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
\[x^2  k^2 = \begin{pmatrix}x2\end{pmatrix}\begin{pmatrix} x + 2 \end{pmatrix}\]
and
\[ax^2  k^2 = \begin{pmatrix}\sqrt{a}xk\end{pmatrix}\begin{pmatrix} \sqrt{a}x + k \end{pmatrix}\]
Tutorial
In the following tutorial we review the ... .
Exercise 1
Write each of the following quadratics in factored form:

\(x^216\)

\(4p^2  25\)

\(x^29y^2\)

\(100m^249n^2\)

\(25p^2  16q^2\)
Answers Without Working

\(x^216 = \begin{pmatrix} x  4 \end{pmatrix}\begin{pmatrix} x + 4 \end{pmatrix}\)

\(4p^2  25 = \begin{pmatrix} 2p  5\end{pmatrix}\begin{pmatrix} 2p + 5 \end{pmatrix}\)

\(x^29y^2 = \begin{pmatrix} x  3y\end{pmatrix} \begin{pmatrix} x + 3y \end{pmatrix}\)

\(100m^249n^2 = \begin{pmatrix} 10m  7n \end{pmatrix} \begin{pmatrix} 10m + 7n \end{pmatrix}\)

\(25p^2  16q^2 = \begin{pmatrix} 5p  4q \end{pmatrix} \begin{pmatrix} 5p + 4a \end{pmatrix}\)
Solution With Working

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:
\[x^2  16 = \begin{pmatrix} x  4 \end{pmatrix} \begin{pmatrix} x + 4 \end{pmatrix} \]

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:
\[4p^2  25 = \begin{pmatrix} 2p  5 \end{pmatrix} \begin{pmatrix} 2p + 5 \end{pmatrix} \]

We factorize \(x^29y^2\) as follows:
This can be written as the difference of two squares by writing:
\[x^29y^2 = x^2  \begin{pmatrix} 3y \end{pmatrix}^2\]
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:
\[\begin{aligned} x^29y^2 & = x^2  \begin{pmatrix} 3y \end{pmatrix}^2 \\
& = \begin{pmatrix} x  3y \end{pmatrix}\begin{pmatrix} x + 3y \end{pmatrix}
\end{aligned}\]

We factorize \(100m^249n^2\) as follows:
This can be written as the difference of two squares by writing:
\[100m^249n^2 = \begin{pmatrix}10m \end{pmatrix}^2  \begin{pmatrix} 7n \end{pmatrix}^2\]
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:
\[\begin{aligned}
100m^249n^2 &= \begin{pmatrix} 10m \end{pmatrix}^2  \begin{pmatrix} 7n \end{pmatrix}^2 \\
& = \begin{pmatrix} 10m  7n \end{pmatrix}\begin{pmatrix} 10m + 7n \end{pmatrix}
\end{aligned}\]

We factorize \(25p^216q^2\) 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:
\[\begin{aligned}
25p^216q^2 &= \begin{pmatrix} 5p \end{pmatrix}^2  \begin{pmatrix} 4q \end{pmatrix}^2 \\
& = \begin{pmatrix} 5p  4q \end{pmatrix}\begin{pmatrix} 5p + 4q \end{pmatrix}
\end{aligned}\]
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
\[x + a = 0\]
\[xa = 0 \quad \]
TwoStep 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:

\(x^2  4 = 0\)

\(x^2  49 = 0\)

\(2x^2  50 = 0\)

\(4x^2  9 = 0\)

\(16x^2  81 = 0\)
Solution Without Working

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

The solutions to \(x^2  49 = 0\) are:
\(x =  7\) and \(x = 7\)

The solutions to \(2x^2  50 = 0\) are:
\(x = 5\) and \(x = 5\)

The solutions to \(4x^2  9 = 0\) are:
\(x =  \frac{3}{2}\) and \(x = \frac{3}{2}\)

The solutions to \(16x^2  81 = 0\) are:
\(x =  \frac{9}{4}\) and \(x =  \frac{9}{4}\)
Solution With Working

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 \(x = 2\).

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 \(x = 7\).

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 \(x = 5\).

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

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

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 \(x =  \sqrt{5}\).
Tutorial
In the following tutorial we learn what a factor is as well as see how to find a number's factors.