Compare Your Answers
Determine the zeros of the following quadratic functions by factoring.

  1. \(y = 2x^2 - 14x + 24\)

    \(\begin{array}{l}
     y = 2x^2 - 14x + 24 \\
     y = 2\left( {x^2 - 7x + 12} \right) \\
     \end{array}\)

    You need a sum of \(-7\) and a product of \(12\).

    \(\begin{array}{l}
     -4 + \left( { -3} \right) = -7 \\
     \left( {- 4} \right)\left( {- 3} \right) = 12 \\
     \end{array}\)

    \(y = 2\left( {x - 3} \right)\left( {x - 4} \right)\)

    \(\begin{align}
     x - 3 &= 0 \\
     x &= 3 \\
     \end{align}\)
    \(\begin{align}
     x - 4 &= 0 \\
     x &= 4 \\
     \end{align}\)

    The zeros are \(3\) and \(4\).

  2. \(f(x) = -3x^2 + 9x + 84\)

    \(\begin{array}{l}
     f\left( x \right) = -3x^2 + 9x + 84 \\
     f\left( x \right) = -3\left( {x^2 - 3x - 28} \right) \\
     \end{array}\)

    You need a sum of \(–3\) and a product of \(–28\).

    \(\begin{array}{l}
     4 + \left( { -7} \right) =  -3 \\
     \left( 4 \right)\left( { -7} \right) =  -28 \\
     \end{array}\)

    \(f\left( x \right) =  -3\left( {x + 4} \right)\left( {x - 7} \right)\)

    \(\begin{align}
     x + 4 &= 0 \\
     x &= -4 \\
     \end{align}\)
    \(\begin{align}
     x - 7 &= 0 \\
     x &= 7 \\
     \end{align}\)


    The zeros are \(–4\) and \(7\).

  3. \(f(x) = 2x^2 - x - 3\)

    No GCF.
    You need a sum of \(–1\) and a product of \(–6\).

    \(\begin{array}{l}
     -3 + 2 = -1 \\
     \left( { - 3} \right)\left( 2 \right) = -6 \\
     \end{array}\)

    Factor by decomposition.

    \(\begin{array}{l}
     f\left( x \right) = 2x^2 - x - 3 \\
     f\left( x \right) = 2x^2 - 3x + 2x - 3 \\
     f\left( x \right) = x\left( {2x - 3} \right) + 1\left( {2x - 3} \right) \\
     f\left( x \right) = \left( {2x - 3} \right)\left( {x + 1} \right) \\
     \end{array}\)

    \(\begin{align}
     2x - 3 &= 0 \\
     2x &= 3 \\
     x &= \frac{3}{2} \\
     \end{align}\)
    \(\begin{align}
     x + 1 &= 0 \\
     x &= -1 \\
     \end{align}\)

    The zeros are \(\frac{3}{2}\) and \(-1\).