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

$f\left(x\right)=x^3+6x^2+6x-4$

The given function $f\left(x\right)=x^3+6x^2+6x-4$is of degree three, so it has at most three real zeros.

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

The factors of the constant term $-4$are

role="math" localid="1646112881906" $p:\pm 1,\pm 2,\pm 4$

The factors of the leading coefficient $1$are

$q:\pm 1$

So the possible rational zeros are

$\frac{p}{q}:\pm 1,\pm 2,\pm 4$

The graph of the function is given as

So the graph has two turning points and thus has three 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 $x^2+4x-2$is the depressed polynomial. So the given function is factored as

$f\left(x\right)=(x+2)(x^2+4x-2)$

Equating the depressing polynomial with zero we get $x^2+4x-2=0$. Now solving it using the quadratic formula we get

$$\begin{gathered} x=\frac{-4\pm \sqrt{4^2-4·1·(-2)}}{2·1} \\ x=\frac{-4\pm \sqrt{24}}{2} \\ x=\frac{-4\pm 2\sqrt{6}}{2} \\ x=-2\pm \sqrt{6} \end{gathered}$$

So the other two factors are $(x-(-2+\sqrt{6}\left)\right),(x-(-2-\sqrt{6})\to (x+2-\sqrt{6}),(x+2+\sqrt{6})$

So the factored form is

$f\left(x\right)=(x+2)(x+2-\sqrt{6})(x+2+\sqrt{6})$