Find the real zeros of f. Use the real zeros to factor f. $f\left(x\right)=2x^3-5x^2-4x+12$

The given function $f\left(x\right)=2x^3-5x^2-4x+12$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 $12$are

$p:\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12$

The factors of the leading coefficient $2$are

$q:\pm 1,\pm 2$

So the possible rational zeros are

$\frac{p}{q}:\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12,\pm \frac{1}{2},\pm \frac{3}{2}$

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 $2x^2-x-6$is the depressed polynomial. So the given function is factored as

$$\begin{gathered} f\left(x\right)=2x^3-5x^2-4x+12 \\ f\left(x\right)=(x-2)(2x^2-x-6) \end{gathered}$$

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

$\begin{array}{rcl}2x^2-x-6& =& 0\\ 2x^2-4x+3x-6& =& 0\\ 2x(x-2)+3(x-2)& =& 0\\ (x-2)(2x+3)& =& 0\end{array}$

So the other zeros are

$\begin{array}{rcl}x-2& =& 0\\ x& =& 2\end{array}$

and

$\begin{array}{rcl}2x+3& =& 0\\ 2x& =& -3\\ x& =& -\frac{3}{2}\end{array}$

All the three real zeros of the function $f\left(x\right)=2x^3-5x^2-4x+12$are $-\frac{3}{2},2,2$.

And it can be written in factored form as $f\left(x\right)=(2x+3)(x-2)(x-2)$.