In Problems 39–56, find the real zeros of f. Use the real zeros to factor f.

$f\left(x\right)=2x^4-3x^3-21x^2-2x+24$

The given function is $f\left(x\right)=2x^4-3x^3-21x^2-2x+24$.

The given function 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 $24$are

role="math" localid="1646129775343" $p:\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 8,\pm 12,\pm 24$

The factors of the leading coefficient $2$are

$q:\pm 1,\pm 2$

So, the possible rational zeros are

role="math" localid="1646129849786" $\frac{p}{q}:\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 8,\pm 12,\pm 24,\pm \frac{1}{2},\pm \frac{3}{2}$

The graph is

From the graph, we can depict that it has four roots.

Since $1$appears to be zero and a potential rational zero also.

By evaluating we get,$f\left(1\right)=0$

Thus, $(x-1)$is a factor of f.

Now, let's use synthetic division to factor f

So, factor fis$(x-1)(2x^3-x^2-22x-24).$

It appears that $4$is a zero and a potential rational zero also.

By evaluating we get, $f\left(4\right)=0$

Thus, $(x-4)$ is a factor of f.

By using synthetic division we get,

So, factor fis$(x-1)(x-4)(2x^2+7x+6).$

Put depressed equation to zero and factor by grouping

$$\begin{gathered} 2x^2+7x+6=0 \\ 2x^2+4x+3x+6=0 \\ 2x(x+2)+3(x+2)=0 \\ (2x+3)(x+2)=0 \end{gathered}$$

Therefore, the other zeros are

localid="1646132579858" $$\begin{gathered} 2x+3=0 \\ x=-\frac{3}{2} \\ andx+2=0 \\ x=-2 \end{gathered}$$

The factored form of f is$(2x+3)(x+2)(x-4)(x-1).$