Math101 · Independent Ontario learning library
Math101learn.math101.caDifference of Squares
A difference of squares is a binomial $a^2-b^2$ that factors as $(a-b)(a+b)$. The identity follows because the middle terms cancel during expansion.
This pattern makes factoring fast, explains conjugates, and supports equation solving, rational-expression simplification, and algebraic identities.
Intuition and core definition
A difference of squares is a binomial $a^2-b^2$ that factors as $(a-b)(a+b)$. The identity follows because the middle terms cancel during expansion. Both terms must be perfect squares and the operation between them must be subtraction; a sum of squares does not factor this way over the real numbers.
Notation, language, and conditions
$a$ and $b$ may themselves be monomials or expressions. For example, $9x^4-25=(3x^2)^2-5^2$. Factoring should continue: $x^4-16=(x^2-4)(x^2+4)=(x-2)(x+2)(x^2+4)$ over the reals.
Why this idea matters
A difference of squares factors through conjugate binomials because the opposite middle terms cancel during multiplication.
A dependable method
- Remove any greatest common factor first.
- Verify that exactly two terms remain with a subtraction sign.
- Write each term as a square, identifying $a$ and $b$.
- Form conjugate factors $(a-b)(a+b)$.
- Factor again if a factor is another difference of squares, then expand to check.
Worked example
Representations and interpretation
An area dissection interprets $a^2-b^2$ as a large square with a smaller square removed; rearranging the remaining pieces forms a rectangle with side lengths $a-b$ and $a+b$.
Reasoning about variations
The expression $a^2+b^2$ has no real linear factorization of this form, while $a^2+b^2=(a-bi)(a+bi)$ over complex numbers. The number system is therefore part of the factorization conditions.
Common mistakes
How to check your work
- Expand the conjugate factors and verify middle-term cancellation.
- Square each identified base and recover the original two terms.
- Check that the final factors share no overlooked common factor.
Practice
- Factor $49y^2-81$.
- Factor $3x^2-75$ completely.
- Can $x^2+16$ be factored by the real difference-of-squares pattern?
Answers and brief solutions
Show answers
- $(7y-9)(7y+9)$ $49y^2=(7y)^2$ and $81=9^2$.
- $3(x-5)(x+5)$ Remove $3$, then factor $x^2-25$.
- No The terms are added, not subtracted.
Synthesis and transfer
Recognizing a geometric area difference as $a^2-b^2$ decomposes a square frame into dimensions $a-b$ and $a+b$, providing a visual verification of the factors.
A square patio of side $a$ with a centred square opening of side $b$ has remaining area $a^2-b^2$. Rearrangement into a rectangle with sides $a-b$ and $a+b$ makes the factorization physical. Multiplying the conjugates confirms that the middle terms $ab$ and $-ab$ cancel. The pattern does not apply to $a^2+b^2$ over the real numbers, and each term must truly be a square before it is invoked. Repeated use can continue factoring, as in $x^4-16=(x^2-4)(x^2+4)=(x-2)(x+2)(x^2+4)$ over the reals.
Related topics
Teaching and accessibility note
Try it yourself
Hints are part of learning. Open one whenever it makes the next step feel possible.
Factor $49y^2-81$.
- $49y^2=(7y)^2$ and $81=9^2$.
End of lesson
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