Difference of Two Squares (Part 1)

In Algebra I, we drill into student’s heads the formula for the difference of two squares: $x^2 - y^2 = (x-y)(x+y)$

While this formula can be confirmed by just multiplying out the right-hand side, innovative teachers can try to get students to do some exploration to guess the formula for themselves. For example, teachers can use some cleverly chosen multiplication problems: $9 \times 11 = 99$ $19 \times 21 = 399$ $29 \times 31 = 899$ $39 \times 41 = 1599$

Students should be able to recognize the pattern (perhaps with a little prompting): $9 \times 11 = 99 = 100 - 1$ $19 \times 21 = 399 = 400 - 1$ $29 \times 31 = 899 = 900 - 1$ $39 \times 41 = 1599 = 1600 - 1$

Students should hopefully recognize the perfect squares: $9 \times 11 = 99 = 10^2 - 1$ $19 \times 21 = 399 = 20^2 - 1$ $29 \times 31 = 899 = 30^2 - 1$ $39 \times 41 = 1599 = 40^2 - 1$,

so that they can guess the answer to something like $59 \times 61$ without pulling out their calculators. Continuing the exploration, students can use a calculator to find $8 \times 12 = 96$ $18 \times 22 = 396$ $28 \times 32 = 896$ $38 \times 42 = 1596$

Students should be able to recognize the pattern: $8 \times 12 = 10^2 - 4$ $18 \times 22 = 20^2 - 4$ $28 \times 32 = 30^2 - 4$ $38 \times 42 = 40^2 -4$,

and perhaps they can even see the next step: $8 \times 12 = 10^2 - 2^2$ $18 \times 22 = 20^2 - 2^2$ $28 \times 32 = 30^2 - 2^2$ $38 \times 42 = 40^2 -2^2$.

From this point, it’s a straightforward jump to $(10-2) \times (10+2) = 10^2 - 2^2$ $(20-2) \times (20+2) = 20^2 - 2^2$ $(30-2) \times (30+2) = 30^2 - 2^2$ $(40-2) \times (40+2) = 40^2 -2^2$,

leading students to guess that $(x-y)(x+y) = x^2 -y^2$.