Find the real zeros of f. Use the real zeros to factor f.

$f\left(x\right)=x^3+2x^2-5x-6$

The given function $f\left(x\right)=x^3+2x^2-5x-6$is of degree three, so it has at most three real zeros.

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

The factors of the constant term $-6$are $p:\pm 1,\pm 2,\pm 3,\pm 6$

The factors of the leading coefficient $1$are

$q:\pm 1$.

So the possible rational zeros are

$\frac{p}{q}:\pm 1,\pm 2,\pm 3,\pm 6$

The graph of the function is given as

The graph has two turning points, so it will have three real zeros.

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 $x^2+4x+3$is the depressed polynomial. So the given function is factored as

$$\begin{gathered} f\left(x\right)=x^3+2x^2-5x-6 \\ f\left(x\right)=(x-2)(x^2+4x+3) \end{gathered}$$

Equating the depressing polynomial with zero and factoring by grouping we get

$\begin{array}{rcl}{x}^2+4x+3& =& 0\\ x^2+3x+x+3& =& 0\\ x(x+3)+1(x+3)& =& 0\\ (x+3)(x+1)& =& 0\end{array}$

So the zeros are

$\begin{array}{rcl}x+3& =& 0\\ x& =& -3\end{array}$

and

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

All the three real zeros of the function $f\left(x\right)=x^3+2x^2-5x-6$are $-3,-1,2$.

And it can be written in factored form as

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