Question

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

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

Short answer

The real zeros of the function are $-1,-\frac{1}{4},1,4$.

The factored form of the function is $f\left(x\right)=(x+1)(4x+1)(x-1)(x-4)$.

Step-by-step solution

Step 1. Find the possible number of zeros  

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.

Step 2. Use the rational zero theorem   

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

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

Step 3. Graph the function 

The graph of the function is given as

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

Step 4. Find the first factor  

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

Step 5. Find the second factor  

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

Step 6. Factor the depressed polynomial  

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

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