Question

what are the zeros of the equation $y = 4(x - 3)(x + 5)(3x - 1)$?
\\(\circ\\) $x = -5, -4, 1$
\\(\circ\\) $x = -5, \frac{1}{3}, 3$
\\(\circ\\) $x = -3, 4, 5$
\\(\circ\\) $x = -\frac{1}{3}, 3, 5$

Explanation:

Step1: Recall Zero Product Property

The Zero Product Property states that if \(ab = 0\), then either \(a = 0\) or \(b = 0\) (or both). For the function \(y = 4(x - 3)(x + 5)(3x - 1)\), we set \(y = 0\), so \(4(x - 3)(x + 5)(3x - 1)=0\). Since \(4
eq0\), we can focus on the factors \((x - 3)\), \((x + 5)\), and \((3x - 1)\).

Step2: Solve for \(x\) in each factor

• For \((x - 3)=0\), adding 3 to both sides gives \(x = 3\).
• For \((x + 5)=0\), subtracting 5 from both sides gives \(x=-5\).
• For \((3x - 1)=0\), add 1 to both sides: \(3x = 1\), then divide by 3: \(x=\frac{1}{3}\).

Answer:

\(x = -5,\frac{1}{3},3\) (corresponding to the option with these values)