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Difference of Two Squares

Factoring & Solving Quadratics

The difference of two squares is often used to factorize expressions involving quadratics.
In particular quadratic equations looking like: \[x^2-k^2 = 0\]

or
\[ax^2-k^2 = 0\] Remember, the difference of two squares formula was: \[\begin{pmatrix}a+b\end{pmatrix}.\begin{pmatrix}a-b\end{pmatrix} = a^2 - b^2\] Now that we've made a note of that let's learn the method.

Factoring & Difference of Two Squares

Say we're asked to factorize the following: \[x^2 - 4\] The trick is to see that, using the difference of two squares formula, this can we written \[x^2 - k^2 = \begin{pmatrix}x-2\end{pmatrix}\begin{pmatrix} x + 2 \end{pmatrix}\]
and
\[ax^2 - k^2 = \begin{pmatrix}\sqrt{a}x-k\end{pmatrix}\begin{pmatrix} \sqrt{a}x + k \end{pmatrix}\]

Tutorial

In the following tutorial we review the ... .

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. \(25p^2 - 16q^2 = \begin{pmatrix} 5p - 4q \end{pmatrix} \begin{pmatrix} 5p + 4a \end{pmatrix}\)

Solution With Working

  1. We factorize \(x^2 - 16\) as follows:

    This can be written as the difference of two squares by writing: \[x^2 - 16 = x^2 - 4^2\] So we can use our formula \(a^2 - b^2 = \begin{pmatrix}a - b \end{pmatrix}\begin{pmatrix} a + b \end{pmatrix}\) with \(a = x\) and \(b = 4\).
    Using this we find the answer: \[x^2 - 16 = \begin{pmatrix} x - 4 \end{pmatrix} \begin{pmatrix} x + 4 \end{pmatrix} \]

  2. We factorize \(4p^2 - 25\) as follows:

    This can be written as the difference of two squares by writing: \[4p^2 - 25 = \begin{pmatrix} 2p \end{pmatrix}^2 - 5^2\] So we can use our formula \(a^2 - b^2 = \begin{pmatrix}a - b \end{pmatrix}\begin{pmatrix} a + b \end{pmatrix}\) with \(a = 2p\) and \(b = 5\).
    Using this we find the answer: \[4p^2 - 25 = \begin{pmatrix} 2p - 5 \end{pmatrix} \begin{pmatrix} 2p + 5 \end{pmatrix} \]

  3. We factorize \(x^2-9y^2\) as follows:

    This can be written as the difference of two squares by writing: \[x^2-9y^2 = x^2 - \begin{pmatrix} 3y \end{pmatrix}^2\] So we can use our formula \(a^2 - b^2 = \begin{pmatrix}a - b \end{pmatrix}\begin{pmatrix} a + b \end{pmatrix}\) with \(a = x\) and \(b = 3y\).
    Using this we find the answer: \[\begin{aligned} x^2-9y^2 & = x^2 - \begin{pmatrix} 3y \end{pmatrix}^2 \\ & = \begin{pmatrix} x - 3y \end{pmatrix}\begin{pmatrix} x + 3y \end{pmatrix} \end{aligned}\]

  4. We factorize \(100m^2-49n^2\) as follows:

    This can be written as the difference of two squares by writing: \[100m^2-49n^2 = \begin{pmatrix}10m \end{pmatrix}^2 - \begin{pmatrix} 7n \end{pmatrix}^2\] So we can use our formula \(a^2 - b^2 = \begin{pmatrix}a - b \end{pmatrix}\begin{pmatrix} a + b \end{pmatrix}\) with \(a = 10m\) and \(b = 7n\).
    Using this we find the answer: \[\begin{aligned} 100m^2-49n^2 &= \begin{pmatrix} 10m \end{pmatrix}^2 - \begin{pmatrix} 7n \end{pmatrix}^2 \\ & = \begin{pmatrix} 10m - 7n \end{pmatrix}\begin{pmatrix} 10m + 7n \end{pmatrix} \end{aligned}\]

  5. We factorize \(25p^2-16q^2\) as follows:

    This can be written as the difference of two squares by writing: \[25p^2 -16q^2= \begin{pmatrix} 5p \end{pmatrix}^2 - \begin{pmatrix} 4q \end{pmatrix}^2\] So we can use our formula \(a^2 - b^2 = \begin{pmatrix}a - b \end{pmatrix}\begin{pmatrix} a + b \end{pmatrix}\) with \(a = 5p\) and \(b = 4q\).
    Using this we find the answer: \[\begin{aligned} 25p^2-16q^2 &= \begin{pmatrix} 5p \end{pmatrix}^2 - \begin{pmatrix} 4q \end{pmatrix}^2 \\ & = \begin{pmatrix} 5p - 4q \end{pmatrix}\begin{pmatrix} 5p + 4q \end{pmatrix} \end{aligned}\]

Solving \(x^2 - k = 0\)

Using the method we can solve any quadratic equation of the type: \[x^2 - k = 0\] For instance, we can solve \(x^2 - 16 = 0\), or eve

Null Factor Law

The null factor law states:

If a product, \(a\times b = 0\), then either \(a=0\), or \(b=0\), or they both equal \(0\).
Consequently, if we're faced with a product of the type: \[\begin{pmatrix} x - a\end{pmatrix}\begin{pmatrix} x + a \end{pmatrix} = 0\] Then: \[x - a = 0\]
or
\[x + a = 0\] \[x-a = 0 \quad \]

Two-Step Method for Solving \(x^2 - k = 0\)

Given a quadratic equation of the type \(x^2 - k = 0\), we can solve this in two steps:

  • Step 1: factor the quadratic using the difference of two squares formula; by the end of this step we should be dealing with an equation looking like \(\begin{pmatrix} x - a\end{pmatrix}\begin{pmatrix} x + a \end{pmatrix} = 0\).

  • Step 2: solve the quadratic equation using the null factor law; we'll find \(x=a\) or \(x = -a\).

Tutorial

In the following tutorial we review the ... .

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. The solutions to \(x^2 - 4=0\) are:

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

  2. The solutions to \(x^2 - 49 = 0\) are:

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

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

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

  4. The solutions to \(4x^2 - 9 = 0\) are:

    \(x = - \frac{3}{2}\) 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 \(x = - \sqrt{5}\).

Generic Title 2

Tutorial

In the following tutorial we learn what a factor is as well as see how to find a number's factors.

Exercise 3

Solution Without Working

Solution With Working

Exercise 4

Solution Without Working

Solution With Working