For an assignment in my Algebra 2 class we have to find the rational zeros of the equation then write it in factored form.
I am a little lost and don't really get it
Equation:
$h(x)=x^3-5x^2+2x+8$
I have tried first getting the rational zeros and got
$1, 2, 4, 8$
Then using synthetic substition
$2$ | $1$ $-5$ $2$ $8$
____2 -6 -8____
From that i got
$1x^3$ $-3$ $-4$ and finally $0$
Then wrote it into a function
$h(x)=(x-2)(x^2-3x-4x)$
Which i don't know if it is correct, i also don't know what they mean by factor it
It would be helpful if you could show all you steps so i can actaully learn how to do it myself.
Thanks in advance!
You're doing everything right so far! Just factor the quadratic $x^2-3x-4$ to completely factor $x^3-5x^2+2x+8$.
Although, you could do it this way: $$\begin{align*} & x^3-5x^2+2x+8\\ & =x^3-4x^2-x^2+2x+8\\ & =x^2(x-4)-(x-4)(x+2)\\ & =(x-4)(x^2-x-2)\\ & =(x-4)(x-2)(x+1)\end{align*}\tag{1}$$
But not many people might've seen the first step...