Factoring cubic and further polynomials

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August 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.

I've hit a roadblock trying to factorize these polynomials. Could anyone please help me with them?

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

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

$[(x-3)x-4]x+12 = 0<br />$

$f(-2) = 0$

∴ $x^3-3x^2-4x+12$ ≡ $(x+2)(ax^2+bx+c)$

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

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

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

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

∴ $x^3-3x^2-4x+12$ ≡ $(x+2)(x^2-5x+6)$

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

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

$f(2) = 0$

∴ $x^2-5x-4$ ≡ $(x-2)(ax+b)$

$x^2-5x-4$ ≡ $ax^2+bx-2ax-2b$

$x^2-5x-4$ ≡ $(a)x^2+(b-2a)x-2b$

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

∴ $x^2-5x-4$ ≡ $(x-2)(x-3)$

∴ $x^3-3x^2-4x+12$ ≡ $(x+2)(x-2)(x-3)$
August 27th 2009, 06:07 AM
pacman
- grouping will do, but hard to guess....1) http://www.mathhelpforum.com/math-he...e460e8dd-1.gif, the factors of 12 are many,
(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)
August 30th 2009, 03:30 PM
Kataangel

Thanks PQR and pacman.

I have just checked again and $x^4+27$ is correct and not a miscopy and the answer to it was $x(x + 3)(x^2 - 3x + 9)$ . Not sure why though.
August 30th 2009, 04:02 PM
eXist

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

Are you sure the problem wasn't something like:

$x^4 + 27x$ ?

Because $x^4 + 27x = x(x + 3)(x^2 - 3x + 9)$
August 31st 2009, 02:16 PM
Kataangel

I have no idea where (x+2) comes from in PQR's post:
$(x+2)(ax^2+bx+c)$

Quote:

Originally Posted by eXist

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

Are you sure the problem wasn't something like:

$x^4 + 27x$ ?

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

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