Question

solve the equation by factoring.
$16x^2 + 2x - 3 = 0$
the solution set is \boxed{}.
(use a comma to separate answers as needed.)

Explanation:

Step1: Factor the quadratic equation

We have the quadratic equation \(16x^{2}+2x - 3=0\). We need to find two numbers \(a\) and \(b\) such that \(a\times b=16\times(- 3)=- 48\) and \(a + b = 2\). The numbers are \(8\) and \(-6\) since \(8\times(-6)=-48\) and \(8+( - 6)=2\).
Rewrite the middle term:
\(16x^{2}+8x-6x - 3 = 0\)
Group the terms:
\((16x^{2}+8x)-(6x + 3)=0\)
Factor out the common factors from each group:
\(8x(2x + 1)-3(2x + 1)=0\)
Now factor out \((2x + 1)\):
\((8x-3)(2x + 1)=0\)

Step2: Solve for \(x\)

Set each factor equal to zero:

• For \(8x-3 = 0\), add \(3\) to both sides: \(8x=3\), then divide by \(8\): \(x=\frac{3}{8}\)
• For \(2x + 1=0\), subtract \(1\) from both sides: \(2x=-1\), then divide by \(2\): \(x=-\frac{1}{2}\)

Answer:

\(\frac{3}{8},-\frac{1}{2}\)