Answer
$p(x)=x^4+2x^3-3x^2-4x+4$
Work Step by Step
The graph is tangent to the x-axis at $x=- 2$ and $1$. It means that the polynomial has zeros of $-2$ and $1$ with even multiplicities.
Since the degree is 4, both have multiplicities of 2.
Then,
$p(x)=A(x+2)^2(x-1)^2$
$p(x)=A(x^2+4x+4)(x^2-2x+1)$
$p(x)=A(x^4+2x^3-3x^2-4x+4)$
$p(x)=Ax^4+2Ax^3-3Ax^2-4Ax+4A$
The graph also intercepts the y-axis at $y=4$, then the constant term is 4.
It gives $A=1$.
Therefore, $p(x)=x^4+2x^3-3x^2-4x+4$.
