless, here are the steps.

Answer 22

Since the coefficient in front of the m^2 is a 12, you multiply that by the last term 5. That gives you 60. Now what you do is find ALL the factors of 60. 1, 60...2, 30...3, 20...4, 15...5, 12. Your goal here is to find the two numbers that, while multiplied together equal 60, they can combine either by addition or subtraction to equal 11. That would be the 4 and the 15. The 4 would have to be positive and the 15 would have to be negative because 4 - 15 = -11 and -11 is the coefficient in front of the m. So combine those two numbers, the 4 and thee -15 in such as way as to equal -11 in your expression. That would look like this:

Answer 23

\[12m ^{2}+4m-15m-5\]But because you need to always have a Plus sign in between the two factors, you would change the sign and then accommodate that sign change within the parenthesis when you group the terms together.

Answer 24

\[(12m ^{2}+4m)+(-15m-5)\]That ensures that the signs will be correct when you go to factor and get your final result. Now factor out what is common within each individual set of parenthesis. In the first one, it is a 4m, and in the second one it is a -5. Here's how you do that and what it looks like:

Answer 25

\[4m(3m+1)-5(3m+1)\]Now you have yet another "thing" common to both terms, and that is the (3m + 1). Factor that out and you get this:

Answer 26

\[(3m+1)(4m-5)\]See that? I hope you come back to look at this. I apologize again for my oversight of the 12 in the beginning. That changed everything!