Un-multiplying: the sum and product hunt
Tonight's idea
Expanding turns \((x + 4)(x + 3)\) into \(x^2 + 7x + 12\). Factoring runs it backwards: given \(x^2 + 7x + 12\), find the brackets. You met this in grade 8. Tonight is why the method works.
Picture \(x^2 + 7x + 12\) as tiles: one square, \(x\) by \(x\); seven strips, \(x\) by 1; twelve unit squares. Can they fill a rectangle? Put the square in a corner. Stand \(p\) of the strips against its right side and lay the other \(q\) underneath. That leaves a corner gap \(p\) wide and \(q\) tall, and the units must fill it exactly.
The strips say \(p + q = 7\). The corner says \(p \times q = 12\). In symbols: \((x + p)(x + q) = x^2 + (p + q)x + pq\).
So the hunt is for two numbers with a given sum and a given product. Here they are 4 and 3. Start with the product, because it has fewer candidates. Negative numbers are allowed. A positive product means \(p\) and \(q\) have the same sign. A negative product means opposite signs.
Watch one: \(x^2 - 2x - 15\)
The product is \(-15\), which is negative, so one number is positive and the other negative. Ignoring signs, the pairs that multiply to 15 are 1 and 15, or 3 and 5.
The sum must be \(-2\). Only 3 and \(-5\) do that: the larger one takes the minus sign.
\[x^2 - 2x - 15 = (x + 3)(x - 5)\]
Check by expanding: \(x^2 - 5x + 3x - 15 = x^2 - 2x - 15\). Always check. Expanding is quick and it catches sign slips.
Slide \(b\) and \(c\) and watch the tiles try to form a rectangle: mathdetours.com/demos/factortiles.html.
Your turn
- Factor: (a) \(x^2 + 9x + 20\); (b) \(x^2 + 13x + 12\); (c) \(x^2 + 10x + 25\).
- Factor, and expand to check: (a) \(x^2 - 8x + 15\); (b) \(x^2 + 2x - 24\); (c) \(x^2 - x - 30\).
- (a) Draw the tile rectangle for \(x^2 + 5x + 6\) and label its two sides. (b) Explain why the tiles for \(x^2 + 5x + 7\) cannot make a rectangle with whole-number sides.
- \(x^2 + bx + 12\) factors as \((x + p)(x + q)\) with \(p\) and \(q\) whole numbers, positive or negative. Find every possible value of \(b\).
Stretch: the biggest corner
\(x^2 + 10x + c\) factors with \(p\) and \(q\) positive whole numbers. List every possible \(c\). Which is largest, and what shape is the block of unit tiles then? Now write \(p = 5 - d\) and \(q = 5 + d\) and explain why that one must win.
Answers
- (a) \((x + 4)(x + 5)\). (b) \((x + 1)(x + 12)\). (c) \((x + 5)(x + 5) = (x + 5)^2\): the two numbers can be equal.
- (a) Product \(+15\), sum \(-8\): both negative. \((x - 3)(x - 5)\). (b) Product \(-24\), sum \(+2\): 6 and \(-4\). \((x + 6)(x - 4)\). (c) Product \(-30\), sum \(-1\): 5 and \(-6\). \((x + 5)(x - 6)\).
- (a) A rectangle \(x + 2\) by \(x + 3\): the square, 2 strips on one side, 3 on the other, and a 2 by 3 block of six units. (b) The 7 units must form a \(p\) by \(q\) block. Since 7 is prime the block is 1 by 7, which needs \(1 + 7 = 8\) strips. There are only 5.
- Pairs with product 12: 1 and 12, 2 and 6, 3 and 4, and the same pairs with both signs negative. Their sums give \(b = 13, 8, 7, -13, -8, -7\).
Stretch. \(p + q = 10\): the pairs 1, 9; 2, 8; 3, 7; 4, 6; 5, 5 give \(c = 9, 16, 21, 24, 25\). The largest is 25, when the unit block is a 5 by 5 square. In general \(c = (5 - d)(5 + d) = 25 - d^2\), a difference of two squares. That is largest when \(d = 0\).