Factoring Quadratic Equations Without a Linear Term

We can use what we have learned in the previous lesson to write quadratic equations of a special form into factored form.

Particularly, we can rewrite:

$x^2-c^2$ into $(x+c)(x-c)$.

Let's look at an example.

Example 1

Write $v^2-36$ in factored form.

We know that the form we are looking for is $(v-$ __ $)(v+$ __ $)$.

Since we know the square root of $v^2$ is $v$ and the square root of 36 is 6, the factored form is $(v-6)(v+6)$.

We can multiply to check our work.

$(v-6)(v+6)$

$$\begin{gathered} \begin{array}{c}=v^2+6v-6v-36 \\ =v^2-36✓\hfill \end{array} \end{gathered}$$

The process is similar when there is a coefficient on the squared term.

$a^2{x}^2-c^2$ factors into $(ax+c)(ax-c)$

Let's look at another example to understand the process.

Example 2

Write $16m^2-81$ in factored form.

The form we are looking for now is ( __ − __ )( __ + __ ).

Taking the square root of the first term gives us $4m$.

So, we have $(4m-$ __ $)(4m+$ __ $)$.

Now, we are in the same place as in the previous example. The square root of 81 is 9.

So, the factored form is $(4m-9)(4m+9)$.

Again, we can multiply to check.

$(4m-9)(4m+9)$

$$\begin{gathered} \begin{array}{ccc}\hfill & \hfill \hfill & =4m^2+36m-36m-81 \\ =4m^2-81✓\hfill \end{array} \end{gathered}$$

Example 3

As we see in the activity, where the variable is located is not important as long as the quadratic binomial is still the difference of two squares.

Write $64-9b^2$ in factored form.

We use the same concept to find the factored form.

$64-9b^2=(8-3b)(8+3b)$

We can multiply to check.

$(8-3b)(8+3b)$

$$\begin{gathered} \begin{array}{ccc}\hfill & \hfill \hfill & =64+24b-24b-9b^2 \\ =64-9b^2✓\hfill \end{array} \end{gathered}$$

Remember that any expression of the form $x^2+c$ or $a{x}^2+c$ does not have a factored form!

• For $x^2+c$, the $c$ is not squared, and the terms are not being subtracted.
• For $a{x}^2+c$, neither the $a$ nor the $c$ is a perfect square, and the terms are not being subtracted.