Factorization

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March 28th 2008, 08:04 AM
Rocher

Factorization

Guys, I hope you can help me solve some of these factor problems... it's really frustrating >_<

Factor
$ 1) (3m+2n)^2-(m-n)^2$
$2) 2ab^3-2ab$
$3) 1-16a^4$
$4) cx-cy+cz$
$5) px-qx-rx$
$6) 15a^3-10a^2$
$7) 12abc-3bc^2$
$8) 4x^2y-xy^2$
$9) 63pq+14pq^2$
$10) 24a^3m-18a^2m^2$
$11) x^6y-x^4z$
$12) x(a+b)-y(a+b)$
$13) x(x-y)^2-y(x-y)$
$14) -ab(a-b)^2+a(b-a)^2-ac(a-b)^2$
$15) x(a-x)(a-y)-y(x-a)(y-a)$
$16) (m+n)(p+q)-(m+n)(p-q)$
$17) a(x-a)+b(a-x)-c(x-a)$
$18) 2a(x+y-z)+3b(z-x-y)+5c(x+y-z)$
$19) p(a^2+b^2)+q(a^2+b^2)-r(a^2+b^2)$

Some of them are alike, if they are, feel free to only do one of them, so I can figure out how to do the rest myself. Thanks in advance!
March 28th 2008, 08:32 AM
iknowone

$ 3) 1-16a^4 $
Is an example of the difference of 2-squares which obeys: $a^2-b^2 = (a-b)(a+b)$

$ 19) p(a^2+b^2)+q(a^2+b^2)-r(a^2+b^2) $
Is an example of the greatest common factor: $p(a^2+b^2)+q(a^2+b^2)-r(a^2+b^2) = (p + q - r)(a^2+b^2)$

$ 2) 2ab^3-2ab $
Is an example of the greatest common factor and then difference of squares: $2ab(b^2-1) = 2ab(b+1)(b-1)$

$ 17) a(x-a)+b(a-x)-c(x-a) $
Observe that $(a-x) = -(x-a)$ . Do you know the various methods for factoring and what it means to be completely factored?
March 28th 2008, 08:36 AM
Rocher

Quote:

Originally Posted by iknowone

$ 3) 1-16a^4 $
Is an example of the difference of 2-squares which obeys: $a^2-b^2 = (a-b)(a+b)$

But that's to the power of 4, which is why I don't get it.

$ 2) 2ab^3-2ab $
Is an example of the greatest common factor and then difference of squares: $2ab(b^2-1) = 2ab(b+1)(b-1)$

Huh?? =(

$ 17) a(x-a)+b(a-x)-c(x-a) $
Observe that $(a-x) = -(x-a)$ . Do you know the various methods for factoring and what it means to be completely factored?

Nope...

Eh...
March 28th 2008, 08:46 AM
iknowone

Learning some basic factoring techniques is obviously essential if you hope to do 19 such problems since it is unlikely anyone will answer these all for you.

Remember that $(a^2)^2 = a^4$ . So for problem 2 it is still the difference of squares, namely:

$1^2 - (4a^2)^2 = (1+4a^2)(1-4a^2)$ Do you see that one of these terms is again the difference of squares?

The greatest common factor technique asks you to find what is common among all the terms in your expression. So for problem 3 the greatest common factor (the thing that appears in every term) is $2ab$ .
Once you have found this term "bring it out front" and multiply it by what is left. $2ab^3 -2ab = 2ab(b^2 - 1)$ . Do you see that if you were to multiply-out the second expression you would get back to your original one? Also notice that $b^2 - 1$ is the difference of squares which now you know how to do.

$b(a-x) = -b(x-a).$ Once you use this fact you have a greatest common factor for problem 17. Your GCF is (x-a), so bring it out front and multiply it by what is left.
March 28th 2008, 08:56 AM
colby2152

Rocher, welcome to MHF! It is great to see you have questions and are interested in learning math! However, please do not post many questions all at once. The idea is not get answers to your - homework, test, quiz, etc... - questions, but to understand how to get a solution and why.
March 28th 2008, 08:56 AM
Rocher

So confusing lol... I'll try it though.

Still need help on the other problems too...
March 28th 2008, 09:47 AM
iknowone

Let's see some of the work you have done on the problems so we can see if you are doing them correctly. As you said many of them are the same. The 2 techniques I told you about should be used to solve almost every one of the problems.
March 28th 2008, 11:28 AM
james_bond

Quote:

Originally Posted by Rocher

Factor
$ 1) (3m+2n)^2-(m-n)^2=(3m+2n-(m-n))(3m+2n+(m-n))=(2m+3n)(4m+n)$
$2) 2ab^3-2ab=2ab(b^2-1)=2ab(b-1)(b+1)$
$3) 1-16a^4=1^2-(4a^2)^2=(1-(4a^2))(1+(4a^2))=$ $(1^2-(2a)^2)(1+4a^2)=((1-2a)(1+2a))(1+4a^2)=(1-2a)(1+2a)(1+4a^2)$
$4) cx-cy+cz=c(x-y+z)$
$5) px-qx-rx=x(p-q-r)$
$6) 15a^3-10a^2=5a^2(3a-2)$
$7) 12abc-3bc^2=3bc(4a-c)$
$8) 4x^2y-xy^2=xy(4x-y)$
$9) 63pq+14pq^2=7pq(9+2q)$
$10) 24a^3m-18a^2m^2=6a^2m(4a-3m)$
$11) x^6y-x^4z=x^4(x^2y-z)$
$12) x(a+b)-y(a+b)=(a+b)(x-y)$
$13) x(x-y)^2-y(x-y)=(x-y)(x(x-y)-y)(=(x-y)(x^2-xy-y))$
$14) -ab(a-b)^2+a(b-a)^2-ac(a-b)^2$
$15) x(a-x)(a-y)-y(x-a)(y-a)$
$16) (m+n)(p+q)-(m+n)(p-q)$
$17) a(x-a)+b(a-x)-c(x-a)$
$18) 2a(x+y-z)+3b(z-x-y)+5c(x+y-z)$
$19) p(a^2+b^2)+q(a^2+b^2)-r(a^2+b^2)$

I (also ;)) make mistakes, please correct me.
March 28th 2008, 11:48 AM
Plato

Quote:

Originally Posted by Rocher

Factor
$14) -ab(a-b)^2+a(b-a)^2-ac(a-b)^2$

In that whole list #14 is the only real interesting one.
It can be done in only one step.
$14) -ab(a-b)^2+a(b-a)^2-ac(a-b)^2 = a(a-b)^2(-b+1-c)$
March 28th 2008, 11:56 AM
colby2152

Quote:

Originally Posted by Plato

In that whole list #14 is the only real interesting one.
It can be done in only one step.
$14) -ab(a-b)^2+a(b-a)^2-ac(a-b)^2 = a(a-b)^2(-b+1-c)$

In case you were wondering... $(a-b)^2 = (b-a)^2$

because when you break either of them down, both look like this: $a^2-2ab+b^2$ . Negative times a negative is a positive!(Yes)
March 28th 2008, 12:16 PM
Plato

Quote:

Originally Posted by colby2152

In case you were wondering... $(a-b)^2 = (b-a)^2$
because when you break either of them down, both look like this: $a^2-2ab+b^2$ .

You had to be a spoilsport didn’t you?
But the reason it works is deeper that you indicate.
$a^2 = b^2$ if and only if $|a| = |b|$ .
March 28th 2008, 12:38 PM
colby2152

Quote:

Originally Posted by Plato

You had to be a spoilsport didn’t you?
But the reason it works is deeper that you indicate.
$a^2 = b^2$ if and only if $|a| = |b|$ .

You had to throw absolute values in there. I tried keeping it simple and to the point! Just kidding...

Yes, $|a-b|=|b-a|$
March 28th 2008, 07:04 PM
Rocher

Please I need help on the ones with three equations.
March 29th 2008, 10:28 AM
james_bond

Quote:

Originally Posted by Rocher

Please I need help on the ones with three equations.

$14) -ab(a-b)^2+a(b-a)^2-ac(a-b)^2=-(-a + ab + ac) (a - b)^2$
$15) x(a-x)(a-y)-y(x-a)(y-a)=(a - x) (a - y) (x - y)$
$16) (m+n)(p+q)-(m+n)(p-q)=2 (m + n) q$
$17) a(x-a)+b(a-x)-c(x-a)=(-a + b + c) (a - x)$
$18) 2a(x+y-z)+3b(z-x-y)+5c(x+y-z)=(2 a - 3 b + 5 c) (x + y - z)$
$19) p(a^2+b^2)+q(a^2+b^2)-r(a^2+b^2)=(a^2 + b^2) (p + q - r)$