Question

use the zero - factor property to solve the equation. (enter your answers as a comma - separated list.)
$(y - 39)(2y + 5)(y + 12)=0$
$y = $

1. - /2 points

solve the equation by factoring. (enter your answers as a comma - separated list.)
$x^{2}+4x = 0$
$x = $

Explanation:

First Equation: Solve \((y - 39)(2y + 5)(y + 12) = 0\)

Step1: Apply Zero - Factor Property

The Zero - Factor Property states that if \(ab = 0\), then either \(a = 0\) or \(b = 0\). For the equation \((y - 39)(2y+5)(y + 12)=0\), we set each factor equal to zero:

• \(y-39 = 0\)
• \(2y + 5=0\)
• \(y + 12=0\)

Step2: Solve for \(y\) in each equation

• For \(y-39 = 0\), add 39 to both sides of the equation: \(y=39\)
• For \(2y + 5=0\), subtract 5 from both sides: \(2y=-5\), then divide both sides by 2: \(y =-\frac{5}{2}\)
• For \(y + 12=0\), subtract 12 from both sides: \(y=-12\)

Second Equation: Solve \(x^{2}+4x = 0\)

Step1: Factor the left - hand side

Factor out the common factor \(x\) from \(x^{2}+4x\). We get \(x(x + 4)=0\)

Step2: Apply Zero - Factor Property

Set each factor equal to zero:

• \(x = 0\)
• \(x+4=0\)

Step3: Solve for \(x\) in each equation

• For \(x = 0\), the solution is \(x = 0\)
• For \(x+4=0\), subtract 4 from both sides: \(x=-4\)

Answer:

For the equation \((y - 39)(2y + 5)(y + 12)=0\), \(y = 39,-\frac{5}{2},-12\)
For the equation \(x^{2}+4x = 0\), \(x = 0,-4\)