Tutor.com Polynomials Session

Feb. 15, 2014

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Session Transcript - Math - College Algebra, 2/3/2014 9:34PM - Tutor.comSession Date: 2/3/2014 9:34PM
Length: 28.6 minute(s)
Subject: Math - College Algebra


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

Guest (Customer)
[00:00:00] *** The student has 1153 minutes remaining. ***
[00:00:02] I have a math problem of adding polynomials and combining like term that involve fractions and I cannot figure out how to do it.

Clifford T (Tutor)
[00:00:09] Hello and welcome to tutor.com !

Guest (Customer)
[00:00:13] Hi

Clifford T (Tutor)
[00:00:21] So what problem were you looking at?

Guest (Customer)
[00:00:39] Would it be easier to put it on the white board?

Clifford T (Tutor)
[00:00:46] probably

Guest (Customer)
[00:01:01] Not good at this but here goes

Clifford T (Tutor)
[00:01:06] typing up equations can be rough
[00:01:15] neither am I :)
[00:01:30] It looks fine so far

Guest (Customer)
[00:02:42] there you go

Clifford T (Tutor)
[00:02:47] ok
[00:03:00] So first, since all of these are added together, we can drop all the parenthesis
[00:03:27] let's just erase

Guest (Customer)
[00:03:27] startie

Clifford T (Tutor)
[00:03:33] So you don't have to write them up again
[00:03:52] ok
[00:04:00] Now we have to collect "like terms"
[00:04:04] we can only add two terms
[00:04:10] if they have the same power of x
[00:04:22] like the two x^2 terms
[00:04:33] and the two x terms

Guest (Customer)
[00:04:36] ok I follow u

Clifford T (Tutor)
[00:04:44] ok
[00:05:23] good!
[00:05:26] and the next two?

Guest (Customer)
[00:06:03] or is the a minus?

Clifford T (Tutor)
[00:06:12] yes
[00:06:24] ok
[00:06:30] Now let's focus on the x^2
[00:06:44] since both terms have an x^2, we can factor it out front
[00:07:01] like that

Guest (Customer)
[00:07:04] ok

Clifford T (Tutor)
[00:07:06] does that make sense?

Guest (Customer)
[00:07:10] yes

Clifford T (Tutor)
[00:07:17] good
[00:07:29] Now we need to add those two fractions together

Guest (Customer)
[00:07:49] how do you navigate this board

Clifford T (Tutor)
[00:08:12] you mean to make more room at the bottom?
[00:08:52] So remember that to add fractions, we need a common denominator first
[00:09:15] So what is a number that both 7 and 3 divide?

Guest (Customer)
[00:09:52] This is where I get lost

Clifford T (Tutor)
[00:10:01] ok, no problem
[00:10:11] So one way that always works to find a common denominator
[00:10:18] is to multiply the two numbers together
[00:10:23] So 7*3 = 21
[00:10:36] this will certainly be a number that both 7 and 3 will divide
[00:10:50] good
[00:10:57] now we have to convert the numerators
[00:11:11] they won't be the same as they were
[00:11:26] So to get 7 to turn into 21, what did we have to multiply by?

Guest (Customer)
[00:12:02] 3 x 7

Clifford T (Tutor)
[00:12:05] right
[00:12:14] we had to multiply 7 by 3
[00:12:19] so for the first fraction
[00:12:25] we will also have to multiply the 6 by 3
[00:12:35] which is a little lower than 36
[00:12:47] right!
[00:12:53] Now for the second fraction
[00:13:01] what did we multiply 3 by to get 21 again?
[00:13:15] perfect!
[00:13:33] now we keep the common denominator
[00:13:36] and add numerators
[00:13:53] You've got it!
[00:14:04] So keep that term in mind for our answer
[00:14:11] but now we have to look at the x's
[00:14:23] So what does it look like when we pull the x out?

Guest (Customer)
[00:14:26] the 25/21 ?

Clifford T (Tutor)
[00:14:30] yup

Guest (Customer)
[00:14:35] ok

Clifford T (Tutor)
[00:14:41] and 25 and 21 don't have common divisors
[00:14:46] so that is as simple as it gets

Guest (Customer)
[00:14:53] ok

Clifford T (Tutor)
[00:15:18] so what does it look like when we pull an x out of 6x-2/3x?

Guest (Customer)
[00:15:32] lost

Clifford T (Tutor)
[00:15:44] so remember
[00:15:49] that both terms here have an x
[00:15:55] so we pull that x out front

Guest (Customer)
[00:15:59] ok

Clifford T (Tutor)
[00:16:10] then what is left over in each term when we divide out the x?
[00:16:36] remember that the x is gone
[00:16:40] it has been divided out
[00:17:11] and there is still a subtraction sign
[00:17:16] between the 6 and the 2/3

Guest (Customer)
[00:17:20] missed neg

Clifford T (Tutor)
[00:17:24] right
[00:17:36] So now we want to subtract these two fractions
[00:17:47] So first, how do we write 6 as a fraction?
[00:18:18] perfect!
[00:18:25] Now again we need a common denominator
[00:18:34] So what is a number that both 1 and 3 divide?
[00:19:21] remember we need a common denominator for addition/subtraction
[00:19:55] so what is a number that both 1 and 3 divide?

Guest (Customer)
[00:20:29] i should know this

Clifford T (Tutor)
[00:20:43] what was the trick we used last time
[00:20:48] to get the common denominator?

Guest (Customer)
[00:21:00] 18

Clifford T (Tutor)
[00:21:05] that will work
[00:21:12] but we can come up with one much smaller

Guest (Customer)
[00:21:11] 9

Clifford T (Tutor)
[00:21:16] smaller still
[00:21:21] 1 divides everything

Guest (Customer)
[00:21:27] ok

Clifford T (Tutor)
[00:21:30] So what is the smallest number that 3 divides?

Guest (Customer)
[00:21:42] 1

Clifford T (Tutor)
[00:21:48] too small :)

Guest (Customer)
[00:21:52] 3

Clifford T (Tutor)
[00:21:57] Exactly!
[00:22:10] and remember our trick, was to multiply the two numbers together
[00:22:15] 1 * 3 = 3
[00:22:28] Now what is the new first numerator?
[00:23:16] (What did we multiply 1 by to get 3?)

Guest (Customer)
[00:23:31] 3

Clifford T (Tutor)
[00:23:34] right
[00:23:38] So multiply 6 by 3
[00:23:42] to get the new numerator
[00:24:02] it will be a little bigger
[00:24:08] that's it!
[00:24:23] Now did we change the second fraction at all?
[00:24:28] perfect!
[00:24:52] good
[00:24:59] and 16 and 3 have no common divisors
[00:25:06] So there is no simplifying here either
[00:25:36] and those two answers put together give us our final answer

Guest (Customer)
[00:25:38] ok

Clifford T (Tutor)
[00:25:53] So any questions about what we did?

Guest (Customer)
[00:26:11] do we still need the exponent in there?

Clifford T (Tutor)
[00:26:17] the 2?

Guest (Customer)
[00:26:21] yes

Clifford T (Tutor)
[00:26:27] yes, we still need it
[00:26:36] because the x^2 is different than just an x

Guest (Customer)
[00:26:43] okay I see now
[00:27:14] this was helpful, should have done this 2 hours ago! ;-}

Clifford T (Tutor)
[00:27:27] haha :) glad to help!
[00:27:32] Well thanks for visiting
[00:27:40] Be sure to fill out the survey and have a great night!

Guest (Customer)
[00:27:48] I would like to print this is that possible

Clifford T (Tutor)
[00:27:54] yes
[00:28:02] the screen should stay up after the session ends
[00:28:07] and you should be able to print it out

Guest (Customer)
[00:28:24] i see it now! Thanks so much!