Standard mathematical notation is shown below. Enhanced typesetting is not ready; you can keep reading and open the practice answers.
Difference of Two Squares
See why opposite signs cancel the mixed terms, then use the identity in reverse.
Same terms, opposite signs
Two brackets with matching terms and opposite signs have a useful shortcut: their mixed terms cancel.
Why the middle terms disappear
The pair (a+b)(a−b)$(a+b)(a-b)$ is called a conjugate pair. Expand all four products to see the cancellation.
(a+b)(a−b)=a2−ab+ab−b2$(a+b)(a-b)=a^2-ab+ab-b^2$
−ab+ab=0$-ab+ab=0$
(a+b)(a−b)=a2−b2$(a+b)(a-b)=a^2-b^2$
Any real values of a$a$ and b$b$ work, including zero and negative values. Only the two mixed terms cancel; the square terms remain.
Expand a conjugate pair
Expand (x+4)(x−4)$(x+4)(x-4)$
Step 1
Identify the matching terms: x$x$ and 4$4$
Step 2
Square each term, then subtract.
(x+4)(x−4)=x2−42$(x+4)(x-4)=x^2-4^2$
x2−42=x2−16$x^2-4^2=x^2-16$
Check by multiplying all four products.
x2−4x+4x−16=x2−16$x^2-4x+4x-16=x^2-16$
Keep the order and square the whole term
Coefficients and two variables
Expand (2x+3y)(2x−3y)$(2x+3y)(2x-3y)$
Step 1
Treat 2x$2x$ and 3y$3y$ as the two whole terms.
Step 2
Square both the coefficient and the variable.
(2x+3y)(2x−3y)=(2x)2−(3y)2$(2x+3y)(2x-3y)=(2x)^2-(3y)^2$
(2x)2−(3y)2=4x2−9y2$(2x)^2-(3y)^2=4x^2-9y^2$
Check the mixed terms: −6xy+6xy=0$-6xy+6xy=0$
A constant comes first
Expand (5−x)(5+x)$(5-x)(5+x)$
Step 1
The subtraction bracket is 5−x$5-x$, so subtract the square of x$x$ from the square of 5$5$
Step 2
Keep that order.
(5−x)(5+x)=52−x2$(5-x)(5+x)=5^2-x^2$
52−x2=25−x2$5^2-x^2=25-x^2$
Swapping the two whole brackets changes nothing. Reversing the subtraction inside a bracket changes the sign.
(x−5)(x+5)=x2−25$(x-5)(x+5)=x^2-25$
Use the identity, then simplify
Simplify (x+3)(x−3)+2x2$(x+3)(x-3)+2x^2$
Step 1
Expand only the conjugate pair.
(x+3)(x−3)+2x2=x2−9+2x2$(x+3)(x-3)+2x^2=x^2-9+2x^2$
Step 2
Collect like terms.
x2−9+2x2=3x2−9$x^2-9+2x^2=3x^2-9$
Not the same as a squared bracket
(a−b)2=a2−2ab+b2$(a-b)^2=a^2-2ab+b^2$
In (a−b)(a−b)$(a-b)(a-b)$ both mixed terms are −ab$-ab$, giving −2ab$-2ab$ when collected. The conjugate pair has mixed terms −ab$-ab$ and +ab$+ab$, which always add to zero. These are algebraic forms, not claims about their numerical signs; ab$ab$ may be positive, negative or zero.
(x−3)2=x2−6x+9$(x-3)^2=x^2-6x+9$
A missing middle term is correct for conjugate pairs, not for a squared bracket. For 25−x2$25-x^2$, keep the constant first rather than switching the subtraction.
Run the identity backwards
Factorising means writing an expression as a product. Check for a subtraction between two square terms, then use their square roots in a conjugate pair.
a2−b2=(a+b)(a−b)$a^2-b^2=(a+b)(a-b)$
Factorise a difference of squares
Factorise 9x2−16$9x^2-16$
Step 1
Write each term as a square.
9x2−16=(3x)2−42$9x^2-16=(3x)^2-4^2$
Step 2
Write a sum bracket and a difference bracket.
(3x)2−42=(3x+4)(3x−4)$(3x)^2-4^2=(3x+4)(3x-4)$
Check by expanding the brackets back to 9x2−16$9x^2-16$
Check the sign before factorising
A sum of squares is not a difference of squares. For example, x2+9$x^2+9$ does not factorise into real linear factors; (x+3)(x−3)$(x+3)(x-3)$ expands to x2−9$x^2-9$, not x2+9$x^2+9$
Can this identity factorise 4x2+25$4x^2+25$?
No. The terms are added. The conjugate pair would give 4x2−25$4x^2-25$ instead. Over the reals, this polynomial has no linear factorisation.
Optional: take out a common factor first
Factorise fully
Factorise 2x2−18$2x^2-18$ fully.
Step 1
Take out the greatest common factor.
2x2−18=2(x2−9)$2x^2-18=2(x^2-9)$
Step 2
Factorise the difference inside the bracket.
2(x2−9)=2(x+3)(x−3)$2(x^2-9)=2(x+3)(x-3)$
Check: 2(x2−9)=2x2−18$2(x^2-9)=2x^2-18$
Stopping at 2(x2−9)$2(x^2-9)$ leaves a difference of squares unfactorised.
Apply the same idea
Mental calculation
Calculate 49×51$49\times51$ without long multiplication.
Step 1
Use numbers equally far from a convenient centre.
49×51=(50−1)(50+1)$49\times51=(50-1)(50+1)$
Step 2
Subtract the small square from the large square.
(50−1)(50+1)=502−12$(50-1)(50+1)=50^2-1^2$
502−12=2499$50^2-1^2=2499$
Find a missing coefficient
The expansion of (x+ky)(x−ky)$(x+ky)(x-ky)$ is x2−9y2$x^2-9y^2$ — Find the positive value of k$k$
Step 1
The subtracted term is the square of ky$ky$ — The number whose square is 9$9$ is 3$3$ or −3$-3$
Step 2
Choose the positive value, 3$3$, then check.
(x+3y)(x−3y)=x2−9y2$(x+3y)(x-3y)=x^2-9y^2$
Try it: use x2−16y2$x^2-16y^2$ instead. What is the positive coefficient?
The coefficient is 4$4$, not 16$16$ — Square it to check: 42=16$4^2=16$
Interpret an area
A rectangle has side lengths x+2$x+2$ cm and x−2$x-2$ cm, with x>2$x>2$ — Write its area in simplified form.
Step 1
Multiply the side lengths.
(x+2)(x−2)=x2−22$(x+2)(x-2)=x^2-2^2$
Step 2
Simplify the expression.
x2−22=x2−4$x^2-2^2=x^2-4$
The area is x2−4$x^2-4$ square centimetres. It also equals the area left when a square of side 2$2$ cm is removed from a square of side x$x$ cm.
A length model using sides a+b$a+b$ and a−b$a-b$ assumes a>b>0$a>b>0$ — That geometric restriction is not needed for the algebraic identity.
Practice Problems
Write your working before opening an answer. Expand to check each factorisation. Questions 6–8 are optional transfer practice.
Practice 1
Expand (x+6)(x−6)$(x+6)(x-6)$ and write the two mixed terms. Explain why they cancel.
Show answer
Use the matching terms. The mixed terms are −6x$-6x$ and 6x$6x$; they add to zero.
(x+6)(x−6)=x2−62$(x+6)(x-6)=x^2-6^2$
x2−36$x^2-36$
The mixed terms have opposite signs, so there is no middle term.
Practice 2
Expand (7−x)(7+x)$(7-x)(7+x)$
Show answer
Keep the subtraction order.
(7−x)(7+x)=72−x2$(7-x)(7+x)=7^2-x^2$
49−x2$49-x^2$
Start with the square of 7, not the square of x.
Practice 3
Expand (3x+2y)(3x−2y)$(3x+2y)(3x-2y)$
Show answer
Square each whole term.
(3x+2y)(3x−2y)=(3x)2−(2y)2$(3x+2y)(3x-2y)=(3x)^2-(2y)^2$
9x2−4y2$9x^2-4y^2$
Square the coefficients as well as the variables.
Practice 4
Simplify (x+4)(x−4)+16$(x+4)(x-4)+16$
Show answer
Apply the identity before collecting terms.
(x+4)(x−4)+16=x2−16+16$(x+4)(x-4)+16=x^2-16+16$
x2$x^2$
The constant terms cancel after the brackets are expanded.
Practice 5
Factorise 49x2−9$49x^2-9$
Show answer
Identify the square roots.
49x2−9=(7x)2−32$49x^2-9=(7x)^2-3^2$
(7x+3)(7x−3)$(7x+3)(7x-3)$
Use 7x and 3 in the brackets, not 49x and 9.
Practice 6
Optional: factorise 3x2−75$3x^2-75$ fully.
Show answer
Take out the common factor first.
3x2−75=3(x2−25)$3x^2-75=3(x^2-25)$
3(x2−25)=3(x+5)(x−5)$3(x^2-25)=3(x+5)(x-5)$
3(x+5)(x−5)$3(x+5)(x-5)$
The factor 3 stays outside both brackets.
Practice 7
Optional: calculate 98×102$98\times102$ mentally.
Show answer
Use 100 as the centre.
98×102=(100−2)(100+2)$98\times102=(100-2)(100+2)$
(100−2)(100+2)=10000−4$(100-2)(100+2)=10000-4$
9996$9996$
Subtract the square of 2, not 2 itself.
Practice 8
Optional: a rectangle has sides x+5$x+5$ cm and x−5$x-5$ cm, with x>5$x>5$ — Give its simplified area.
Show answer
Multiply the side lengths.
(x+5)(x−5)=x2−52$(x+5)(x-5)=x^2-5^2$
x2−25$x^2-25$
The area is in square centimetres. The restriction keeps both lengths positive.
Interactive Quiz
1. Expand (4x+y)(4x−y)$(4x+y)(4x-y)$
2. Factorise x2−64$x^2-64$
3. Which expression equals (x−3)2$(x-3)^2$?
Review the exit quiz
Question 1: B. Square the whole term 4x$4x$ to get 16x2$16x^2$ — The mixed terms cancel.
Question 2: C. The square root of 64$64$ is 8$8$ — Opposite signs give a difference of squares.
Question 3: A. Two negative mixed terms combine to give −6x$-6x$ — They do not cancel. Explain this contrast aloud before moving on.