1. ## factoring

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

2. ## Re: factoring

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?

3. ## Re: factoring

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.

4. ## Re: factoring

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.

5. ## 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 ...

$x^2 + 16$

6. ## Re: factoring

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?

7. ## 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 $x$?

8. ## Re: factoring

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?

9. ## 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.