Expand and factor
Tonight's idea
A rectangle \(a + b\) wide and \(c + d\) tall has area \((a + b)(c + d)\). Cut it along the joins into four smaller rectangles: \(ac\), \(bc\), \(ad\) and \(bd\). The pieces make the whole, so
\((a + b)(c + d) = ac + ad + bc + bd\): each part of the first bracket times each part of the second.
That is expanding, and the picture is why nothing is missed or counted twice. The picture uses positive lengths; the algebraic rule also works with zero and negative numbers. Put \(c = a\) and \(d = b\): the rectangle is a square and \((a + b)^2 = a^2 + 2ab + b^2\).
Can you turn a square with a missing corner into a rectangle? Take \(a > b > 0\). Remove a \(b \times b\) corner from an \(a \times a\) square. Cut the L-shape into pieces P and Q, turn Q a quarter turn, and slide it beside P:
The same area is now a rectangle \(a+b\) long and \(a-b\) tall. So \(a^2-b^2=(a+b)(a-b)\): the difference of two squares. Algebra confirms it, even when \(a\) or \(b\) is zero or negative:
\[(a + b)(a - b) = a^2 - ab + ab - b^2 = a^2 - b^2.\]
The cross terms cancel, and the identity works in both directions.
Factoring is expanding run backwards: given the four pieces added up, find the rectangle. For \(x^2 + 8x + 15\), look for two numbers that multiply to 15 and add to 8. Try the factor pairs: 1 and 15 add to 16, but 3 and 5 add to 8. So \(x^2 + 8x + 15 = (x + 3)(x + 5)\). Expanding the answer is the check, and it never lies.
Watch one: \(51 \times 49\) in your head
The two numbers sit one either side of 50, so they are \(50 + 1\) and \(50 - 1\):
\[51 \times 49 = (50 + 1)(50 - 1) = 50^2 - 1^2 = 2500 - 1 = 2499.\]
Same trick, bigger gap: \(103 \times 97 = 100^2 - 3^2 = 9991\), and \(32 \times 28 = 900 - 4 = 896\). Any two numbers equally spaced around a round number. Now run it backwards. Is 9991 prime? It is \(100^2 - 3^2 = (100 - 3)(100 + 3) = 97 \times 103\). Not prime, and found without trying a single division.
Your turn
- Expand: (a) \((x + 3)(x + 5)\) and (b) \((2x + 1)(x - 4)\). Draw the area model for (a) and label all four pieces.
- Compute in your head, writing down only the identity you used: (a) \(45 \times 35\), (b) \(99^2\) (hint: \((100 - 1)^2\)), (c) \(1001 \times 999\).
- Factor: (a) \(x^2 - 49\), (b) \(4x^2 - 9\), (c) \(x^2 - x - 6\). Expand each answer to check it.
- Take two consecutive whole numbers, \(n\) and \(n + 1\). (a) Show that the difference of their squares is always odd. (b) Show it equals their sum. (c) Hence find \(1000^2 - 999^2\) without squaring anything.
Stretch: a number with 10 digits
Is \(2^{32} - 1 = 4\,294\,967\,295\) prime? Write it as a difference of two squares and factor; repeat on the piece that allows it. Then, using squares of whole numbers (including zero): 1, 3, 4 and 5 are differences of two squares (\(4 = 2^2 - 0^2\)), but 2 and 6 are not. Which whole numbers can never be written as \(a^2 - b^2\) with whole numbers \(a\) and \(b\)?
Answers
- (a) \(x^2 + 5x + 3x + 15 = x^2 + 8x + 15\); the pieces are \(x^2\), \(5x\), \(3x\) and \(15\). (b) \(2x^2 - 8x + x - 4 = 2x^2 - 7x - 4\).
- (a) \(40^2 - 5^2 = 1600 - 25 = 1575\). (b) \(100^2 - 2 \cdot 100 + 1 = 9801\). (c) \(1000^2 - 1 = 999\,999\).
- (a) \((x - 7)(x + 7)\). (b) \((2x - 3)(2x + 3)\), since \(4x^2 = (2x)^2\). (c) Two numbers multiplying to \(-6\) and adding to \(-1\): \(-3\) and \(2\), so \((x - 3)(x + 2)\).
- \((n + 1)^2 - n^2 = n^2 + 2n + 1 - n^2 = 2n + 1\). (a) \(2n + 1\) is odd. (b) \(2n + 1 = n + (n + 1)\), the sum. Or factor: \((n+1)^2 - n^2 = (n + 1 - n)(n + 1 + n) = 1 \times (2n + 1)\). (c) \(1000 + 999 = 1999\).
Stretch. \(2^{32} - 1 = (2^{16})^2 - 1^2 = (2^{16} - 1)(2^{16} + 1)\). Then \(2^{16} - 1 = (2^8 - 1)(2^8 + 1)\) and \(2^8 - 1 = (2^4 - 1)(2^4 + 1) = 15 \times 17\). Altogether \(2^{32} - 1 = 15 \times 17 \times 257 \times 65537 = 3 \times 5 \times 17 \times 257 \times 65537\). Not prime. For the second part, \(a^2 - b^2 = (a - b)(a + b)\), and \(a - b\) and \(a + b\) differ by \(2b\), so they are both odd or both even. Their product is therefore odd, or a multiple of 4. The numbers that are even but not multiples of 4, that is \(2, 6, 10, 14, \ldots\), can never be a difference of two squares. Every other whole number can: \(0 = 0^2 - 0^2\); odd \(2k + 1 = (k+1)^2 - k^2\) for \(k \ge 0\); and positive multiples of 4 satisfy \(4k = (k+1)^2 - (k-1)^2\) for \(k \ge 1\).