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Factoring Quadratics - Part 1

Difference of Two Squares

We now learn how to factor quadratics using the difference of two squares formula.

In particular we can factor quadratics looking like: \[ax^2-k^2 = 0\] Notice: comparing this to the generic quadratic \(ax^2 + bx +c\) we can see that there is no \(x\) term, in other words \(b = 0\).

Remember, the difference of two squares formula was: \[a^2-b^2 =\begin{pmatrix}a+b\end{pmatrix}.\begin{pmatrix}a-b\end{pmatrix} \] This formula can be used to show, for instance, that: \[x^2-25 = \begin{pmatrix}x-5\end{pmatrix}.\begin{pmatrix}x+5\end{pmatrix}\] or that: \[4x^2 - 9 = \begin{pmatrix}2x-3\end{pmatrix}.\begin{pmatrix}2x+3\end{pmatrix}\] We learn the method in the tutorial below.

Tutorial 1: Factoring Quadratics - Difference of Two Squares

In the following tutorial we learn how to use the difference of two squares to factorize quadratics.

Exercise 1

Write each of the following quadratics in factored form:

  1. \(x^2-16\)
  2. \(4p^2 - 25\)
  3. \(x^2-9y^2\)
  4. \(100m^2-49n^2\)
  5. \(25p^2 - 16q^2\)

Answers Without Working

  1. \(x^2-16 = \begin{pmatrix} x - 4 \end{pmatrix}\begin{pmatrix} x + 4 \end{pmatrix}\)
  2. \(4p^2 - 25 = \begin{pmatrix} 2p - 5\end{pmatrix}\begin{pmatrix} 2p + 5 \end{pmatrix}\)
  3. \(x^2-9y^2 = \begin{pmatrix} x - 3y\end{pmatrix} \begin{pmatrix} x + 3y \end{pmatrix}\)
  4. \(100m^2-49n^2 = \begin{pmatrix} 10m - 7n \end{pmatrix} \begin{pmatrix} 10m + 7n \end{pmatrix}\)
  5. \(25p^2 - 16q^2 = \begin{pmatrix} 5p - 4q \end{pmatrix} \begin{pmatrix} 5p + 4a \end{pmatrix}\)

Solving Quadratic Equations Using the Difference of Two Squares

The method for factoring quadratics using the difference of two squares can be used to solve quadratic equations that can be written: \[ax^2 - k^2 = 0\] This is illustrated in the following tutorial.

Tutorial 2: Factoring & Solving Quadratic Equations - Difference of Two Squares

In the following tutorial we learn how to use the difference of two squares to factorize quadratics.

Exercise 2

Use factorization to solve each of the following quadratic equations:

  1. \(x^2 - 4 = 0\)
  2. \(x^2 - 49 = 0\)
  3. \(2x^2 - 50 = 0\)
  4. \(4x^2 - 9 = 0\)
  5. \(16x^2 - 81 = 0\)

Solution Without Working

  1. In factored form we write: \[x^2 - 4 = \begin{pmatrix}x - 2\end{pmatrix}.\begin{pmatrix}x + 2\end{pmatrix}\] The solutions to \(x^2 - 4=0\) are therefore:

    \(x = 2\) and \(x = -2\)

  2. In factored form we write: \[x^2 - 49 = \begin{pmatrix}x - 7\end{pmatrix}.\begin{pmatrix}x + 7\end{pmatrix}\] The solutions to \(x^2 - 49 = 0\) are therefore:

    \(x = 7\) and \(x = -7\)

  3. The solutions to \(2x^2 - 50 = 0\) are:

    \(x = -5\) and \(x = 5\)

  4. In factored form we write: \[4x^2 - 9 = \begin{pmatrix}2x - 3\end{pmatrix}.\begin{pmatrix}2x + 3\end{pmatrix}\] The solutions to \(4x^2 - 9 = 0\) are therefore:

    \(x = \frac{3}{2}\) and \(x = - \frac{3}{2}\)

  5. The solutions to \(16x^2 - 81 = 0\) are:

    \(x = - \frac{9}{4}\) and \(x = - \frac{9}{4}\)

Solution With Working

  1. We factor and solve \(x^2 - 4=0\) in two steps:

    • Step 1: we notice that \(x^2 - 4\) is the difference of two squares.

      Indeed \(x^2 - 4 = x^2 - 2^2\)

      So we can write it in factored form: \[x^2 - 4 = \begin{pmatrix} x - 2 \end{pmatrix}\begin{pmatrix} x + 2 \end{pmatrix}\]

    • Step 2: we use the null product law to solve the factored equation \(\begin{pmatrix} x - 2 \end{pmatrix}\begin{pmatrix} x + 2 \end{pmatrix} = 0 \).

      That's:

      \(\begin{pmatrix} x - 2 \end{pmatrix}\begin{pmatrix} x + 2 \end{pmatrix} = 0\) if and only if:

      \(x - 2 = 0\) or \(x+2 = 0\), which leads to:

      \(x = 2\) or \(x = -2\).

      Finally we can state that this quadratic equation has two solutions:

      \(x = 2\) and \(x = -2\).

  2. We factor and solve \(x^2 - 49=0\) in two steps:

    • Step 1: we notice that \(x^2 - 49\) is the difference of two squares.

      Indeed \(x^2 - 49 = x^2 - 7^2\).

      So we can write it in factored form: \[x^2 - 49 = \begin{pmatrix} x - 7 \end{pmatrix}\begin{pmatrix} x + 7 \end{pmatrix}\]

    • Step 2: we use the null product law to solve the factored equation \(\begin{pmatrix} x - 7 \end{pmatrix}\begin{pmatrix} x + 7 \end{pmatrix} = 0 \).

      That's:

      \(\begin{pmatrix} x - 7 \end{pmatrix}\begin{pmatrix} x + 7 \end{pmatrix} = 0\) if and only if:

      \(x - 7 = 0\) or \(x+7 = 0\), which leads to:

      \(x = 7\) or \(x = -7\).

      Finally we can state that this quadratic equation has two solutions:

      \(x = 7\) and \(x = -7\).

  3. We factor and solve \(2x^2 - 50=0\) in two steps:

    Trick: We can start by dividing both sides of this equation by \(2\). Dividing \(2x^2 - 50 = 0\) by \(2\) it leads to: \[x^2 - 25 = 0\] We can now solve it in two steps:
    • Step 1: we ss that \(x^2 - 25\) is the difference of two squares.

      Indeed \(x^2 - 25 = x^2 - 5^2\) so we can write it in factored form: \[x^2 - 25 = \begin{pmatrix} x - 5 \end{pmatrix}\begin{pmatrix} x + 5 \end{pmatrix}\]

    • Step 2: we use the null product law to solve the factored equation \(\begin{pmatrix} x - 5 \end{pmatrix}\begin{pmatrix} x + 5 \end{pmatrix} = 0 \).

      That's:

      \(\begin{pmatrix} x - 5 \end{pmatrix}\begin{pmatrix} x + 5 \end{pmatrix} = 0\) if and only if:

      \(x - 5 = 0\) or \(x+5 = 0\), which leads to:

      \(x = 5\) or \(x = -5\).

      Finally we can state that this quadratic equation has two solutions:

      \(x = 5\) and \(x = -5\).


  4. We factor and solve \(4x^2 - 9=0\) in two steps:

    • Step 1: we notice that \(4x^2 - 9\) is the difference of two squares.

      Indeed \(x^2 - 49 = \begin{pmatrix}2x\end{pmatrix}^2 - 3^2\).

      So we can write it in factored form: \[x^2 - 49 = \begin{pmatrix} 2x - 3 \end{pmatrix}\begin{pmatrix} 2x + 3 \end{pmatrix}\]

    • Step 2: we use the null product law to solve the factored equation \(\begin{pmatrix} 2x - 3 \end{pmatrix}\begin{pmatrix} 3x + 3 \end{pmatrix} = 0 \).

      That's:

      \(\begin{pmatrix} 2x - 3 \end{pmatrix}\begin{pmatrix} 2x + 3 \end{pmatrix} = 0\) if and only if:

      \(2x - 3 = 0\) or \(2x+3 = 0\), which leads to:

      \(2x = 3\) or \(2x = -3\).

      Which, in turn, leads to: \(x = \frac{3}{2}\) or \(x = -\frac{3}{2}\).

      Finally we can state that this quadratic equation has two solutions:

      \(x = \frac{3}{2}\) and \(x = -\frac{3}{2}\).

  5. We factor and solve \(16x^2 - 81=0\) in two steps:

    • Step 1: we notice that \(16x^2 - 81\) is the difference of two squares.

      Indeed \(16x^2 - 81 = \begin{pmatrix}4x\end{pmatrix}^2 - 9^2\).

      So we can write it in factored form: \[16x^2 - 81 = \begin{pmatrix} 4x - 9 \end{pmatrix}\begin{pmatrix} 4x + 9 \end{pmatrix}\]

    • Step 2: we use the null product law to solve the factored equation \(\begin{pmatrix} 2x - 3 \end{pmatrix}\begin{pmatrix} 3x + 3 \end{pmatrix} = 0 \).

      That's:

      \(\begin{pmatrix} 2x - 3 \end{pmatrix}\begin{pmatrix} 2x + 3 \end{pmatrix} = 0\) if and only if:

      \(2x - 3 = 0\) or \(2x+3 = 0\), which leads to:

      \(2x = 3\) or \(2x = -3\).

      Which, in turn, leads to: \(x = \frac{3}{2}\) or \(x = -\frac{3}{2}\).

      Finally we can state that this quadratic equation has two solutions:

      \(x = \frac{3}{2}\) and \(x = -\frac{3}{2}\).

  6. We factor and solve \(x^2 -5 = 0\) in two steps:

    • Step 1: Using the fact that \(5 = \begin{pmatrix} \sqrt{5} \end{pmatrix}^2\), we can write \(x^2 - 5\) as the difference of two squares:

      Indeed \(x^2 - 5 = x^2 - \begin{pmatrix} \sqrt{5} \end{pmatrix}^2\).

      So we can write it in factored form: \[x^2 - 5 = \begin{pmatrix} x - \sqrt{5} \end{pmatrix} \begin{pmatrix} x + \sqrt{5} \end{pmatrix}\]

    • Step 2: we use the null product law to solve the factored equation \(\begin{pmatrix} x - \sqrt{5} \end{pmatrix} \begin{pmatrix} x + \sqrt{5} \end{pmatrix} = 0\).

      That's: \(\begin{pmatrix} x - \sqrt{5} \end{pmatrix} \begin{pmatrix} x + \sqrt{5} \end{pmatrix} = 0\) if and only if:

      \(x - \sqrt{5} = 0\) or \(x + \sqrt{5} = 0\).

      This leads to: \(x = \sqrt{5}\) or \(x = - \sqrt{5}\).

      Finally we can state that this quadratic equation has two solutions:

      \(x = \sqrt{5}\) and \(x = - \sqrt{5}\).


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