Question

how many x-intercepts would the function shown below have?

$f(x) = 2(x + 9)^2(6x + 1)(x^2 - 5)$

Explanation:

Step1: Find roots from each factor

To find the x - intercepts, we set \(f(x)=0\). So we solve \(2(x + 9)^{2}(6x+1)(x^{2}-5)=0\). Since the product of factors is zero when at least one factor is zero, we consider each factor:

• For \((x + 9)^{2}=0\), we have \(x=-9\) (a repeated root, but we count the distinct root here for x - intercepts, but we will check the multiplicity later for counting the number of x - intercepts considering distinct roots and roots from quadratic).
• For \(6x + 1=0\), we solve for \(x\): \(6x=-1\), so \(x=-\frac{1}{6}\).
• For \(x^{2}-5=0\), we solve for \(x\): \(x^{2}=5\), so \(x=\pm\sqrt{5}\).

Step2: Count the number of distinct x - intercepts

The factor \((x + 9)^{2}\) gives a single distinct x - intercept \(x=-9\) (even though it has multiplicity 2, it is one x - intercept on the graph). The factor \(6x + 1\) gives \(x =-\frac{1}{6}\) (one x - intercept). The factor \(x^{2}-5\) is a quadratic, and using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\) (here \(a = 1\), \(b=0\), \(c=-5\)), we get two distinct real roots \(x=\sqrt{5}\) and \(x=-\sqrt{5}\).

Now we count the number of distinct x - intercepts: from \((x + 9)^{2}\): 1, from \(6x + 1\): 1, from \(x^{2}-5\): 2. So in total, \(1 + 1+2=4\).

Answer:

4