Question

how many x-intercepts would the function shown below have?
$f(x) = -4(x^2 + 25)(x + 2)^2(x^2 - 2)$

Explanation:

Step1: Find roots by setting \( f(x) = 0 \)

To find the \( x \)-intercepts, we set \( f(x) = 0 \). Since \( -4
eq 0 \), we solve the equations from the other factors:

1. \( x^2 + 25 = 0 \)
2. \( (x + 2)^2 = 0 \)
3. \( x^2 - 2 = 0 \)

Step2: Solve \( x^2 + 25 = 0 \)

For \( x^2 + 25 = 0 \), we have \( x^2 = -25 \). The solutions are \( x = \pm 5i \) (complex roots, not real \( x \)-intercepts).

Step3: Solve \( (x + 2)^2 = 0 \)

For \( (x + 2)^2 = 0 \), taking the square root of both sides gives \( x + 2 = 0 \), so \( x = -2 \) (a real root, and it has multiplicity 2, but it's still one distinct \( x \)-intercept from this factor).

Step4: Solve \( x^2 - 2 = 0 \)

For \( x^2 - 2 = 0 \), we can rewrite it as \( x^2 = 2 \), so \( x = \pm \sqrt{2} \) (two distinct real roots).

Step5: Count real \( x \)-intercepts

From the factors:

• \( (x + 2)^2 \) gives 1 distinct real \( x \)-intercept (\( x = -2 \)).
• \( x^2 - 2 \) gives 2 distinct real \( x \)-intercepts (\( x = \sqrt{2} \) and \( x = -\sqrt{2} \)).
• \( x^2 + 25 \) gives no real \( x \)-intercepts.

Adding these, we have \( 1 + 2 = 3 \) distinct real \( x \)-intercepts.

Answer:

The function has 3 \( x \)-intercepts.