I am working on being better at factor but I could use some help.

this (6-3x)^4 + x[4(6-3x)^3(-3)] factors to (6-3x)^3[(6-3x)-12x] to 3(6-3x)^3(2-5x)

looking for some help on how this is done. thank you.

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- Sep 25th 2013, 06:27 PMJonroberts74factoring help
I am working on being better at factor but I could use some help.

this (6-3x)^4 + x[4(6-3x)^3(-3)] factors to (6-3x)^3[(6-3x)-12x] to 3(6-3x)^3(2-5x)

looking for some help on how this is done. thank you. - Sep 25th 2013, 06:55 PMchen09Re: factoring help

From here you should notice that you could extract coefficients(common factors) out of the brackets, along with their respective exponents. - Sep 25th 2013, 08:59 PMJonroberts74Re: factoring help
I notice a bit of how you are getting that but not entirely. I downloaded an older text book that deals with nothing but factoring. Maybe it'll help but maybe you could explain a little bit. Sorry, I have been stuck teaching everything to myself and sometimes it is a bit difficult. Thank you!

- Sep 25th 2013, 10:23 PMchen09Re: factoring help
See that .

So from where I left off in my last post,

Let me know if you still need some clarification. - Sep 26th 2013, 11:03 AMJonroberts74Re: factoring help
hey! I think I got it! thanks so much

let me go over my steps:

I start with (6-3x)^4 + x[4(6-3x)^3(-3)]

take out the binomial (6-3x)^3 giving me (6-3x)^3[(6-3x) + x(4(-3))

next I have (6-3x)^3[6-3x-12x]

then, (6-3x)^3[6-15x]

factor out 3 of the right binomial for 3(6-3x)^3(2-5x)

then factor out the coefficient of the the left binomial with the exponent to 3^1(3^3)(2-x)(2-5x)

and a final answer of

3^4(2-x)(2-5x) - Sep 26th 2013, 11:21 AMJonroberts74Re: factoring help
before I get ahead of myself though now I have to factor:

3(x+4)^2(x-3)^6 + 6(x-3)^5(x+4)^3 down to 3(x+4)^2(x+3)^5(3x+5) - Sep 26th 2013, 11:30 AMJonroberts74Re: factoring help
nevermind, thank you again!