Algebraic Identities

Expansions worth knowing by sight

An identity is an equation that is true for every value of its letters. You have already met some without the name: 3(x + 2) = 3x + 6 is true whatever x is. A few expansions come up so often that it pays to know them by sight, so that you can write them down, or spot them, without multiplying out four products every time.

The first is the square of a sum. ${\left(a+b\right)}^2$(a + b)² means (a + b)(a + b), a square whose side is a + b.

Cut each side of the square into a and b. The square holds an a by a square, a b by b square, and two a by b rectangles.

Where the 2ab comes from

The square of side a + b is made of four pieces: $a^2$a² in one corner, $b^2$b² in the opposite corner, and two rectangles, each a by b. So ${\left(a+b\right)}^2=a^2+2ab+b^2$(a + b)² = a² + 2ab + b².

The middle term is 2ab because there are two rectangles, one on each side of the diagonal. Writing ${\left(a+b\right)}^2=a^2+b^2$(a + b)² = a² + b² leaves both of them out, and that is the most common mistake with this identity. For example, ${\left(x+3\right)}^2=x^2+2\times 3\times x+3^2=x^2+6x+9$(x + 3)² = x² + 2 × 3 × x + 3² = x² + 6x + 9, not $x^2+9$x² + 9.

The square of a difference

Expand ${\left(a-b\right)}^2$(a − b)² as two brackets: $\left(a-b\right)\left(a-b\right)=a^2-ab-ab+b^2=a^2-2ab+b^2$(a − b)(a − b) = a² − ab − ab + b² = a² − 2ab + b². The middle term is now −2ab, and the last term is $+b^2$+b², because $-b\times -b=+b^2$−b × −b = +b².

So ${\left(x-5\right)}^2=x^2-10x+25$(x − 5)² = x² − 10x + 25. Compare it with ${\left(x+5\right)}^2=x^2+10x+25$(x + 5)² = x² + 10x + 25: only the sign of the middle term changes.

Squaring numbers in your head

The identities work for numbers too. To square 31, write it as 30 + 1: $31^2=30^2+2\times 30\times 1+1^2=900+60+1=961$31² = 30² + 2 × 30 × 1 + 1² = 900 + 60 + 1 = 961. To square 29, write it as 30 − 1: $29^2=900-60+1=841$29² = 900 − 60 + 1 = 841.

The difference of two squares

The third identity is a difference of two squares: $a^2-b^2=\left(a-b\right)\left(a+b\right)$a² − b² = (a − b)(a + b). Expanding the right side shows it: $\left(a-b\right)\left(a+b\right)=a^2+ab-ab-b^2=a^2-b^2$(a − b)(a + b) = a² + ab − ab − b² = a² − b², because the two middle terms cancel.

A picture shows why the product must come out that way. Start with a square of side a and cut a smaller square of side b out of one corner. What is left has an area of $a^2-b^2$a² − b². Move one strip of what is left and it becomes a rectangle whose sides are a + b and a − b.

A 10 by 10 square with a d by d corner cut out. The bottom strip stands on its end beside the top part, and the two pieces make a rectangle 10 + d long and 10 − d wide. Nothing is added or lost, so $\left(10+d\right)\left(10-d\right)=100-d^2$(10 + d)(10 − d) = 100 − d².

The same cut on a square of side x. Take away the b by b corner, and each of the three pieces left has one side of length x − b.

Reading it both ways

In the square of side x, the three pieces left are ${\left(x-b\right)}^2$(x − b)², b(x − b) and b(x − b). Each has a side of x − b, so together they make one rectangle that is x − b wide and (x − b) + b + b = x + b long. So $x^2-b^2=\left(x-b\right)\left(x+b\right)$x² − b² = (x − b)(x + b).

Read from left to right, the identity factors: $x^2-49=x^2-7^2=\left(x-7\right)\left(x+7\right)$x² − 49 = x² − 7² = (x − 7)(x + 7). Read from right to left, it multiplies: $97\times 103=\left(100-3\right)\left(100+3\right)=100^2-3^2=10000-9=9991$97 × 103 = (100 − 3)(100 + 3) = 100² − 3² = 10000 − 9 = 9991.

A difference of two squares is not the square of the difference. ${\left(53-47\right)}^2=6^2=36$(53 − 47)² = 6² = 36, but $53^2-47^2=\left(53-47\right)\left(53+47\right)=6\times 100=600$53² − 47² = (53 − 47)(53 + 47) = 6 × 100 = 600.