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

$f\left(x\right)=3x^3-7x^2+12x-28$

The given function $f\left(x\right)=3x^3-7x^2+12x-28$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 $-28$are

role="math" localid="1646111022250" $p:\pm 1,\pm 2,\pm 4,\pm 7,\pm 14,\pm 28$

The factors of the leading coefficient $3$are

$q:\pm 1,\pm 3$

So the possible rational zeros are

$\frac{p}{q}:\pm 1,\pm 2,\pm 4,\pm 7,\pm 14,\pm 28,\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{4}{3},\pm \frac{7}{3},\pm \frac{14}{3},\pm \frac{28}{3}$

The graph of the function is given as

The graph resembles a cubic function and from the graph it can be seen that it has a real root which is greater than and near to the$2$.

The number $\frac{7}{3}$is close to $2$.

So finding $f\left(\frac{7}{3}\right)$we get

$$\begin{gathered} f\left(\frac{7}{3}\right)=3·{\left(\frac{7}{3}\right)}^3-7·{\left(\frac{7}{3}\right)}^2+12·\frac{7}{3}-28 \\ f\left(\frac{7}{3}\right)=\frac{343}{9}-\frac{343}{9}+28-28 \\ f\left(\frac{7}{3}\right)=0 \end{gathered}$$

As the function value is zero, so $\frac{7}{3}$is a zero of the function and $3x-7$is a factor.

Using synthetic division divide $3x^3-7x^2+12x-28$by$3x+7$

So the quotient $x^2+4$is the depressed polynomial and it is a prime polynomial, so it cannot be factored further. Thus the function has only one real zero and itis written in factored form as

$f\left(x\right)=(3x+7)(x^2+4)$