Factorization of Quadratic Equations – SS1 Mathematics Lesson Note (First Term)

Factorising a quadratic equation means writing ax² + bx + c = 0 as a product of two linear factors, then setting each factor to zero to find the roots.

Behavioural objectives

By the end of the lesson, students should be able to:

1. factorise a quadratic expression by the common factor, difference of two squares and splitting the middle term methods;
2. solve a quadratic equation by factorisation;
3. state the sum and product of the roots of ax² + bx + c = 0 in terms of a, b and c;
4. show how these results come from the equation (x – α)(x – β) = 0;
5. find α + β and αβ for a given quadratic equation;
6. find the value of expressions such as α/β + β/α and α – β using the sum and product of the roots.

What is a quadratic equation?

A quadratic equation is an equation whose highest power of the unknown is 2.

• General form: ax² + bx + c = 0, where a, b and c are numbers and a ≠ 0.
• Coefficient: a is the coefficient of x², b is the coefficient of x.
• Constant: c is the term without x.
• Roots: the values of x that make the equation true. A quadratic equation has two roots, usually called α and β.
• Example: in 3x² – 4x – 1 = 0, a = 3, b = -4 and c = -1.

How do we solve a quadratic equation by factorisation?

Factorisation solves a quadratic equation by writing it as (px + q)(rx + s) = 0, so that px + q = 0 or rx + s = 0.

1. Write the equation in the form ax² + bx + c = 0.
2. Factorise the left-hand side into two brackets.
3. Set each bracket equal to zero.
4. Solve each linear equation to get the two roots.
5. Check by substituting each root into the original equation.

What are the methods of factorising a quadratic expression?

There are three common methods: common factor, difference of two squares and splitting the middle term.

Method 1: Common factor

• When to use: when the equation has no constant term, c = 0.
• Example: x² – 4x = 0 gives x(x – 4) = 0.
• Roots: x = 0 or x = 4.

Method 2: Difference of two squares

• When to use: when the equation has only x² and a number, both being perfect squares, with a minus sign between them.
• Example: x² – 9 = 0 gives (x – 3)(x + 3) = 0.
• Roots: x = 3 or x = -3.

Method 3: Splitting the middle term

This method works for any quadratic expression that can be factorised.

1. Multiply a by c.
2. Find two numbers whose product is ac and whose sum is b.
3. Replace bx with two terms using these numbers.
4. Group the terms in pairs and take out the common factor of each pair.
5. Take out the common bracket.

Example A: x² – 5x + 6 = 0

1. ac = 1 × 6 = 6 and b = -5.
2. The numbers are -2 and -3, since (-2)(-3) = 6 and -2 + (-3) = -5.
3. x² – 2x – 3x + 6 = 0
4. x(x – 2) – 3(x – 2) = 0
5. (x – 2)(x – 3) = 0
6. x = 2 or x = 3.

Example B: 2x² + 7x + 3 = 0

1. ac = 2 × 3 = 6 and b = 7.
2. The numbers are 6 and 1, since 6 × 1 = 6 and 6 + 1 = 7.
3. 2x² + 6x + x + 3 = 0
4. 2x(x + 3) + 1(x + 3) = 0
5. (2x + 1)(x + 3) = 0
6. x = -1/2 or x = -3.

How do we find the sum and product of the roots?

The sum of the roots of ax² + bx + c = 0 is -b/a and the product is c/a.

The roots are linked to the coefficients. It is usual to call the roots α and β. These two results let us find expressions such as α/β + β/α without solving the equation.

How are the sum and product of the roots obtained?

1. Start with ax² + bx + c = 0 …….. (1)
2. If the roots are α and β, equation (1) is equivalent to (x – α)(x – β) = 0.
3. Expand: x² – αx – βx + αβ = 0.
4. Collect terms: x² – (α + β)x + αβ = 0 …….. (2)
5. Divide equation (1) by a: x² + (b/a)x + c/a = 0 …….. (3)
6. Compare equations (2) and (3): α + β = -b/a and αβ = c/a.

Result

For any quadratic equation ax² + bx + c = 0 with roots α and β:

• Sum of roots: α + β = -b/a
• Product of roots: αβ = c/a
• Equation from its roots: x² – (α + β)x + αβ = 0

Checking with a factorised equation

• Equation: x² – 5x + 6 = 0 has roots 2 and 3.
• Sum: 2 + 3 = 5, and -b/a = -(-5)/1 = 5.
• Product: 2 × 3 = 6, and c/a = 6/1 = 6.

What are useful expressions in α and β?

Some expressions in α and β can be rewritten using only α + β and αβ.

• Sum of squares: α² + β² = (α + β)² – 2αβ
• Sum of reciprocals: α/β + β/α = (α² + β²)/αβ = ((α + β)² – 2αβ)/αβ
• Difference of roots: (α – β)² = (α + β)² – 4αβ, so α – β = ±√((α + β)² – 4αβ)

Worked examples

1. If the roots of 3x² – 4x – 1 = 0 are α and β, find α + β and αβ.
2. If α and β are the roots of the equation 3x² – 4x – 1 = 0, find the value of (a) α/β + β/α and (b) α – β.

Solution to example 1

1. Compare 3x² – 4x – 1 = 0 with ax² + bx + c = 0.
2. a = 3, b = -4, c = -1
3. α + β = -b/a = -(-4)/3 = 4/3 = 1 1/3
4. αβ = c/a = -1/3

Solution to example 2(a)

1. The equation is the same as in example 1, so α + β = 4/3 and αβ = -1/3.
2. α/β + β/α = (α² + β²)/αβ
3. = ((α + β)² – 2αβ)/αβ
4. = ((4/3)² – 2(-1/3)) ÷ (-1/3)
5. = (16/9 + 2/3) ÷ (-1/3)
6. = (16/9 + 6/9) ÷ (-1/3)
7. = 22/9 × (-3/1)
8. = -22/3

Solution to example 2(b)

1. (α – β)² = α² + β² – 2αβ
2. But α² + β² = (α + β)² – 2αβ.
3. So (α – β)² = (α + β)² – 2αβ – 2αβ = (α + β)² – 4αβ.
4. α – β = ±√((α + β)² – 4αβ)
5. = ±√((4/3)² – 4(-1/3))
6. = ±√(16/9 + 4/3)
7. = ±√((16 + 12)/9)
8. = ±√28/√9 = ±√28/3
9. Therefore α – β = ±√28/3, which can also be written as ±2√7/3.

Evaluation

1. Solve by factorisation: x² – 7x + 12 = 0.
2. Write down the sum and the product of the roots of ax² + bx + c = 0 in terms of a, b and c.
3. Starting from (x – α)(x – β) = 0, show how the sum and product of the roots are obtained.
4. Write α/β + β/α in terms of α + β and αβ.
5. Write (α – β)² in terms of α + β and αβ.

Assignment

If α and β are the roots of 2x² – 5x + 3 = 0, find:

1. α + β and αβ;
2. α/β + β/α.