Question

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

$f\left(x\right)=4x^5+12x^4-x-3$

Short answer

The real zeros offare$-3,\frac{1}{\sqrt{2}},and-\frac{1}{\sqrt{2}}.$

The factored form off is$(2x^2+1)(\sqrt{2}x-1)(\sqrt{2}x+1)(x+3).$

Step-by-step solution

Step 1. Finding the possible number of zeros

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

The given function is of degree five, so it has at most five 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$-3$ are

$p:\pm 1,\pm 3$

The factors of the leading coefficient $4$are

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

So, the possible rational zeros are

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

Step 3. Graph the polynomial function 

The graph is

From the graph, we conclude that it has three roots.

Step 4. Finding the factor 

Since $-3$appears to be zero and a potential rational zero also.

By evaluating we get,$f(-3)=0$

Thus, $(x+3)$ is a factor of f.

Use synthetic division to factor f

So, factor fis$(x+3)(4x^4-1)$

Step 5. Factor the depressed polynomial 

Put depressed equation to zero and factor by grouping

$$\begin{gathered} 4x^4-1=0 \\ 4x^4=1 \\ {x}^4=\frac{1}{4} \\ x=\pm \frac{1}{\sqrt{2}} \end{gathered}$$

Therefore, the other zeros are$\frac{1}{\sqrt{2}}and-\frac{1}{\sqrt{2}}.$

The factored form of f is$(2x^2+1)(\sqrt{2}x+1)\left(\sqrt{2}x-1\right)(x+3).$