# factoring

• Nov 18th 2011, 10:24 AM
cheshirecat
factoring
how do i factor 0= x^5-x^4+14x^3-16x^2-32x?
• Nov 18th 2011, 10:40 AM
Also sprach Zarathustra
Re: factoring
Quote:

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

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, 12:46 PM
TheChaz
Re: factoring
Quote:

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

Theory suggests using the "rational root theorem" to narrow down the possible roots to those that AsZ gave.
Then "synthetic division" will make it rather painless to determine if a given contender is indeed a root.
• Nov 18th 2011, 02:34 PM
cheshirecat
Re: factoring
Quote:

Originally Posted by Also sprach Zarathustra
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?

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, 02:47 PM
skeeter
Re: factoring
Code:

```-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```
note the coefficients of the final depressed polynomial ...

$x^2 + 16$
• Nov 18th 2011, 06:42 PM
cheshirecat
Re: factoring
Quote:

Originally Posted by skeeter
Code:

```-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```
note the coefficients of the final depressed polynomial ...

$x^2 + 16$

But how did you know to pick those two coefficients from that particular root? How did you know using synthetic division there were 2 complex roots?
• Nov 18th 2011, 07:44 PM
Quacky
Re: factoring
How I would approach: use an appropriate theorem to find the first three roots.

Then, we have:

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

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

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

we can then equate coefficients.

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

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

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

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

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

$bx^4-x^4=-x^4$

$b=0$

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

$-2c=-32$

$c=16$

Thus making the factors:

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

At this point, does common sense not dictate that $x^2+16\neq{0}$ for real $x$?
• Nov 19th 2011, 08:39 AM
cheshirecat
Re: factoring
Quote:

Originally Posted by cheshirecat
But how did you know to pick those two coefficients from that particular root? How did you know using synthetic division there were 2 complex roots?

how do u do this part?
• Nov 19th 2011, 10:03 AM
TheChaz
Re: 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.