1. ## Factorize

How does:

x^3 - 9x^2 + 24x -16

Become:

(x - 1) (x^2 - 8x - 16)

And what are the values of x?

What would be good study material for this type of algebra?

2. Originally Posted by anon_404
How does:

x^3 - 9x^2 + 24x -16

Become:

(x - 1) (x^2 - 8x - 16)

And what are the values of x?

What would be good study material for this type of algebra?
The most efficent way to do this I would presume would be to use the method of synthetic division for dividing polynomials. Have you learned this method?

3. Originally Posted by TheMasterMind
The most efficent way to do this I would presume would be to use the method of synthetic division for dividing polynomials. Have you learned this method?
I havn't used that method.

4. Synthetic Division

There is a good example on synthetic division found here.

5. Originally Posted by anon_404
How does:

x^3 - 9x^2 + 24x -16

Become:

(x - 1) (x^2 - 8x - 16)

And what are the values of x?

What would be good study material for this type of algebra?
Are you sure that's a correct factorisation? Look at the last terms: -1 x -16 = +16 not -16 as stated in the question

6. Originally Posted by anon_404
How does:

x^3 - 9x^2 + 24x -16

Become:

(x - 1) (x^2 - 8x - 16)
It doesn't! It is easy to see quickly that the constant term in that product is (-1)(-16)= +16, not -16 as in the first expression. It is true that
$x^3- 9x^2+ 24x+ 16= (x-1)(x^2- 8x- 16)$
The best way to see this is to multiply it back:
$(x-1)(x^2- 8x- 16)= (x)(x^2- 8x- 16)- (x^2- 8x- 16)$
$= x^3- 8x^2- 16x- x^2+ 8x+ 16= x^2- (8+1)x^2- (16-8)x+ 16$
[tex]= x^3- 9x^2- 8x+ 16[/itex]

And what are the values of x?
??? x can be any number! Did you mean "what are the values of x that make this 0" or "what values of x solve $x^3- 9x^2- 8x+ 16= 0$?

In this case you can also factor $x^2- 8x- 16= (x- 4)^2$ so
$x^3- 8x- 16= (x- 1)(x^2- 8x- 16)= (x-1)(x-4)^2= 0$ and you can use the fact that "if ab= 0 then either a= 0 or b= 0".

What would be good study material for this type of algebra?[/QUOTE]