If a product is zero
Tonight's idea
Two numbers multiply to 6. What are they? You cannot tell: \(6 \times 1\), \(3 \times 2\), \(4 \times 1.5\), \((-12) \times (-0.5)\) and endlessly many more. Now two numbers multiply to 0. This time you know something for certain.
If \(a \times b = 0\), then \(a = 0\) or \(b = 0\) (or both).
Why? Suppose \(a\) is not 0. Then you may divide both sides by \(a\), which leaves \(b = 0 \div a = 0\). So if \(a\) is not zero, \(b\) has to be.
This is the key to a quadratic equation, one containing \(x^2\). Take \(x^2 - 7x + 12 = 0\). The sum and product hunt factors the left side:
\[(x - 3)(x - 4) = 0.\]
A product is zero, so \(x - 3 = 0\) or \(x - 4 = 0\). That gives \(x = 3\) or \(x = 4\). One hard equation has become two easy ones, and both answers are correct.
The zero must stand alone on one side first. \((x - 3)(x - 4) = 6\) tells you nothing about either bracket.
Watch one: \(x^2 + 3x = 10\)
Get zero on one side: \(x^2 + 3x - 10 = 0\).
Factor. Product \(-10\), sum \(3\): the numbers are 5 and \(-2\). So \((x + 5)(x - 2) = 0\).
Set each bracket to zero: \(x + 5 = 0\) gives \(x = -5\), and \(x - 2 = 0\) gives \(x = 2\).
Check both in the original equation. \((-5)^2 + 3(-5) = 25 - 15 = 10\). \(2^2 + 3(2) = 4 + 6 = 10\).
See the left side as a rectangle of tiles before you solve: mathdetours.com/demos/factortiles.html.
Your turn
- Solve: (a) \((x - 4)(x + 9) = 0\); (b) \(x(x - 6) = 0\); (c) \((2x - 1)(x + 3) = 0\).
- Solve by factoring: (a) \(x^2 + 8x + 15 = 0\); (b) \(x^2 - 3x - 28 = 0\); (c) \(x^2 = 5x\).
- A rectangle is 3 cm longer than it is wide, and its area is 40 cm². Call the width \(x\). Write an equation, solve it, and give the rectangle's measurements. One solution makes no sense here. Say which and why.
- Sam solves \((x - 1)(x - 4) = 10\) by writing "\(x - 1 = 10\) or \(x - 4 = 10\), so \(x = 11\) or \(x = 14\)". (a) Check Sam's answers. (b) What went wrong? (c) Solve the equation properly.
Stretch: three factors
Solve \(x^3 = 9x\). A classmate divides both sides by \(x\) and gets \(x^2 = 9\). Which solution did that throw away, and why was dividing not allowed? Find all three solutions by moving everything to one side and factoring fully.
Answers
- (a) \(x = 4\) or \(x = -9\). (b) \(x = 0\) or \(x = 6\). (c) \(2x - 1 = 0\) gives \(x = \tfrac12\); or \(x = -3\).
- (a) \((x + 3)(x + 5) = 0\): \(x = -3\) or \(-5\). (b) \((x - 7)(x + 4) = 0\): \(x = 7\) or \(-4\). (c) \(x^2 - 5x = 0\), so \(x(x - 5) = 0\): \(x = 0\) or \(5\). Dividing by \(x\) would lose \(x = 0\).
- \(x(x + 3) = 40\), so \(x^2 + 3x - 40 = 0\) and \((x + 8)(x - 5) = 0\). \(x = -8\) or \(x = 5\). A width cannot be negative, so the rectangle is 5 cm by 8 cm.
- (a) \(x = 11\): \(10 \times 7 = 70\). \(x = 14\): \(13 \times 10 = 130\). Neither is 10. (b) The rule only works for zero. Two numbers can multiply to 10 without either being 10. (c) Expand: \(x^2 - 5x + 4 = 10\), so \(x^2 - 5x - 6 = 0\) and \((x - 6)(x + 1) = 0\). \(x = 6\) or \(x = -1\). Check: \(5 \times 2 = 10\) and \((-2) \times (-5) = 10\).
Stretch. Dividing by \(x\) is only allowed when \(x\) is not 0, and \(x = 0\) is a solution: \(0^3 = 9 \times 0\). Instead, \(x^3 - 9x = 0\), so \(x(x^2 - 9) = 0\), and by the difference of two squares \(x(x - 3)(x + 3) = 0\). A product of three factors is zero only if one of them is: \(x = 0\), \(3\) or \(-3\).