Hello, and welcome to today's mathematics lesson. Today, we are going to explore factorisation. This is a powerful technique that helps us break down complex algebraic expressions into simpler building blocks. By the end of this lesson, you will understand what factors are, learn different methods to factorise expressions, and discover how to apply these skills to simplify algebraic fractions.
Let us begin with the fundamental idea. When we multiply numbers or expressions together, each part of that multiplication is called a factor of the final product.
For example, if we consider the product 5x, the numbers 5 and x are its factors. Similarly, in the expression (2x - 5)(3x + 2), the two parts (2x - 5) and (3x + 2) are factors.
We can verify this by expanding: (2x - 5) multiplied by (3x + 2) gives us 6x² - 11x - 10. Therefore, we say that (2x - 5) and (3x + 2) are factors of 6x² - 11x - 10.
Factorisation, then, is the reverse process of expansion — we start with a polynomial and express it as a product of its factors.
The first method we will learn is taking out the common factor. This is often the simplest and most direct approach.
Look at every term in the expression and identify what is common to all of them. Then, we factor out this common element using the distributive property in reverse.
Consider the expression 6x² + 9x. Both terms contain 3, and both contain x. So we factor out 3x, giving us 3x(2x + 3).
Always check your answer by expanding it back — 3x × 2x is 6x², and 3x × 3 is 9x, which matches our original expression.
Next, we turn to factorising by grouping. This method is useful when there is no single common factor across all terms, but we can group terms that do share factors.
Take the expression ax + ay + bx + by. The first two terms have a in common, giving us a(x + y). The last two terms have b in common, giving us b(x + y).
Now we notice that (x + y) is common to both groups. Factoring this out, we obtain (a + b)(x + y).
The key insight is to arrange terms so that when we group them, a common binomial factor emerges.
Now we come to one of the most important techniques: factorising quadratic expressions of the form x² + bx + c.
Here, we need to find two numbers that multiply to give the constant term and add to give the coefficient of x.
Let us factorise x² + 7x + 12. We need two numbers that multiply to 12 and add to 7. Those numbers are 3 and 4.
Therefore, x² + 7x + 12 equals (x + 3)(x + 4).
What if the middle term is negative? Consider x² - 5x + 6. We need numbers that multiply to 6 and add to negative 5. Those are negative 2 and negative 3. So we write (x - 2)(x - 3).
For quadratics of the form ax² + bx + c, where the coefficient of x² is not 1, we use a systematic approach.
Multiply the coefficient of x² and the constant term together, then find two numbers that multiply to this product and add to the coefficient of x. Rewrite the middle term using these two numbers, then factorise by grouping.
Let us try 6x² - 11x - 10. Here, the coefficient of x² is 6, the coefficient of x is negative 11, and the constant term is negative 10. Multiplying 6 and negative 10 gives negative 60. We need factors of negative 60 that add to negative 11. Those are negative 15 and positive 4.
We rewrite: 6x² - 15x + 4x - 10. Grouping: the first two terms give 3x(2x - 5), and the last two give 2(2x - 5). The common factor is (2x - 5), so our answer is (2x - 5)(3x + 2).
We now move to special factorisation patterns that you should recognise instantly.
The first is the difference of squares.
The precise statement is: a² - b² equals (a + b)(a - b).
This only works for a difference, never a sum of squares. For example, 9x² - 16 becomes (3x + 4)(3x - 4), since 9x² is (3x)² and 16 is 4².
The second special pattern is the perfect square trinomial.
We have two forms: a² + 2ab + b² equals (a + b)², and a² - 2ab + b² equals (a - b)².
Notice the pattern: the first and last terms are perfect squares, and the middle term is twice the product of their square roots. For instance, x² + 6x + 9 fits this pattern — it is (x + 3)². Similarly, 4x² - 12x + 9 equals (2x - 3)².
The third special pattern is the sum and difference of cubes.
The sum of cubes: a³ + b³ equals (a + b)(a² - ab + b²).
The difference of cubes: a³ - b³ equals (a - b)(a² + ab + b²).
Pay careful attention to the signs. In the sum formula, the binomial factor has a plus, but the trinomial has a minus in the middle. In the difference formula, the binomial has a minus, but the trinomial has all plus signs. For example, x³ - 27 is (x - 3)(x² + 3x + 9), since 27 is 3³.
Finally, let us see how factorisation helps us simplify algebraic fractions. When numerator and denominator share common factors, we can cancel them to reduce the fraction to its simplest form.
Consider the fraction with numerator x² - 9 and denominator x² + 6x + 9. The numerator factors as (x + 3)(x - 3) using difference of squares. The denominator factors as (x + 3)² using perfect square.
We can cancel one factor of (x + 3) from numerator and denominator, leaving (x - 3)/(x + 3).
Always state the condition: this simplification is valid only when x is not equal to negative 3, since that would make the original denominator zero.
Let us recap the key takeaways from today's lesson.
First, factorisation is the process of writing a polynomial as a product of its factors, the reverse of expansion. Second, always look for a common factor first before trying other methods. Third, for quadratics, find two numbers that multiply to the constant term and add to the coefficient of x. Fourth, memorise the three special patterns: difference of squares, perfect square trinomials, and sum or difference of cubes. Fifth, factorisation allows us to simplify algebraic fractions by cancelling common factors. Sixth, always check your factorisation by expanding your answer.
Factorisation is a skill that improves with practice. Start each problem by asking yourself what pattern you recognise, and work systematically through the methods we have discussed. You now have a powerful toolkit for breaking down complex expressions into their essential components.
Thank you for your attention, and I look forward to seeing you apply these techniques with confidence.