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.
Printable View
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.
Why is this incorrect? In any case what you can do is factorize it further, something like
4x2(x2 - 6x + 8)
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.Quote:
Originally posted by Dilenger4
It should end up being 4x2(x-4)(x-2)
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
I didn't get it. The answer came from another source.
Further clarification is needed. :D
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)?
(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)
Couldn't be clearer. :D
Thanks bugzpodder