Answer
$P(x)=x^{2}(x-1)(x-2)$
Zeros: $ 0\quad$ ( double), $\quad 1,\ 2 \quad$ ( single)
Work Step by Step
Factor out $x^{2}$
$P(x)=x^{2}(x^{2}-3x+2)$
( $-1$ and $-2$ are factors of 2 whose sum is $-3)$
$P(x)=x^{2}(x-1)(x-2)$
Zeros: $ 0\quad$ ( double), $1,\ 2 \quad$ ( single)
The degre is 4, even. Leading coefficient is positive.
The end behavior is as for $x^{4}$: $+\infty$ on both far sides.
The graph
falls from the right left,
touches the x-axis at $(0,0)$ and turns up,
turns before x=1 and crosses the x-axis at $(1,0),$
rises and turns before x=2,
turns to cross the x-axis at $92,0)$ and continues to fall to the far right.
