Factoring{resolved}

[Dillinger4]

Once again i am having trouble factoring. Problem is that somtimes i don't know how to first approach a problem.

4x⁴ - 24x³ + 32x²

Now is this a product of two bionominals? I keep wanting to do this. 4x(x³ - 6x² + 8x) Which is incorrect.

[krtxmrtz]

Why is this incorrect? In any case what you can do is factorize it further, something like
4x²(x² - 6x + 8)

[Dillinger4]

It should end up being 4x²(x-4)(x-2)

[krtxmrtz]

Originally posted by Dilenger4
It should end up being 4x²(x-4)(x-2)

Well, you got it already. What else are you supposed to do? The way to arrive at this final factorization is by solving -calculating the roots- of the 2nd degree equation.

I suppose you know how to solve an equation such as

ax² + bx + c = 0

Once you've found the roots (let these be p and q) you can rewrite it as

(x - p)(x - q) = 0

[Dillinger4]

I didn't get it. The answer came from another source.

Further clarification is needed.

This is how i end up solving this problem which is incorrect.

6a² - 18a -24

(2a - 8)(3a + 3)
6a² + 6a -24a -24
6a² -18a -24

It should be.

6a² -18a -24

6(a - 4)(a + 1)
6(a² + a -4a -4)
6(a² -3a -4)
6a² - 18a -24

So my question is. If i end up getting (2a - 8)(3a + 3) can i somehow figure out how to get 6(a - 4)(a + 1)?

[bugzpodder]

(2a - 8)(3a + 3)

2a-8=2(a-4)
3a+3=3(a+1)

hence (2a-8)(3a+3)=(2(a-4))(3(a+1))=2(a-4)3(a+1)=6(a-4)(a+1)

[Dillinger4]

Couldn't be clearer.



Thanks bugzpodder