# 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:

$8 \times 12 = 10^2 - 2^2$

$18 \times 22 = 20^2 - 2^2$

$28 \times 32 = 30^2 - 2^2$

$38 \times 42 = 40^2 -2^2$.

$(10-2) \times (10+2) = 10^2 - 2^2$

$(20-2) \times (20+2) = 20^2 - 2^2$

$(30-2) \times (30+2) = 30^2 - 2^2$

$(40-2) \times (40+2) = 40^2 -2^2$,

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

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