Question

1. solve for the values of x by factoring: ( x^2 + 20x + 64 = 0 ).

( \bigcirc ) ( x = -16, x = -4 )
( \bigcirc ) ( x = -8, x = 8 )
( \bigcirc ) ( x = -14, x = -3 )
( \bigcirc ) ( x = -32, x = -2 )

Explanation:

Step1: Factor the quadratic equation

We need to find two numbers that multiply to \( 64 \) (the constant term) and add up to \( 20 \) (the coefficient of \( x \)). The numbers \( 16 \) and \( 4 \) satisfy this because \( 16\times4 = 64 \) and \( 16 + 4 = 20 \). So, we can factor the quadratic equation \( x^{2}+20x + 64 = 0\) as \((x + 16)(x + 4)=0\).

Step2: Solve for \( x \)

Using the zero - product property, if \( ab = 0\), then either \( a = 0\) or \( b = 0\).
For \((x + 16)(x + 4)=0\), we set each factor equal to zero:

• When \( x+16 = 0\), we solve for \( x\) by subtracting \( 16\) from both sides of the equation: \(x=-16\).
• When \( x + 4=0\), we solve for \( x\) by subtracting \( 4\) from both sides of the equation: \(x=-4\).

Answer:

A. \( x = -16, x = -4 \)