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

$f\left(x\right)=4x^4-15x^3-8x^2+15x+4$

The given function$f\left(x\right)=4x^4-15x^3-8x^2+15x+4$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 $4$are

$p:\pm 1,\pm 2,\pm 4$

The factors of the leading coefficient $4$are

$q:\pm 1,\pm 2,\pm 4$

So the possible rational zeros are

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

The graph of the function is given as

So the graph has three turning points and thus has four real roots.

From the graph, it appears that $-1$is a zero of the function. On performing synthetic division we get

As the remainder is zero so $-1$is a zero of the function.

And the quotient $4x^3-19x^2+11x+4$is the depressed polynomial. So the given function is factored as

role="math" localid="1646119174856" $f\left(x\right)=(x+1)(4x^3-19x^2+11x+4)$

From the graph, it can be seen that $1$may also be a zero. On performing synthetic division with the depressed polynomial we get

As the remainder is zero so $1$is a zero of the function.

And the quotient $4x^2-15x-4$is the depressed polynomial. So the given function is factored as

role="math" localid="1646119299998" $$\begin{gathered} f\left(x\right)=(x+1)(4x^3-19x^2+11x+4) \\ f\left(x\right)=(x+1)(x-1)(4x^2-15x-4) \end{gathered}$$

Equate the depressed polynomial equal to zero and factor by grouping.

$\begin{array}{rcl}4x^2-15x-4& =& 0\\ 4x^2-16x+x-4& =& 0\\ 4x(x-4)+(x-4)& =& 0\\ (x-4)(4x+1)& =& 0\end{array}$

So the other zeros are

role="math" $\begin{array}{rcl}x-4& =& 0\\ x& =& 4\end{array}$

and

$\begin{array}{rcl}4x+1& =& 0\\ 4x& =& -1\\ x& =& -\frac{1}{4}\end{array}$

So the final factored form is

$$\begin{gathered} f\left(x\right)=(x+1)(x-1)(4x^2-15x-4) \\ f\left(x\right)=(x+1)(x-1)(x-4)(4x+1) \end{gathered}$$