Question

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

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

Short answer

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

The factored form off is$(2x+1)(2x-1)(x^2+4).$

Step-by-step solution

Step 1. Finding the possible number of zeros

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

The given function 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 polynomial function 

The graph is

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

Step 4. Using the real zeros to factor f 

From the graph, we get to know that there are only two real zeros, thus the other two zeros are imaginary.

The zeros are$-\frac{1}{2}and\frac{1}{2}.$

The factored form offis$(2x+1)(2x-1)(x^2+4).$