Find the real zeros of f. Use the real zeros to factor f. $f\left(x\right)=x^4+x^3-3x^2-x+2$

The given function $f\left(x\right)=x^4+x^3-3x^2-x+2$is of degree four, so it has at most four real zeros.

As all the coefficients are integers so we use the rational zeros theorem.

The factors of the constant term $2$are

$p:\pm 1,\pm 2$

The factors of the leading coefficient $1$are

$q:\pm 1$

So the possible rational zeros are

$\frac{p}{q}:\pm 1,\pm 2$

The graph of the function is given as

So the graph has three turning points and thus has four real roots.

From the graph, it appears that $-2$is a zero of the function. On performing synthetic division we get

As the remainder is zero so $-2$is a zero of the function.

And the quotient $x^3-x^2-x+1$is the depressed polynomial. So the given function is factored as

$f\left(x\right)=(x+2)(x^3-x^2-x+1)$

Equate the depressed polynomial equal to zero and factor by grouping.

$\begin{array}{rcl}{x}^3-x^2-x+1& =& 0\\ x^2(x-1)-(x-1)& =& 0\\ (x-1)(x^2-1)& =& 0\\ (x-1)(x-1)(x+1)& =& 0\end{array}$

So the other roots are

$\begin{array}{rcl}x-1& =& 0\\ x& =& 1\end{array}$

and

$$\begin{gathered} x+1=0 \\ x=-1 \end{gathered}$$

And the complete factored form is$f\left(x\right)=(x+2)(x+1)(x-1)(x-1)$