x^5-5x^3+4x
...I don't know exactly how to do that :-|
$\displaystyle x^5 - 5x^3 + 4x = x(x^4 - 5x^2 + 4)$.
Now you have a quadratic in $\displaystyle x^2$. So let $\displaystyle X = x^2$.
$\displaystyle = x(X^2 - 5X + 4)$
$\displaystyle = x(X - 1)(X - 4)$
$\displaystyle = x(x^2 - 1)(x^2 - 4)$
Now use difference of two squares
$\displaystyle = x(x^2 - 1^2)(x^2 - 2^2)$
$\displaystyle = x(x - 1)(x + 1)(x - 2)(x + 2)$.
Or.
Take the possible zero's using the rational rooth theorem.
$\displaystyle \pm1,\pm2,\pm3,\pm4
$
1.Then used synthetic division a few times.
2.Once done, you will have found all the zero's (where r=0).
3.Then just put it in factored form.
1.
1) 1,0,-5,0,4
$\displaystyle 1(1)+0=1(1)-5=-4*1+0=-4+4=0$
$\displaystyle r=0$
-1) 1,1,-4,-4
$\displaystyle 1(-1)+1=0(1)+(-4)=-4+4=0$
$\displaystyle r=0$
2) 1,0,-4,0
$\displaystyle 1(2)+0=2(2)+(-4)=0(2)+0=0$
$\displaystyle r=0$
-2) 1,2,0,0
$\displaystyle 1(-2)+2=0$
$\displaystyle r=0$
3.Now look at the coefficients left.
1,0,0,0
subtract leading exponent by number of divisions made::
5-4=1
therefore the coefficient is x.
place the zero's in factored form
$\displaystyle (x+1)(x-1)(x+2)(x-2)$
multiply by x
$\displaystyle x(x+1)(x-1)(x+2)(x-2)$
Seems harder & longer, but sometimes this method can be a life saver >.> for me at least