Standard Algebraic Identities: Worked Examples

This lesson applies the standard algebraic identities to expand expressions and to evaluate numbers quickly. Each example uses one of the four identities below and is worked out in full.

Square of a sum and a difference

The two square identities are:

$(a+b)^2 = a^2 + 2ab + b^2$

$(a-b)^2 = a^2 - 2ab + b^2$

Expand $(2x+3y)^2$

Here $a = 2x$ and $b = 3y$:

$(2x+3y)^2 = (2x)^2 + 2(2x)(3y) + (3y)^2 = 4x^2 + 12xy + 9y^2$

Expand $(4p-3q)^2$

Here $a = 4p$ and $b = 3q$:

$(4p-3q)^2 = (4p)^2 - 2(4p)(3q) + (3q)^2 = 16p^2 - 24pq + 9q^2$

Expand $(2x-7)^2$

Here $a = 2x$ and $b = 7$:

$(2x-7)^2 = (2x)^2 - 2(2x)(7) + 7^2 = 4x^2 - 28x + 49$

Difference of two squares

The identity is:

$(a+b)(a-b) = a^2 - b^2$

Expand $\left(\tfrac{x}{2}+\tfrac{3y}{4}\right)\left(\tfrac{x}{2}-\tfrac{3y}{4}\right)$

With $a = \tfrac{x}{2}$ and $b = \tfrac{3y}{4}$:

$\left(\tfrac{x}{2}\right)^2 - \left(\tfrac{3y}{4}\right)^2 = \tfrac{x^2}{4} - \tfrac{9y^2}{16}$

Evaluate $(1.1m-0.4)(1.1m+0.4)$

$(1.1m)^2 - (0.4)^2 = 1.21m^2 - 0.16$

Expand $\left(\tfrac{3}{2}p+\tfrac{2}{3}q\right)\left(\tfrac{3}{2}p-\tfrac{2}{3}q\right)$

$\left(\tfrac{3}{2}p\right)^2 - \left(\tfrac{2}{3}q\right)^2 = \tfrac{9}{4}p^2 - \tfrac{4}{9}q^2$

Product of binomials with a common term

The identity is:

$(x+a)(x+b) = x^2 + (a+b)x + ab$

Expand $(x+8)(x+9)$

With $a = 8$ and $b = 9$:

$x^2 + (8+9)x + (8)(9) = x^2 + 17x + 72$

Expand $(x+10)(x-8)$

With $a = 10$ and $b = -8$:

$x^2 + (10-8)x + (10)(-8) = x^2 + 2x - 80$

Expand $\left(\tfrac{1}{2}p+5\right)\left(\tfrac{1}{2}p+6\right)$

Here the common term is $\tfrac{1}{2}p$, with $a = 5$ and $b = 6$:

$\left(\tfrac{1}{2}p\right)^2 + (5+6)\left(\tfrac{1}{2}p\right) + (5)(6) = \tfrac{1}{4}p^2 + \tfrac{11}{2}p + 30$

Expand $(2x-4)(2x+10)$

Here the common term is $2x$, with $a = -4$ and $b = 10$:

$(2x)^2 + (-4+10)(2x) + (-4)(10) = 4x^2 + 12x - 40$

The product of two sums

The identity is:

$(a+b)(c+d) = ac + ad + bc + bd$

Expand $(x+3)(y+2)$

$xy + 2x + 3y + 6$

Expand $(m+5)(n-3)$

$mn - 3m + 5n - 15$

Expand $(p+2q)(r-s)$

$pr - ps + 2qr - 2qs$

Evaluating numbers using identities

Writing each number around a round figure lets the same identities do the arithmetic.

$102^2$

$(100+2)^2 = 100^2 + 2(100)(2) + 2^2 = 10000 + 400 + 4 = 10404$

$99^2$

$(100-1)^2 = 100^2 - 2(100)(1) + 1^2 = 10000 - 200 + 1 = 9801$

$105 \times 95$

$(100+5)(100-5) = 100^2 - 5^2 = 10000 - 25 = 9975$

$205 \times 202$

$(200+5)(200+2) = 200^2 + (5+2)(200) + (5)(2) = 40000 + 1400 + 10 = 41410$

$75 \times 68$

$(70+5)(70-2) = 70^2 + (5-2)(70) + (5)(-2) = 4900 + 210 - 10 = 5100$

$98 \times 97$

$(100-2)(100-3) = 100^2 + (-2-3)(100) + (-2)(-3) = 10000 - 500 + 6 = 9506$

Differences of squares of numbers

$45^2 - 15^2$

$(45+15)(45-15) = 60 \times 30 = 1800$

$7.5^2 - 1.5^2$

$(7.5+1.5)(7.5-1.5) = 9 \times 6 = 54$

Key takeaways

• The square identities give $(a+b)^2 = a^2 + 2ab + b^2$ and $(a-b)^2 = a^2 - 2ab + b^2$.
• The product $(a+b)(a-b)$ collapses to the difference of squares $a^2 - b^2$, and $(x+a)(x+b) = x^2 + (a+b)x + ab$.
• Writing a number as a sum or difference around a round figure turns these identities into fast mental arithmetic.