My Math Forum Help on some college math problems please?

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 April 30th, 2012, 05:44 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^2-8x^5 _________ -4x^4 simplify: -5xy(x2-x-2y) Factor: 9xy-12x^5y Simplify: (6ab)(5ac^2) Factor: 8-2x^2 Simplify: 2-(3x-y) Simplify: X^2-9x+20 _________ 8-2x Simplify: 3a^5(3a-a^4b) Simplify: -14-14 Simplify: 5(dn)^3 Simplify: (5xy/x^3y^^4 Simplify: (2a-5b) ^2 Simplify: 2/3a-(5a-3/5)-1/3 Simplify: (a-5)(3a-b-1) Factor: 81/49n^2-p^20 Simplify: a^2(a+3b)+a-(a^3-a) Factor: 8a^3b-18ab^3 Alrighty, those are the ones I'm having trouble with. Any help on them is much appreciated.
 April 30th, 2012, 11:47 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 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.) $\frac{6x^2-8x^5}{-4x^4}$ 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. $\frac{8x^5-6x^2}{4x^4}$ 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 $2x^2$ as the greatest common factor. So, rewrite the expression as: $\frac{2x^2$$4x^3-3$$}{2x^2$$2x^2$$}=\frac{\cancel{2x^2}$$4x^3-3$$}{\cancel{2x^2}$$2x^2$$}=\frac{4x^3-3}{2x^2}$ 2.) $-5xy$$x^2-x-2y$$$ 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. $5xy$$-x^2+x+2y$$$ Next, we want to distribute the factor outside of the parentheses to each term within the parentheses: $5xy$$-x^2$$+5xy(x)+5xy(2y)=-5x^3y+5x^2y+10xy^2$ 3.) Factor: $9xy-12x^5y$ We want to find the greatest factor common to both terms. We see that both terms have $3xy$ as the greatest common factor, so we may write: $3xy$$3-4x^4$$$ Notice that each term within the parentheses is found by dividing the original terms by the common factor: $\frac{9xy}{3xy}=3,\,\frac{12x^5y}{3xy}=4x^4$ Now, try problems 4 - 7 using these techniques. Post your work, and if you have troubles, we will be glad to help. 8.) $\frac{x^2-9x+20}{8-2x}$ 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: $\frac{(x-4)(x-5)}{2(4-x)}=\frac{-(4-x)(x-5)}{2(4-x)}=\frac{-\cancel{(4-x)}(x-5)}{2\cancel{(4-x)}}=\frac{-(x-5)}{2}=\frac{5-x}{2}$ 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, 08:30 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 4-7: 4) Simplify: (6ab)(5ac^2) ac^2+bc^2+ab+11a+5b+30 ^my work was distribution. 5) 8-2x^2 GCF: 2(4-x) 6) 2-(3x-y) 2-1(3x-y) 1(3x-y) Answer: 3x+y The other 10: Simplify: -14-14 -28? Simplify: 5(dn)^3 (5dn)(5dn)(5dn) 15d^3 n^3 Simplify: (5xy/x^3y^^4 (5x2y^7)^4 Simplify: (2a-5b) ^2 (2a-5b)(2a-5b) 4a^2-10ab-10ab+25b 4a^2-20ab+25b Simplify: 2/3a-(5a-3/5)-1/3 2/3x5= 3 1/3 3 1/3a-2/9a-1/3 3 1/9a-1/3 Simplify: (a-5)(3a-b-1) 3a^2-b+5 Factor: 81/49n^2-p^20 (never mind I figured it out) Simplify: a^2(a+3b)+a-(a^3-a) a^3+5ab+ab+4ab+a^4+a^7 Okay there's the work/answers shown. How did I do? Factor: 8a^3b-18ab^3
 May 1st, 2012, 08:32 AM #4 Newbie   Joined: Apr 2012 Posts: 3 Thanks: 0 Re: Help on some college math problems please? Almost forgot: Factor: 8a^3b-18ab^3 (I saw this as 'can't be solved')
May 1st, 2012, 09:21 AM   #5
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Re: Help on some college math problems please?

Quote:
 Originally Posted by Libragirl9 Factor: 8a^3b-18ab^3 (I saw this as 'can't be solved')
8a^3b - 18ab^3
= 2ab(4a^2 - 9b^2)
= 2ab(2a + 3b)(2a - 3b)

By the way, factoring is NOT solving.

May 1st, 2012, 11:05 AM   #6
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Math Focus: Calculus/ODEs
Re: Help on some college math problems please?

Quote:
 Originally Posted by Libragirl9 4) Simplify: (6ab)(5ac^2) ac^2+bc^2+ab+11a+5b+30
You may think of the given expression as:

$6\cdot a\cdot b\cdot5\cdot a\cdot c^2$

By the associative and commutative properties of multiplication, we may write this as:

$$$6\cdot5$$$$a\cdot a$$$$b$$$$c^2$$=30a^2bc^2$

Quote:
 Originally Posted by Libragirl9 5) 8-2x^2 GCF: 2(4-x)
First, we determine that both terms have a two as a factor, so we write:

$8-2x^2=2$$4-x^2$$=2$$2^2-x^2$$$

Now, we observe that the second factor is the difference of two squares, so we have:

$2$$2^2-x^2$$=2(2+x)(2-x)$

Quote:
 Originally Posted by Libragirl9 6) 2-(3x-y) 2-1(3x-y) 1(3x-y) Answer: 3x+y
You first step is okay, but you cannot combine the 2 and the -1 because the -1 is attached to a factor. You want to distribute the -1 as follows:

$2-1(3x-y)=2-1(3x)-1(-y)=2-3x+y$

Quote:
 Originally Posted by Libragirl9 Simplify: -14-14 -28?
Correct.

$-14-14=-(14+14)=-28$

Quote:
 Originally Posted by Libragirl9 Simplify: 5(dn)^3 (5dn)(5dn)(5dn) 15d^3 n^3
The 5 is not being cubed, you see it's not within the parentheses. And if it was you would use 5·5·5 = 125, not 5 + 5 + 5 = 15.

So, using the property $(ab)^c=a^c\cdot b^c$ we would have:

$5(dn)^3=5d^3n^3$

Quote:
 Originally Posted by Libragirl9 Simplify: (5xy/x^3y^^4 (5x2y^7)^4
We are given:

$$$\frac{5xy}{x^3y^8}$$^4$

First, we want to simplify within the parentheses:

$$$\frac{5}{x^2y^7}$$^4$

Now we may distribute the exponent on the outside to each factor on the inside both in the numerator and denominator:

$\frac{5^4}{x^8y^{28}}=\frac{625}{x^8y^{28}}$

Quote:
 Originally Posted by Libragirl9 Simplify: (2a-5b) ^2 (2a-5b)(2a-5b) 4a^2-10ab-10ab+25b 4a^2-20ab+25b
The instructions should read "Expand" not "Simplify" in my opinion.

You are close, but you forgot to square the b in the last term.

We could use the formula $(x-y)^2=x^2-2xy+y^2$ with $x=2a,\,y=5b$ to get:

$(2a-5b) ^2=(2a)^2-2(2a)(5b)+(5b)^2=4a^2-20ab+25b^2$

Quote:
 Originally Posted by Libragirl9 Simplify: 2/3a-(5a-3/5)-1/3 2/3x5= 3 1/3 3 1/3a-2/9a-1/3 3 1/9a-1/3
We are given:

$\frac{2}{3}a-$$5a-\frac{3}{5}$$-\frac{1}{3}$

First, distribute the negative sign within the parentheses:

$\frac{2}{3}a-5a+\frac{3}{5}-\frac{1}{3}$

Now, combine like terms:

$a$$\frac{2}{3}-5$$+$$\frac{3}{5}-\frac{1}{3}$$$

We now must add fractions, so we need common denominators:

$a$$\frac{2}{3}-5\cdot\frac{3}{3}$$+$$\frac{3}{5}\cdot\frac{3}{3}-\frac{1}{3}\cdot\frac{5}{5}$$$

$a$$\frac{2-15}{3}$$+$$\frac{9-5}{15}$$$

$-\frac{13}{3}a+\frac{4}{15}$

Quote:
 Originally Posted by Libragirl9 Simplify: (a-5)(3a-b-1) 3a^2-b+5
We are given:

$(a-5)(3a-b-1)$

We want to distribute each term in the first factor to each term in the second factor, so we have:

$a(3a)+a(-b)+a(-1)-5(3a)-5(-b)-5(-1)=3a^2-ab-a-15a+5b+5=3a^2-ab-16a+5b+5$

Quote:
 Originally Posted by Libragirl9 Factor: 81/49n^2-p^20 (never mind I figured it out)
Just in case you didn't figure it out:

$\frac{81}{49}n^2-p^{20}=$$\frac{9}{7}n$$^2-$$p^{10}$$^2=$$\frac{9}{7}n+p^{10}$$$$\frac{9}{7}n-p^{10}$$$

Quote:
 Originally Posted by Libragirl9 Simplify: a^2(a+3b)+a-(a^3-a) a^3+5ab+ab+4ab+a^4+a^7
We are given:

$a^2(a+3b)+a-(a^3-a)$

First, distribute:

$a^3+3a^2b+a-a^3+a$

Combining like terms, we are left with:

$3a^2b+2a$

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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