Lynn Marecek; MaryAnne Anthony-Smith; Andrea Honeycutt Mathis; Jay Abramson; Sharon North; Amy Baldwin; Alyssa Howell

Activity

In questions 1 - 2, each row of the table contains a pair of equivalent expressions.

1.

Convert the factored form expression to the standard form.

If you get stuck, consider drawing a diagram.

Factored form	Standard form
$(x - 10) (x + 10)$
$(2 x + 1) (2 x - 1)$
$(4 - x) (4 + x)$
$(c + \frac{2}{5}) (c - \frac{2}{5})$
$(x + 5) (x + 5)$

Compare your answers:

Factored form	Standard form
$(x - 10) (x + 10)$	$x^{2} - 100$
$(2 x + 1) (2 x - 1)$	$4 x^{2} - 1$
$(4 - x) (4 + x)$	$16 - x^{2}$
$(c + \frac{2}{5}) (c - \frac{2}{5})$	$c^{2} - \frac{4}{25}$
$(x + 5) (x + 5)$	$x^{2} + 10 x + 25$

2.

Convert the standard form expression to the factored form.

If you get stuck, consider drawing a diagram. (HINT: One of the expressions does not convert.)

Factored form	Standard form
	$x^{2} - 81$
	$49 - y^{2}$
	$9 z^{2} - 16$
	$25 t^{2} - 81$
	$\frac{49}{16} - d^{2}$
	$x^{2} - 6$
	$x^{2} + 100$

Compare your answers:

Factored form	Standard form
$(x - 9) (x + 9)$	$x^{2} - 81$
$(7 - y) (7 + y)$	$49 - y^{2}$
$(3 z + 4) (3 z - 4)$	$9 z^{2} - 16$
$(5 t + 9) (5 t - 9)$	$25 t^{2} - 81$
$(\frac{7}{4} - d) (\frac{7}{4} + d)$	$\frac{49}{16} - d^{2}$
$(x - \sqrt{6}) (x + \sqrt{6})$	$x^{2} - 6$
prime	$x^{2} + 100$

Video: Factoring Quadratic Equations Without a Linear Term

Watch the following video to learn more about factoring a quadratic equation that is written in standard form and has no linear term.

Access multimedia content

Self Check

Find the factored form of the expression

16p^2 − 49

.

$p(16p - 49)$
$(4p - 7)(4p - 7)$
$(4p + 7)(4p + 7)$
$(4p + 7)(4p - 7)$

Additional Resources

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.

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

$\begin{matrix} = v^{2} + 6 v - 6 v - 36 = v^{2} - 36 ✓ \end{matrix}$

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

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

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

Example 2

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

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

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

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

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

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

Again, we can multiply to check.

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

$\begin{matrix} = 4 m^{2} + 36 m - 36 m - 81 = 4 m^{2} - 81 ✓ \end{matrix}$

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 $64 - 9 b^{2}$ in factored form.

We use the same concept to find the factored form.

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

We can multiply to check.

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

$\begin{matrix} = 64 + 24 b - 24 b - 9 b^{2} = 64 - 9 b^{2} ✓ \end{matrix}$

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.

Try it

Try It: Factoring Quadratic Equations Without a Linear Term

1.

Write $x^{2} - 4$ in factored form.

Here is how to write the expressions in factored form.

$(x - 2) (x + 2)$

2.

Write $100 x^{2} - 121$ in factored form.