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?
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.
note the coefficients of the final depressed polynomial ...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
$\displaystyle x^2 + 16$
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 $\displaystyle x$?
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.