1. factoring 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.

2. Re: factoring help

$\displaystyle (6-3x)^4 + x(4(6-3x)^3(-3)) = (6-3x)^3(6-3x-12x) = (6-3x)^3(6-15x) = \cdots$

From here you should notice that you could extract coefficients(common factors) out of the brackets, along with their respective exponents.

3. Re: 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!

4. Re: factoring help

See that $\displaystyle (6-3x) = 3(2-x) \text{ and } (6-15x) = 3 (2-5x)$.

So from where I left off in my last post,

$\displaystyle (6-3x)^3(6-15x)= (3(2-x))^3(3)(2-5x) =3^33^1(2-x)^3(2-5x) = 3^4(2-x)^3(2-5x)=\cdots$

Let me know if you still need some clarification.

5. Re: factoring help

hey! I think I got it! thanks so much

let me go over my steps:

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)

3^4(2-x)(2-5x)

6. Re: 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)

7. Re: factoring help

nevermind, thank you again!