The question:
Factorise $\displaystyle 2(2x^3 - 3x^2 - x + 1)$
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.
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?
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:
$\displaystyle \frac{1\pm\sqrt{5}}{2}$
Which doesn't appear to produce the right result. What have I done wrong? Thanks.
Sorry guys, apparently it's supposed to be (according to my text):
$\displaystyle 2(2x^3 - 3x^2 - x^2 + 1)$
My text has a worked solution I was trying to follow, which looks like this:
$\displaystyle 2(2x^3 - 3x^2 - x^2 + 1)$
= $\displaystyle 2(x - 1)( 2x^2 - x - 1)$
=$\displaystyle 2(x - 1)(x - 1)(2x + 1)$
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:
$\displaystyle 2(2x^3 - 3x^2 + 1)$
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.