first off, let's notice the graph touches the x-axis at -1 and 3, namely, those are the zeros/solutions/roots of the polynomial and therefore, the factors come from those points.
now, at -1, the graph doesn't cross the x-axis, instead it simply bounces off of it, that means the zero of x = -1, has an even multiplicity, could be 4 or 2 or 6, but let's go with 2.
at x = 3, the graph does cross the x-axis, meaning it has an odd multiplicity, could be 3 or 1, or 7 or 9, but let's use 1.
[tex]\bf \begin{cases} x=-1\implies &x+1=0\\ x=3\implies &x-3=0 \end{cases}~\hspace{5em}\stackrel{\textit{even multiplicity}}{(x+1)^2}\qquad \stackrel{\textit{odd multiplicity}}{(x-3)^1}=\stackrel{y}{0} \\\\\\ (x^2+2x+1)(x-3)=y\implies x^3+2x^2+x-3x^2-6x-3=y \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill x^3-x^2-5x-3=y~\hfill[/tex]
y = x³ - x² - 5x - 3
the graph has roots at x = - 1 and x = 3
there is a turning point at x = - 1 indicating a double root
given x = a and x = b are the roots of a polynomial then
(x - a) and (x - b ) are factors and the polynomial is the product of the factors
here x = - 1 ( double root ) ⇔ (x + 1)² is it's factors and
x = 3 hence (x + 3) is a factor
f(x) = (x + 1)²(x - 3)
= (x² + 2x + 1)(x - 3)
= x³ - x² - 5x - 3 ← is a possible polynomial
