Hello, and welcome to your mathematics lesson for Class Nine. Today, we are diving into Chapter Five: Factorisation. By the end of this session, you will understand what factorisation means, master five powerful methods to break down algebraic expressions, and learn how to recognise when an expression can or cannot be factorised.
Let us begin with the fundamental idea. When a polynomial—that is, an algebraic expression—is written as the product of two or more expressions, each of those expressions is called a factor of the polynomial.
Consider the expression x² + 5x + 6. This can be rewritten as (x + 3)(x + 2). Therefore, (x + 3) and (x + 2) are factors of the original expression.
The process of writing an expression as a product of terms or brackets is called factorisation. Think of it as the reverse of multiplication. If multiplication builds expressions up, factorisation breaks them down.
Now, let us explore the first method: taking out common factors.
When every term in an expression shares a common factor, we divide each term by that factor and place the quotient inside brackets, keeping the common factor outside. The key is to find the highest common factor, or H.C.F., of all terms.
Take 6a² – 3ax. The terms are 6a² and –3ax. The H.C.F. is 3a. So we write: 6a² – 3ax = 3a(2a – x).
Sometimes, expressions disguise their common factors. Look at x(a – 5) + y(5 – a). Notice that (5 – a) equals –(a – 5). So we rewrite: x(a – 5) – y(a – 5), which gives (a – 5)(x – y).
The second method is factorisation by grouping.
This works when an expression has an even number of terms. We arrange them into groups where each group has its own common factor, then extract a common factor from the resulting groups.
Consider ab + bc + ax + cx. Group as (ab + bc) + (ax + cx). From the first group, b is common; from the second, x is common. This yields b(a + c) + x(a + c). Now (a + c) itself is common, giving (a + c)(b + x).
Another example: a² + 1/a² + 2 – 5a – 5/a. The first three terms form a perfect square: (a + 1/a)². The last two terms give –5(a + 1/a). Factor out (a + 1/a) to get (a + 1/a)(a + 1/a – 5).
The third method handles trinomials of the form ax² + bx + c, where a, b, and c are real numbers.
Here, we split the middle term strategically. We break b into two parts whose sum is b and whose product equals a × c. Then we apply grouping.
Take x² + 5x + 6. We need two numbers that add to 5 and multiply to 6. Those numbers are 3 and 2. So: x² + 3x + 2x + 6, which becomes x(x + 3) + 2(x + 3), and finally (x + 3)(x + 2).
For 2x² – 7x + 6, we need numbers summing to –7 with product 2 × 6 = 12. These are –4 and –3. Thus: 2x² – 4x – 3x + 6, giving 2x(x – 2) – 3(x – 2), so (x – 2)(2x – 3).
There is also a test for factorisability. For ax² + bx + c, calculate b² – 4ac, known as the discriminant. If this is a perfect square, the trinomial factorises; otherwise, it does not. For 5x² + 17x + 6, we get 289 – 120 = 169, which is 13², a perfect square, so factorisation is possible. But for 3x² – 8x – 15, we obtain 64 + 180 = 244, not a perfect square, so this trinomial cannot be factorised.
The fourth method exploits the difference of two squares.
Since (x + y)(x – y) = x² – y², we can reverse this: x² – y² = (x + y)(x – y).
Consider 9(x – y)² – (x + 2y)². This is [3(x – y)]² – (x + 2y)². Apply the formula: [3x – 3y + x + 2y][3x – 3y – x – 2y], simplifying to (4x – y)(2x – 5y).
For 48x³ – 27x, first extract 3x: 3x(16x² – 9). Now 16x² – 9 is (4x)² – 3², giving 3x(4x + 3)(4x – 3).
Some expressions need creative rearrangement. Take x⁴ + x²y² + y⁴. Add and subtract x²y²: (x² + y²)² – (xy)², which becomes (x² + y² + xy)(x² + y² – xy).
The fifth method deals with the sum and difference of two cubes.
Through algebraic expansion, we derive two essential identities.
First: a³ + b³ = (a + b)(a² – ab + b²).
Second: a³ – b³ = (a – b)(a² + ab + b²).
For a³ + 27b³, recognise this as a³ + (3b)³. Apply the sum of cubes: (a + 3b)(a² – 3ab + 9b²).
For 8a³ – 27b³, write as (2a)³ – (3b)³. Using the difference of cubes: (2a – 3b)(4a² + 6ab + 9b²).
Sometimes we combine techniques. Consider 2a⁷ – 128a. First, extract 2a: 2a(a⁶ – 64). Now a⁶ – 64 is (a³)² – 8², a difference of squares: (a³ + 8)(a³ – 8). Each of these is a sum and difference of cubes, giving four factors in total.
These identities also reveal divisibility properties. Since a³ – b³ = (a – b)(a² + ab + b²), the expression is always divisible by (a – b). Thus 15³ – 8³ is divisible by 15 – 8 = 7. Similarly, a³ + b³ is always divisible by (a + b), so 15³ + 8³ is divisible by 15 + 8 = 23.
Let us recap the key takeaways from this chapter.
First, factorisation is the process of expressing a polynomial as a product of simpler expressions, essentially the reverse of multiplication.
Second, always look for common factors first—this is the most straightforward method and often simplifies other approaches.
Third, for expressions with even numbers of terms, try grouping; rearrange terms strategically to create common factors within groups.
Fourth, for trinomials ax² + bx + c, split the middle term using two numbers whose sum is b and product is ac; use the discriminant b² – 4ac to test factorisability.
Fifth, recognise difference of squares: x² – y² = (x + y)(x – y); this applies even to complex expressions and can be applied repeatedly.
Sixth, master the sum and difference of cubes formulas; they are powerful tools that often combine with other factorisation techniques.
Remember, factorisation is as much an art as a science. With practice, you will develop an intuition for which method to apply. Keep exploring, keep practising, and you will master these techniques. Until next time, happy learning!