##### Feb. 21, 2014

Session Transcript - Math - Intermediate Algebra, 2/16/2014 1:43PM - Tutor.comSession Date: 2/16/2014 1:43PM
Length: 20.8 minute(s)
Subject: Math - Intermediate Algebra

System Message
[00:00:00] *** Please note: All sessions are recorded for quality control. ***

Guest (Customer)
[00:00:00] *** The student has 673 minutes remaining. ***
[00:00:00] I need some help with factoring 4r^2 - 15r +9 please
[00:00:18] Hi Liz.

Liz A (Tutor)
[00:00:22] Hi, how are you?

Guest (Customer)
[00:00:26] good and you?

Liz A (Tutor)
[00:01:02] Good, thanks :) let's see if we can figure this problem out!

Guest (Customer)
[00:01:35] yep that is it.

Liz A (Tutor)
[00:01:36] OK. Do you have any idea where we should start?

Guest (Customer)
[00:01:58] well I am seeing this: (2r )(2r )
[00:02:10] good start right? LOL
[00:02:23] the middle term and the last term are throwing me.

Liz A (Tutor)
[00:02:36] sure. that's one way this could look

Guest (Customer)
[00:02:36] and I've already erased a hole in my paper

Liz A (Tutor)
[00:03:01] the key here is that the coefficients in front of the "x" terms in our factoring
[00:03:03] should multiply to 4
[00:03:07] so we could have 2,2
[00:03:09] what else could we have?

Guest (Customer)
[00:03:20] 1 and 4

Liz A (Tutor)
[00:03:34] yep, exactly!

Guest (Customer)
[00:04:12] I had not tried 1 and 4 yet

Liz A (Tutor)
[00:04:14] That's a good start. what else do we know here?

Guest (Customer)
[00:04:14] :-)
[00:04:35] I see that 3 and 5 go into 15
[00:04:45] and 3 and 3 go into 9
[00:04:51] so we have 3
[00:05:04] as a common factor between the middle and last term

Liz A (Tutor)
[00:05:24] good, that sounds about right. We need our last terms in our factoring
[00:05:34] to multiply to 9
[00:05:35] right?

Guest (Customer)
[00:05:44] yes that was a question I had.

Liz A (Tutor)
[00:06:09] yes, that's the case here. they don't need to add to 15, though, since we have coefficients that aren't 1 in front of the "r" terms

Guest (Customer)
[00:06:09] so using 1 and 4 for our "r^2

Liz A (Tutor)
[00:06:49] what are the combinations for our constant terms, first?
[00:06:50] 3,3
[00:06:51] what else?

Guest (Customer)
[00:07:10] 1,9 for 9

Liz A (Tutor)
[00:07:20] yep, those are both positive
[00:07:23] we could also have the negatives though, right?

Guest (Customer)
[00:07:30] yes

Liz A (Tutor)
[00:07:47] Can you tell based on the expression above which we'll have? both positive or both negative

Guest (Customer)
[00:08:06] I am going to say negative

Liz A (Tutor)
[00:08:11] yep. why?

Guest (Customer)
[00:08:18] becase 2 negatives = a positive
[00:08:32] and two negatives added are negative

Liz A (Tutor)
[00:08:46] exactly! We still get positive 9, but the only way we can get that -15 is if both our constants are negative

Guest (Customer)
[00:08:53] yes

Liz A (Tutor)
[00:09:05] OK. So, from here it's just a matter of testing combinations until we get to the right one
[00:10:16] Do you agree that these are all the possibilities we can have here?

Guest (Customer)
[00:10:26] yes. I am working on paper also

Liz A (Tutor)
[00:10:27] note that since 1,4 are different coefficients, the order in which we use -1, -9 for instance matters
[00:10:30] so we get two possibilities there

Guest (Customer)
[00:11:06] and the terms are -3 and -3

Liz A (Tutor)
[00:11:23] aha! yep, taht looks like it

Guest (Customer)
[00:11:33] so inside our groups we have (4r-3)(r-3)

Liz A (Tutor)
[00:11:44] Yes! Good work :)

Guest (Customer)
[00:12:22] ok great can you hang with me a second for maybe another

Liz A (Tutor)
[00:12:31] Sure

Guest (Customer)
[00:13:12] ok here is another
[00:13:34] x^2 +22x + 121
[00:13:49] similar set up it looks like
[00:14:16] we don't have a GCF

Liz A (Tutor)
[00:14:21] OK, since we just have an x^2 out front, this makes our job a bit easier
[00:14:26] we know that our factoring is going to look something like

Guest (Customer)
[00:14:30] yay!
[00:14:31] LOL
[00:14:46] and everything is positive.

Liz A (Tutor)
[00:14:48] So let's figure out what combinations of that term we can have

Guest (Customer)
[00:14:49] woohoo

Liz A (Tutor)
[00:14:51] yep!
[00:14:58] Everything is positive here too, so we know that those constants have to be positive
[00:15:06] what combinations of constants can we have here?

Guest (Customer)
[00:15:08] amen!
[00:15:17] ok for 22 we have 2,11
[00:15:22] 11, 1
[00:15:32] -2, -11
[00:15:43] sorry, not 11, 1

Liz A (Tutor)
[00:15:49] careful. I think you want to look at the 121 term first
[00:15:55] remember that our constants multiply to 121

Guest (Customer)
[00:16:05] oh duh. right. sorry
[00:16:07] ok

Liz A (Tutor)
[00:16:09] no problem!

Guest (Customer)
[00:16:29] so 1, 121
[00:16:44] -1, -121

Liz A (Tutor)
[00:16:56] think you're missing one pair :)

Guest (Customer)
[00:17:10] yes. looking now
[00:18:01] 0 and 121
[00:18:29] 11
[00:18:40] yes

Liz A (Tutor)
[00:18:41] yep!

Guest (Customer)
[00:18:50] all prime numbers

Liz A (Tutor)
[00:18:52] Can you see which of the four it needs to be here? they need to add to 22

Guest (Customer)
[00:19:19] 11,11

Liz A (Tutor)
[00:19:53] Yep! Looks like you're getting the hang of this :)

Guest (Customer)
[00:20:01] awesome.

Liz A (Tutor)
[00:20:10] Do you have any questions on anything we've done here?

Guest (Customer)
[00:20:18] no and that was an easy one too
[00:20:22] thank you so much

Liz A (Tutor)
[00:20:35] No problem. Please fill out the survey at the end if you have a moment :) have a nice day!