how do i factor 0= x^5-x^4+14x^3-16x^2-32x?

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- Nov 18th 2011, 09:24 AMcheshirecatfactoring
how do i factor 0= x^5-x^4+14x^3-16x^2-32x?

- Nov 18th 2011, 09:40 AMAlso sprach ZarathustraRe: factoring

x^5-x^4+14x^3-16x^2-32x=

x(x^4-x^3+14x^2-16x-32)=

one root is obviously x_1=0, the second is one from the group : {1,-1,2,-2,4,-4,8,-8,16,-16,32,-32}. It's easy to see that x_2=2 and x_3=-1 are roots.

Now, divide x^4-x^3+14x^2-16x-32 by (x-2)(x+1), what would you get? - Nov 18th 2011, 11:46 AMTheChazRe: factoring
- Nov 18th 2011, 01:34 PMcheshirecatRe: factoring
I don't know how to divide (x-2)(x+1) into x^4-x^3+14x^2-16x-32. I know how to get the 3 real roots by using synthetic division which gives x(x+1)(x-2), but I cannot figure out how to get the (x^2 + 16) part of the answer. I know that gives the 2 complex solutions 4i and -4i and I also don't know how they knew there were 2 complex solutions.

- Nov 18th 2011, 01:47 PMskeeterRe: factoringCode:
`-1]...1...-1...14...-16...-32`

..........-1....2...-16....32

-----------------------------

......1...-2...16...-32.....0

2]....1...-2...16...-32

...........2....0....32

-----------------------

......**1....0...16**.....0

$\displaystyle x^2 + 16$ - Nov 18th 2011, 05:42 PMcheshirecatRe: factoring
- Nov 18th 2011, 06:44 PMQuackyRe: factoring
How I would approach: use an appropriate theorem to find the first three roots.

Then, we have:

$\displaystyle 0=x^5-x^4+14x^3-16x^2-32x$

$\displaystyle =x(x+1)(x-2)(ax^2+bx+c)$

$\displaystyle =x(x^2-x-2)(ax^2+bx+c)$

we can then equate coefficients.

In order to have $\displaystyle 1x^5$, we can quickly see that $\displaystyle a=1$

$\displaystyle =x(x^2-x-2)(x^2+bx+c)$

$\displaystyle =(x^3-x^2-2x)(x^2+bx+c)$

Now we can look at the $\displaystyle x^4$ terms. If we expand this, but only by looking at the terms that will give us an $\displaystyle x^4$, we get:

$\displaystyle x^3(bx)-x^2(x^2)=-x^4$

$\displaystyle bx^4-x^4=-x^4$

$\displaystyle b=0$

Then, looking at the terms involving just $\displaystyle x$:

$\displaystyle -2c=-32$

$\displaystyle c=16$

Thus making the factors:

$\displaystyle 0=x(x+1)(x-2)(x^2+16)$

At this point, does common sense not dictate that $\displaystyle x^2+16\neq{0}$ for real $\displaystyle x$? - Nov 19th 2011, 07:39 AMcheshirecatRe: factoring
- Nov 19th 2011, 09:03 AMTheChazRe: factoring
You just GUESS. We determined that there are only a handful of potential rational roots. I like to start small, with +/-1. Then move onto the larger ones. It takes two successful trials to reduce a 4th degree polynomial to a quadratic, at which point you can use the quadratic formula.