All courses Algebra I · A-SSE.2 56 of 76
Use expression structure to identify useful rewrites, such as difference-of-squares factoring.

Factor a difference of squares

Problem
Factor \(x^{2}-49\) as a difference of squares. Identify both square roots, apply the conjugate identity, and expand the factors to verify the original expression.
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Hint

Write each term as an explicit square.

a²−b²=(a−b)(a+b). Both terms must be squares and the operation must be subtraction.

Solution walkthrough

01

Recognize both square terms

\[x^2-49=x^2-7^2\]

The first term is the square of x, and 49 is the square of 7. The subtraction between them makes this a difference of squares.

02

Apply the conjugate identity

\[\begin{aligned} a^2-b^2=(a-b)(a+b) \\ a=x,~b=7 \end{aligned}\]

The identity applies because both terms are perfect squares and their operation is subtraction.

03

Write the factorization

\[x^2-7^2=(x-7)(x+7)\]

Substituting the two square roots into the conjugate factors gives one difference and one sum.

04

Expand to verify

\[(x-7)(x+7)=x^2+7x-7x-49=x^2-49\]

The opposite middle terms cancel and the constants multiply to -49, exactly reproducing the original expression.

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Another way

  1. Find two binomial constants whose sum is 0 and product is -49: -7 and 7.

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Common mistake

Do not use two identical binomials. Conjugate signs are essential because their cross terms cancel.

Solution walkthrough video