# Factorize

• May 13th 2009, 03:44 AM
anon_404
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?
• May 13th 2009, 05:41 AM
TheMasterMind
Quote:

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?
• May 13th 2009, 05:46 AM
anon_404
Quote:

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.
• May 14th 2009, 06:02 AM
craig
Synthetic Division

There is a good example on synthetic division found here.
• May 14th 2009, 07:18 AM
e^(i*pi)
Quote:

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
• May 14th 2009, 12:23 PM
HallsofIvy
Quote:

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]

Quote:

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]