Example  1
Solve \(\left( {x - 3} \right)\left( {2x + 1} \right) = 0\), and verify the solution.

This is a quadratic equation with one side equal to zero, and it has already been factored. You can now determine which values of \(x\) make each factor equal to \(0\).

\(\left( {x - 3} \right)\left( {2x + 1} \right) = 0\)

\(\begin{align} x - 3 &= 0 \\
 x &= 3 \\
 \end{align}\)
\(\begin{align}
 2x + 1 &= 0 \\
 2x &= -1 \\
 x &= -\frac{1}{2} \\
 \end{align}\)
The solutions to the equation \(\left( {x - 3} \right)\left( {2x + 1} \right) = 0\) are \(3\) and \(-\frac{1}{2}\).

Verify the solutions by substituting them back into the original equation.

For \(x = 3\),

Left Side
 Right Side
\(\begin{array}{r}
 \left( {x - 3} \right)\left( {2x + 1} \right) \\
 \left( {3 - 3} \right)\left[ {2\left( 3 \right) + 1} \right] \\
 0\left( 7 \right) \\
 0 \end{array}\)
 \(0\)
               LS = RS

The two sides are equal, so \(3\) is a solution.
For \(x = -\frac{1}{2}\),

Left Side
 Right Side
\(\begin{array}{r}
 \left( {x - 3} \right)\left( {2x + 1} \right) \\
 \left( { - \frac{1}{2} - 3} \right)\left[ {2\left( { - \frac{1}{2}} \right) + 1} \right] \\
 \left( { - \frac{7}{2}} \right)\left( 0 \right) \\
 0 \end{array}\)
 \(0\)
                              LS = RS

The two sides are equal, so \(-\frac{1}{2}\) is also a solution.