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April 30th, 2012, 06:44 AM  #1 
Newbie Joined: Apr 2012 Posts: 3 Thanks: 0  Help on some college math problems please?
I get a few other questions but these do not make any sense. Simplify: 6x^28x^5 _________ 4x^4 simplify: 5xy(x2x2y) Factor: 9xy12x^5y Simplify: (6ab)(5ac^2) Factor: 82x^2 Simplify: 2(3xy) Simplify: X^29x+20 _________ 82x Simplify: 3a^5(3aa^4b) Simplify: 1414 Simplify: 5(dn)^3 Simplify: (5xy/x^3y^^4 Simplify: (2a5b) ^2 Simplify: 2/3a(5a3/5)1/3 Simplify: (a5)(3ab1) Factor: 81/49n^2p^20 Simplify: a^2(a+3b)+a(a^3a) Factor: 8a^3b18ab^3 Alrighty, those are the ones I'm having trouble with. Any help on them is much appreciated. 
April 30th, 2012, 12:47 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs  Re: Help on some college math problems please?
I have moved this from abstract algebra. While it is algebra and may seem abstract if you are struggling with it, abstract algebra is a few years down the road from this. You have posted 18 problems with no work shown. Not trying to be critical, but most folks appreciate seeing an effort, and no more than a few problems per topic. I know you are new here and so I'm just giving friendly advice, one Libra to another. Let's look at the first few, and then hopefully you can apply the same techniques to the others. 1.) My first step would be to rewrite, getting rid of the negative sign in the denominator. If we multiply both the numerator and denominator by 1, we are in effect multiplying the expression by (1)/(1) = 1 and so leaving the value unchanged. Now, let's look at all of the terms in both the numerator and the denominator and find the greatest factor common to all. We see that they all have as the greatest common factor. So, rewrite the expression as: 2.) My first step would be to get rid of that leading negative sign. If we multiply both factors by 1, then we are in effect multiplying by (1)(1) = 1, and so leaving the value unchanged. Next, we want to distribute the factor outside of the parentheses to each term within the parentheses: 3.) Factor: We want to find the greatest factor common to both terms. We see that both terms have as the greatest common factor, so we may write: Notice that each term within the parentheses is found by dividing the original terms by the common factor: Now, try problems 4  7 using these techniques. Post your work, and if you have troubles, we will be glad to help. 8.) In order to simplify, we need to factor first. The numerator is a quadratic, and we see we need two factors of 20 whose sum is 9. Those are 4 and 5. The terms in the denominator have a 2 as a common factor, so we may write the expression as: Okay, before we proceed it's time for you to attempt what I haven't done for the first 10 problems. Once you have those done and understood, we'll move on to the remaining 8 problems. 
May 1st, 2012, 09:30 AM  #3 
Newbie Joined: Apr 2012 Posts: 3 Thanks: 0  Re: Help on some college math problems please?
Okay. I was rushing to post the problems and forgot to show my work. Sorry about that. Here's my work for 47: 4) Simplify: (6ab)(5ac^2) ac^2+bc^2+ab+11a+5b+30 ^my work was distribution. 5) 82x^2 GCF: 2(4x) 6) 2(3xy) 21(3xy) 1(3xy) Answer: 3x+y The other 10: Simplify: 1414 28? Simplify: 5(dn)^3 (5dn)(5dn)(5dn) 15d^3 n^3 Simplify: (5xy/x^3y^^4 (5x2y^7)^4 Simplify: (2a5b) ^2 (2a5b)(2a5b) 4a^210ab10ab+25b 4a^220ab+25b Simplify: 2/3a(5a3/5)1/3 2/3x5= 3 1/3 3 1/3a2/9a1/3 3 1/9a1/3 Simplify: (a5)(3ab1) 3a^2b+5 Factor: 81/49n^2p^20 (never mind I figured it out) Simplify: a^2(a+3b)+a(a^3a) a^3+5ab+ab+4ab+a^4+a^7 Okay there's the work/answers shown. How did I do? Factor: 8a^3b18ab^3 
May 1st, 2012, 09:32 AM  #4 
Newbie Joined: Apr 2012 Posts: 3 Thanks: 0  Re: Help on some college math problems please?
Almost forgot: Factor: 8a^3b18ab^3 (I saw this as 'can't be solved') 
May 1st, 2012, 10:21 AM  #5  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039  Re: Help on some college math problems please? Quote:
= 2ab(4a^2  9b^2) = 2ab(2a + 3b)(2a  3b) By the way, factoring is NOT solving.  
May 1st, 2012, 12:05 PM  #6  
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs  Re: Help on some college math problems please? Quote:
By the associative and commutative properties of multiplication, we may write this as: Quote:
Now, we observe that the second factor is the difference of two squares, so we have: Quote:
Quote:
Quote:
So, using the property we would have: Quote:
First, we want to simplify within the parentheses: Now we may distribute the exponent on the outside to each factor on the inside both in the numerator and denominator: Quote:
You are close, but you forgot to square the b in the last term. We could use the formula with to get: Quote:
First, distribute the negative sign within the parentheses: Now, combine like terms: We now must add fractions, so we need common denominators: Quote:
We want to distribute each term in the first factor to each term in the second factor, so we have: Quote:
Quote:
First, distribute: Combining like terms, we are left with: Okay, you asked how you did. Frankly, you really need to review the techniques used to work these problems. Coming here is a good start. I appreciate you taking the time to post your work, because it gives us a better idea where you are going wrong in working the problems. If you have any questions about what I did, please feel free to ask, and ask as many questions as you need until you feel like you have gotten it.  

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