Question

how many x-intercepts would the function shown below have?
$f(x) = (4x + 5)(x^2 + 4)$

Explanation:

Step1: Recall x-intercept definition

To find the \(x\)-intercepts, we set \(f(x) = 0\) and solve for \(x\). So we have the equation \((4x + 5)(x^2 + 4)=0\).

Step2: Apply zero - product property

The zero - product property states that if \(ab = 0\), then either \(a = 0\) or \(b = 0\). So we set each factor equal to zero:

• Case 1: \(4x+5 = 0\)

Solve for \(x\): \(4x=-5\), then \(x =-\frac{5}{4}\). This is a real solution.

• Case 2: \(x^2 + 4=0\)

Solve for \(x\): \(x^2=-4\). Since the square of a real number is non - negative, there are no real solutions for \(x\) from this equation (because for real \(x\), \(x^2\geq0\), and \(- 4<0\)).

Answer:

The function \(f(x)=(4x + 5)(x^2 + 4)\) has 1 \(x\)-intercept.