# Factoring cubic and further polynomials

• Aug 27th 2009, 02:11 AM
RelevantComment
Factoring cubic and further polynomials
Hi, just stopped lurking on mhf and made my first post.

Ok I can factor quadratics and the differences of squares. I'm basically using the factor pairs method (finding factor pairs that add up to the b term) for quadratics as it almost always works. I can expand cubic and further polynomial brackets but I cannot do it the other way around by factorizing.

$\displaystyle x^3-3x^2-4x+12$ and
$\displaystyle x^4+27$
$\displaystyle 3x^(3/2)-9x^(1/2)+6x^(-1/2)$ and finally
$\displaystyle x^3y-4xy$
• Aug 27th 2009, 02:34 AM
PQR

$\displaystyle x^3-3x^2-4x+12 = 0$

$\displaystyle [(x-3)x-4]x+12 = 0$

$\displaystyle f(-2) = 0$

$\displaystyle x^3-3x^2-4x+12$ $\displaystyle (x+2)(ax^2+bx+c)$

$\displaystyle x^3-3x^2-4x+12$$\displaystyle ax^3+bx^2+cx+2ax^2+2bx+2c \displaystyle x^3-3x^2-4x+12$$\displaystyle ax^3+bx^2+2ax^2+cx+2bx+2c$

$\displaystyle x^3-3x^2-4x+12$ $\displaystyle ax^3+(b+2a)x^2+(c+2b)x+2c$

$\displaystyle a =1$; $\displaystyle b+2a = -3$, $\displaystyle b = -5$; $\displaystyle c+2b -4$, $\displaystyle c = 6$.

$\displaystyle x^3-3x^2-4x+12$ $\displaystyle (x+2)(x^2-5x+6)$

$\displaystyle x^2-5x-6 = 0$

$\displaystyle (x-5)x-6 = 0$

$\displaystyle f(2) = 0$

$\displaystyle x^2-5x-4$ ≡ $\displaystyle (x-2)(ax+b)$

$\displaystyle x^2-5x-4$ ≡ $\displaystyle ax^2+bx-2ax-2b$

$\displaystyle x^2-5x-4$ ≡ $\displaystyle (a)x^2+(b-2a)x-2b$

$\displaystyle a = 1$; $\displaystyle b-2a$ = $\displaystyle -5$, $\displaystyle b = -3$.

$\displaystyle x^2-5x-4$ ≡ $\displaystyle (x-2)(x-3)$

$\displaystyle x^3-3x^2-4x+12$ ≡ $\displaystyle (x+2)(x-2)(x-3)$

• Aug 27th 2009, 06:07 AM
pacman
(x^3 - 3x^2) - (4x -12) = x^2(x - 3) - 4(x - 3), you see (x - 3)?

we can factor that, (x - 3)(x^2 - 4), ohh, (x^2 - 4) is the difference of 2

squares,

http://www.mathhelpforum.com/math-he...e460e8dd-1.gif = (x - 3)(x - 2)(x + 2)

2) http://www.mathhelpforum.com/math-he...c9ef4c2d-1.gif, seems miscopied, it myt be x^3 + 27 = (x + 3)(x^2 - 3x + 9)
• Aug 30th 2009, 03:30 PM
Kataangel
Thanks PQR and pacman.

I have just checked again and $\displaystyle x^4+27$ is correct and not a miscopy and the answer to it was $\displaystyle x(x + 3)(x^2 - 3x + 9)$. Not sure why though.
• Aug 30th 2009, 04:02 PM
eXist
It has to be a miscopy because:
$\displaystyle \big[x^4+27\big] \ne \big[x(x + 3)(x^2 - 3x + 9)\big]$

Are you sure the problem wasn't something like:

$\displaystyle x^4 + 27x$?

Because $\displaystyle x^4 + 27x = x(x + 3)(x^2 - 3x + 9)$
• Aug 31st 2009, 02:16 PM
Kataangel
I have no idea where (x+2) comes from in PQR's post:
$\displaystyle (x+2)(ax^2+bx+c)$

Quote:

Originally Posted by eXist
It has to be a miscopy because:
$\displaystyle \big[x^4+27\big] \ne \big[x(x + 3)(x^2 - 3x + 9)\big]$

Are you sure the problem wasn't something like:

$\displaystyle x^4 + 27x$?

Because $\displaystyle x^4 + 27x = x(x + 3)(x^2 - 3x + 9)$

Yep so actually $\displaystyle x^4 + 27x$ is right then.