Thread: 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?

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?

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.

Synthetic Division

There is a good example on synthetic division found here.

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

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
$\displaystyle x^3- 9x^2+ 24x+ 16= (x-1)(x^2- 8x- 16)$
The best way to see this is to multiply it back:
$\displaystyle (x-1)(x^2- 8x- 16)= (x)(x^2- 8x- 16)- (x^2- 8x- 16)$
$\displaystyle = x^3- 8x^2- 16x- x^2+ 8x+ 16= x^2- (8+1)x^2- (16-8)x+ 16$
[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 $\displaystyle x^3- 9x^2- 8x+ 16= 0$?

In this case you can also factor $\displaystyle x^2- 8x- 16= (x- 4)^2$ so
$\displaystyle x^3- 8x- 16= (x- 1)(x^2- 8x- 16)= (x-1)(x-4)^2= 0$ 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]