When a product is zero
Two numbers can multiply to 12 in endless ways: 3 × 4, 2 × 6, 24 × 0.5, and many more. They can multiply to 0 in only one way: one of them must be 0. If neither number is 0, their product is not 0 either. This is the zero product property.
Use it on the equation (x − 2)(x − 3) = 0. The two brackets are two numbers whose product is 0, so one of them is 0: either x − 2 = 0 or x − 3 = 0. Each of these is a one-step equation. x − 2 = 0 gives x = 2, and x − 3 = 0 gives x = 3.
The word "or" matters. Either bracket being 0 is enough to make the product 0, so each bracket gives a solution of its own.
The product (x − 2)(x − 3) is 0 at x = 2, where the first bracket is 0, and at x = 3, where the second bracket is 0. At every other x in the table, neither bracket is 0, and the product is not 0.
Two solutions
So the equation (x − 2)(x − 3) = 0 has two solutions, x = 2 and x = 3. A linear equation has one solution, but a quadratic equation can have two, and here each bracket gives one of them.
Check each one in the equation. When x = 2, the product is 0 × (−1) = 0. When x = 3, it is 1 × 0 = 0.
The two solutions of (x − 2)(x − 3) = 0.
Factor first
Most quadratic equations do not arrive in brackets. To solve , factor the left side first, as in Factoring Quadratics: the two numbers that multiply to 6 and add to 5 are 2 and 3, so the equation is (x + 2)(x + 3) = 0.
Now set each bracket equal to 0. x + 2 = 0 gives x = −2, and x + 3 = 0 gives x = −3. Notice the signs: the bracket x + 2 is zero when x is −2, not 2.
Check x = −2 in the original equation: .
Get zero on one side
The method needs a product equal to 0, so move every term to one side before factoring. Take . Writing it as x(x + 1) = 12 proves nothing, because two numbers can multiply to 12 in endless ways.
Subtract 12 from both sides instead: . The numbers that multiply to −12 and add to 1 are 4 and −3, so (x + 4)(x − 3) = 0, and x = −4 or x = 3. Check both in the original equation: , and .
When x is a common factor
Solve . There is no number term, and both terms have a common factor of x, so the equation is x(x − 5) = 0. The first factor is x itself, so x = 0 or x − 5 = 0, which gives x = 0 or x = 5.
Do not divide both sides by x. Dividing by x leaves x − 5 = 0 and loses the solution x = 0. Factoring keeps both.
The usual mistakes
Taking the numbers in the brackets as the solutions. (x + 2)(x + 3) = 0 has the solutions −2 and −3, because each bracket is zero when x is the opposite of the number in it.
Factoring before the equation equals 0. From x(x + 1) = 12 you cannot say x = 12 or x + 1 = 12, because a product of 12 does not need either factor to be 12.
Giving one solution. A quadratic that factors into two different brackets has two solutions, and the question wants both.
Equations from words
In an application, the quadratic has to be written before it can be solved. Let a letter stand for the unknown, write every other quantity in terms of it, and use the fact the question gives to make an equation.
For example, a number multiplied by 2 more than itself is 15. Let the number be x. Then 2 more than it is x + 2, and the fact gives x(x + 2) = 15. Expand and subtract 15 from both sides: . The numbers 5 and −3 multiply to −15 and add to 2, so (x + 5)(x − 3) = 0, and x = −5 or x = 3.
Both answers fit the equation, but check each one against the question. If x were a length in meters, x = −5 would be rejected, because a length cannot be negative.
Worked example: A Rectangular Patio with a Known Area and a Length 3 m More Than Its Width
Question A rectangular patio is 3 m longer than it is wide, and its area is 40 m2. (a) Find the width and the length of the patio. (b) An edging strip runs all the way round the patio. How long is the strip?
1.Let the width be x m. The length is 3 m more, so it is (x + 3) m. The area is length times width, so x(x + 3) = 40.
The width is x m and the length is (x + 3) m, so the area gives x(x + 3) = 40. 2.Expand the bracket: x2 + 3x = 40. Subtract 40 from both sides, so that one side is zero: x2 + 3x − 40 = 0.
Expand the bracket and subtract 40 from both sides: x2 + 3x − 40 = 0. 3.Factorize. Two numbers with a product of −40 and a sum of 3 are 8 and −5, so (x + 8)(x − 5) = 0.
The numbers 8 and −5 have a product of −40 and a sum of 3: (x + 8)(x − 5) = 0. 4.A product is zero only when one of its factors is zero, so x = −8 or x = 5. A width cannot be negative, so x = −8 is rejected. (a) The width is 5 m and the length is 5 + 3 = 8 m. Check: 5 × 8 = 40.
(a) x = −8 or x = 5. A width cannot be negative, so the width is 5 m and the length is 8 m. 5.(b) The strip is the perimeter of the patio: 2 × (5 + 8) = 26 m.
(b) The strip is 2 × (5 + 8) = 26 m long.
Answer: (a) The width is 5 m and the length is 8 m; (b) 26 m
Common mistakes
- Solving x(x + 3) = 40 by writing x = 40 or x + 3 = 40. Only a product of zero forces one of its factors to be zero, so the equation must be rearranged to x2 + 3x − 40 = 0 before it is factorized.
- Giving both x = −8 and x = 5 as widths. Both numbers satisfy the equation, but x is a length in meters, and a length cannot be negative.
Worked example: A Right-Angled Shade Sail with Edges x and x + 7 and a Known Longest Edge
Question A shade sail is a right-angled triangle. The two edges that meet at the right angle are x m and (x + 7) m long, and the longest edge is 13 m long. (a) Find x and the lengths of the two shorter edges. (b) Find the area of the sail.
1.By Pythagoras' theorem, the squares of the two shorter edges add up to the square of the hypotenuse: x2 + (x + 7)2 = 132.
By Pythagoras' theorem, x2 + (x + 7)2 = 132. 2.Expand: (x + 7)2 = x2 + 14x + 49 and 132 = 169, so 2x2 + 14x + 49 = 169.
Expand both squares: 2x2 + 14x + 49 = 169. 3.Subtract 169 from both sides: 2x2 + 14x − 120 = 0. Divide both sides by 2: x2 + 7x − 60 = 0.
Subtract 169 from both sides, then divide both sides by 2: x2 + 7x − 60 = 0. 4.Factorize. Two numbers with a product of −60 and a sum of 7 are 12 and −5, so (x + 12)(x − 5) = 0, which gives x = −12 or x = 5. A length cannot be negative, so x = −12 is rejected. (a) x = 5, and the edges are 5 m and 5 + 7 = 12 m long. Check: 52 + 122 = 25 + 144 = 169 = 132.
(a) (x + 12)(x − 5) = 0. A length cannot be negative, so x = 5, and the edges are 5 m and 12 m long. 5.(b) The two shorter edges meet at the right angle, so one is the base and the other is the height: the area is 12 × 5 × 12 = 30 m2.
(b) The area of the sail is 12 × 5 × 12 = 30 m2.
Answer: (a) x = 5, so the edges are 5 m and 12 m long; (b) 30 m2
Common mistakes
- Expanding (x + 7)2 as x2 + 49. The bracket is multiplied by itself, (x + 7)(x + 7), and that gives the middle term 14x as well.
- Writing x + (x + 7) = 13. Pythagoras' theorem relates the squares of the sides, not the sides themselves, and in any triangle the two shorter sides add up to more than the longest side.