Factoring Using Difference of Squares

When you learned how to multiply binomials we talked about two special products.

\(\begin{align*} \text{The sum and difference formula:} \quad (a + b)(a - b) & = a^2 - b^2\\ \text{The square of a binomial formulas:} \qquad \quad \ \ (a + b)^2 & = a^2 + 2ab + b^2\\ (a - b)^2 & = a^2 - 2ab + b^2\end{align*}\)

In this section we'll learn how to recognize and factor these special products.

Source: cK-12, https://www.ck12.org/algebra/factor-difference-of-squares/lesson/Factorization-using-Difference-of-Squares-ALG-I/
This work is licensed under a Creative Commons Attribution-NonCommercial 3.0 License.

We use the sum and difference formula to factor a difference of two squares. A difference of two squares is any quadratic polynomial in the form \(\begin{align*}a^2 - b^2\end{align*}\), where \(\begin{align*}a\end{align*}\) and \(\begin{align*}b\end{align*}\) can be variables, constants, or just about anything else. The factors of \(\begin{align*}a^2 - b^2\end{align*}\) are always \(\begin{align*}(a + b)(a - b)\end{align*}\); the key is figuring out what the \(\begin{align*}a\end{align*}\) and \(\begin{align*}b\end{align*}\) terms are.

1. Factor the difference of squares:

a) \(\begin{align*}x^2 - 9\end{align*}\)

Rewrite \(\begin{align*}x^2 - 9\end{align*}\) as \(\begin{align*}x^2 - 3^2\end{align*}\). Now it is obvious that it is a difference of squares.

The difference of squares formula is:

\(\begin{align*} a^2 - b^2 = (a + b)(a - b)\end{align*}\)

Let's see how our problem matches with the formula:

\(\begin{align*}x^2 - 3^2 = (x + 3)(x - 3)\end{align*}\)

The answer is:

\(\begin{align*}x^2 - 9 = (x + 3)(x - 3)\end{align*}\)

We can check to see if this is correct by multiplying \(\begin{align*}(x + 3)(x - 3)\end{align*}\):

\(\begin{align*}& \quad \quad \ \ x + 3\\ & \underline{\;\;\;\;\;\;\;\;\;x - 3}\\ & \quad -3x - 9\\ & \underline{x^2 + 3x\;\;\;\;\;\;}\\ & x^2 + 0x - 9\end{align*}\)

The answer checks out.

Note: We could factor this polynomial without recognizing it as a difference of squares. With the methods we learned in the last section we know that a quadratic polynomial factors into the product of two binomials:

\(\begin{align*}(x\;\;\;\;)(x\;\;\;\;)\end{align*}\)

We need to find two numbers that multiply to -9 and add to 0 (since there is no \(\begin{align*}x-\end{align*}\)term, that's the same as if the \(\begin{align*}x-\end{align*}\)term had a coefficient of 0). We can write -9 as the following products:

\(\begin{align*}& -9 = -1 \cdot 9 && \text{and} && -1 + 9 = 8\\ & -9 = 1 \cdot (-9) && \text{and} && 1 + (-9) = -8\\ & -9 = 3 \cdot (-3) && \text{and} && 3 + (-3) = 0 \qquad These \ are \ the \ correct \ numbers.\end{align*}\)

We can factor \(\begin{align*}x^2 - 9\end{align*}\) as \(\begin{align*}(x + 3)(x - 3)\end{align*}\), which is the same answer as before. You can always factor using the methods you learned in the previous section, but recognizing special products helps you factor them faster.

b) \(\begin{align*}x^2 - 100\end{align*}\)

Rewrite \(\begin{align*}x^2 - 100\end{align*}\) as \(\begin{align*}x^2 - 10^2\end{align*}\). This factors as \(\begin{align*}(x + 10)(x - 10)\end{align*}\).

c) \(\begin{align*}x^2 - 1\end{align*}\)

Rewrite \(\begin{align*}x^2 - 1\end{align*}\) as \(\begin{align*}x^2 - 1^2\end{align*}\). This factors as \(\begin{align*}(x + 1)(x - 1)\end{align*}\).

2. Factor the difference of squares:

a) \(\begin{align*}16x^2 - 25\end{align*}\)

Rewrite \(\begin{align*}16x^2 - 25\end{align*}\) as \(\begin{align*}(4x)^2 - 5^2\end{align*}\). This factors as \(\begin{align*}(4x + 5)(4x - 5)\end{align*}\).

b) \(\begin{align*}4x^2 - 81\end{align*}\)

Rewrite \(\begin{align*}4x^2 - 81\end{align*}\) as \(\begin{align*}(2x)^2 - 9^2\end{align*}\). This factors as \(\begin{align*}(2x + 9)(2x - 9)\end{align*}\).

c) \(\begin{align*}49x^2 - 64\end{align*}\)

Rewrite \(\begin{align*}49x^2 - 64\end{align*}\) as \(\begin{align*}(7x)^2 - 8^2\end{align*}\). This factors as \(\begin{align*}(7x + 8)(7x - 8)\end{align*}\).

3. Factor the difference of squares:

a) \(\begin{align*}x^2 - y^2\end{align*}\)

\(\begin{align*}x^2 - y^2\end{align*}\) factors as \(\begin{align*}(x + y)(x - y)\end{align*}\).

b) \(\begin{align*}9x^2 - 4y^2\end{align*}\)

Rewrite \(\begin{align*}9x^2 - 4y^2\end{align*}\) as \(\begin{align*}(3x)^2 - (2y)^2\end{align*}\). This factors as \(\begin{align*}(3x + 2y)(3x - 2y)\end{align*}\).

c) \(\begin{align*} x^2 y^2 - 1\end{align*}\)

Rewrite \(\begin{align*} x^2 y^2 - 1\end{align*}\) as \(\begin{align*}(xy)^2 - 1^2\end{align*}\). This factors as \(\begin{align*}(xy + 1)(xy - 1)\end{align*}\).

Factor the difference of squares:

Factor the following differences of squares.

1. \(x^2 - 4\)

2. \(x^2 - 36\)

3. \(-x^2 + 100\)

4. \(x^2 -400\)

5. \(9x^2 - 4\)

1. \((x-2)(x+2)\)

2. \((x-6)(x+6)\)

3. \(-(x-10)(x+10)\)

4. \((x+20)(x-20)\)

5. \((3 x+2)(3 x-2)\)