# Factorisation Help

• October 31st 2010, 02:26 AM
Glitch
Factorisation Help
The question:
Factorise $2(2x^3 - 3x^2 - x + 1)$

I have no idea where to start. :( Is there a method of factorising cubics and other high degree polynomials, preferably without remembering massive formulas?

Thanks.
• October 31st 2010, 02:34 AM
Unknown008
To find the factors, it's basically trial and error.

You should use the factor theorem, where if you a have a function:f(x), then if (x-p) is a factor of f(x), f(p) will be equal to zero.

In your example, trying (x-1) gives:

f(1) = 2 - 3 - 1 + 1 = -1

So, (x-1) is not a factor. Let's try (x+1). [You normally try out the factors of the coefficient of the highest power of x and the term with the smallest power of x]

f(-1) = -2 -3 + 1 + 1 = -3

So, (x+1) is not a factor.

Let's try (x-2)

f(2) = 16 - 12 - 2 + 1 = 3

(x-2) is not a factor. Let's try (x+2)

f(-2) = -16 - 3 + 2 + 1 = -16

(x+2) is not a factor either...

Well, now, let's take (x - 0.5).

f(0.5) = 0.25 - 0.75 - 0.5+1 = 0

So, we get (x - 0.5) as a factor.

Or,

(2x - 1) as a factor.

Do you know how to use long division to continue the factoring?
• October 31st 2010, 03:05 AM
Glitch
Hmm, I just tried and I made a mess. Could you walk me though the division? Thanks.
• October 31st 2010, 03:15 AM
Unknown008
Yep, it will be a little difficult though through the forum... and some time.

Code:

Long Division:       x^2 -  x  -  1    2x - 1|2x^3 - 3x^2 -  x + 1       2x^3 -  x^2       0x^3 - 2x^2 -  x + 1             - 2x^2 +  x               0x^2 - 2x + 1                   - 2x + 1                     0x + 0
• October 31st 2010, 03:47 AM
Glitch
OK, I got the solution you did. However factorising the result looks different to the solution my text states, which is:

2(x - 1)(x - 1)(2x + 1)

When I solve x^2 - x - 1 I get:
$\frac{1\pm\sqrt{5}}{2}$

Which doesn't appear to produce the right result. What have I done wrong? Thanks.
• October 31st 2010, 03:48 AM
Unknown008
Well, the question you posted have the solution I posted...

See: http://www.wolframalpha.com/input/?i=(2x^3+-+3x^2+-+x+%2B+1)+%3D+0
• October 31st 2010, 03:50 AM
Glitch
Right, but I want to factorise it further.
• October 31st 2010, 03:50 AM
mr fantastic
Quote:

Originally Posted by Glitch
OK, I got the solution you did. However factorising the result looks different to the solution my text states, which is:

2(x - 1)(x - 1)(2x + 1)

When I solve x^2 - x - 1 I get:
$\frac{1\pm\sqrt{5}}{2}$

Which doesn't appear to produce the right result. What have I done wrong? Thanks.

Since 2(x - 1)(x - 1)(2x + 1) = 4x^3 - 6x^2 + 2, either the book is wrong or you have not posted the correct question.
• October 31st 2010, 04:28 AM
Glitch
Sorry guys, apparently it's supposed to be (according to my text):
$2(2x^3 - 3x^2 - x^2 + 1)$

My text has a worked solution I was trying to follow, which looks like this:
$2(2x^3 - 3x^2 - x^2 + 1)$
= $2(x - 1)( 2x^2 - x - 1)$
= $2(x - 1)(x - 1)(2x + 1)$

However, expanding it back out doesn't seem to produce the original question. I looked at it and worked out that the textbook actually meant:
$2(2x^3 - 3x^2 + 1)$

So my original question was wrong because I tried to fix it in my notes (which I wrote months ago), and that fix was wrong too! So my text and my notes were both wrong. -_-

Anyway, thanks for your assistance. Your factorization technique is pretty darn cool, and will assist me in future. Cheers.