# Math Help - Factorization problem

1. ## Factorization problem

Hi Guys,

Nearly there with this problem, but need a nudge to the finish!

Prove that is a factor of and write down two other factors of the expression. Hence factorize the expression completely.

My solution,

Let,

Then since is a factor(Factor theorem)

And substituting , proves is a factor.

Similarly (a - b) and (c - a) are the other factors.

So by factor theorem,

How do I find out Factor 4?

2. Originally Posted by mathguy80
Hi Guys,

Nearly there with this problem, but need a nudge to the finish!

Prove that is a factor of and write down two other factors of the expression. Hence factorize the expression completely.

My solution,

Let,

Then since is a factor(Factor theorem)

And substituting , proves is a factor.

Similarly (a - b) and (c - a) are the other factors.

So by factor theorem,

How do I find out Factor 4?

to find factor-4 i guess you can do the following:
let factor-4 = f(a,b,c).
then f(0,b,c)= -(b+c), f(a,0,c)= -(c+a), f(a,b,0)= -(a+b).
so i anticipate that factor-4 = f(a,b,c)=(a+b+c). this also suggested by the fact that factor-4 cannot have second order terms.
although this isn't a sure shot way to say anything about factor-4 but still it must have helped.
another way could be to take partial derivatives w.r.t a, b, c and then compare...
did this help??

3. correct factor is (b-c)(a-c)(a-b)(a+b+c)

you got one wrong (c-a) should be (a-c)

4. Originally Posted by skoker
correct factor is (b-c)(a-c)(a-b)(a+b+c)

you got one wrong (c-a) should be (a-c)
that's right thanks!
how did you factorize it? check my post too.

5. Hello, mathguy80!

This one requires some serious factoring . . .

Prove that (b - c) is a factor of: .a^3(b - c) + b^3(c - a) + c^3(a - b)

Factor the expression completely.

We have:

. a^3(b - c) + b^3 c - ab^3 + ac^3 - bc^3

. . = .a^3(b - c) + b^3 c - bc^3 - ab^3 + ac^3

. . = . a^3(b - c) + bc(b^2 - c^2) - a(b^3 - c^3)

. . = . a^3(b - c) + bc(b - c)(b + c) - a(b - c)(b^2 + bc + c^2)

. . = . (b - c) [a^3 + bc(b + c) - a(b^2 + bc + c^2)]

. . = . (b - c) [a^3 + b^2 c + bc^2 - ab^2 - abc - ac^2]

. . = . (b - c) [a^3 - ab^2 - abc + b^2 c - ac^2 + bc^2]

. . = . (b - c) [a(a^2 - b^2) - bc(a - b) - c^2(a - b)]

. . = . (b - c) [a(a - b)(a + b) - bc(a - b) + c^2(a - b)]

. . = . (a - b) (b - c) [a(a + b) - bc - c^2]

. . = . (a - b) (b - c) [a^2 + ab - bc - c^2]

. . = . (a - b) (b - c) [a^2 - c^2 + ab - bc]

. . = . (a - b) (b - c) [(a - c)(a + c) + b( a - c)]

. . = . (a - b) (b - c) (a - c) [a + c + b]

. . = . (a - b) (b - c) (a - c) (a + b + c)

6. It is far more time to do latex then actual problem... but I can't show this with text-style display.

[edit2] I update one more time with original step.

7. Originally Posted by skoker
It is far more time to do latex then actual problem... but I can't show this with text-style display.

I think you'd better recheck your work. (a + b + c)(a - b)^3 * (a - c)^3 * (b - c)^3 is not the same as a(c - b)^3 + b(a - c)^3 + c(b - a)^3. Here is the result when you subtract the two.

-Dan

8. after head smash with latex for quite some time I re-post.

9. Thanks guys that was very helpful. @abhishekkgp still working my way up to calculus, but f(a,b,c) was a good idea to get an idea of the required result. @skoker, yup I messed up one of the factors.

@Soroban that was some gorgeous factoring there! Very cool and looks great, Thanks!

10. Originally Posted by topsquark
I think you'd better recheck your work. (a + b + c)(a - b)^3 * (a - c)^3 * (b - c)^3 is not the same as a(c - b)^3 + b(a - c)^3 + c(b - a)^3. Here is the result when you subtract the two.

-Dan
you are right. this was bugging me so I go back and check. I actually do not need that step.