Factorization: Class 9 Mathematics

Key points

To factorize an expression is to write it as a product of simpler expressions called factors. Factorization is the reverse of expansion. For example x² + 5x + 6 = (x + 2)(x + 3). The expression is factorized completely when no factor can be broken down further. Always check by multiplying the factors back.

First method: take out the highest common factor (HCF). Find the HCF of the numerical coefficients and the lowest power of each common letter, then divide it out of every term. For 12x³ − 18x², the HCF is 6x², so 12x³ − 18x² = 6x²(2x − 3). Always try this first, before any other method.

Second method: factorization by grouping. When four terms have no single common factor, arrange them in pairs that do. For ax + ay + bx + by, group as a(x + y) + b(x + y) = (a + b)(x + y). Sometimes the terms must be rearranged first to get the same bracket in both groups.

Third method: use the square formulas.

• Perfect square trinomial: a² + 2ab + b² = (a + b)² and a² − 2ab + b² = (a − b)². Test it by checking that the first and last terms are squares and the middle term is twice the product of their roots.
• Difference of two squares: a² − b² = (a + b)(a − b). For example 49x² − 25 = (7x + 5)(7x − 5).
• A sum of two squares such as a² + b² cannot be factorized over the real numbers.
• Repeat the formula when needed: x⁴ − 81 = (x² − 9)(x² + 9) = (x − 3)(x + 3)(x² + 9).

Fourth method: factorize a quadratic trinomial x² + bx + c. Find two numbers whose product is c and whose sum is b. Then x² + bx + c = (x + p)(x + q) where p × q = c and p + q = b. For x² − 7x + 12 we need two numbers with product 12 and sum −7, namely −3 and −4, so it equals (x − 3)(x − 4). The signs of p and q follow the signs of b and c.

For a trinomial ax² + bx + c with a not equal to 1, use the splitting method. Multiply a and c, find two numbers with that product and with sum b, and split the middle term using them. For 6x² + 7x − 3: a × c = −18, and the numbers 9 and −2 have product −18 and sum 7. So 6x² + 9x − 2x − 3 = 3x(2x + 3) − 1(2x + 3) = (3x − 1)(2x + 3).

Fifth method: sum and difference of cubes.

• a³ + b³ = (a + b)(a² − ab + b²).
• a³ − b³ = (a − b)(a² + ab + b²).
• The signs follow the pattern: same sign as the cube sign in the first bracket, then opposite sign for the ab term, then always plus for b².
• Example: 8x³ + 27 = (2x)³ + 3³ = (2x + 3)(4x² − 6x + 9).

Some expressions need a small trick before factoring. To factorize a⁴ + a²b² + b⁴, add and subtract a²b² to make a perfect square: (a² + b²)² − a²b² = (a² + b² + ab)(a² + b² − ab). Spotting a difference of squares is the key to most such questions.

The remainder theorem says: when a polynomial p(x) is divided by x − a, the remainder is p(a). To find the remainder, substitute x = a in the polynomial. For x + 2, use a = −2. The remainder theorem works without carrying out the division.

The factor theorem is a special case: x − a is a factor of p(x) if and only if p(a) = 0. If the remainder is zero, the divisor is a factor. To factorize a cubic, try small values such as ±1, ±2, ±3 (divisors of the constant term) until p(a) = 0, then divide by x − a to get a quadratic and factorize that.

In synthetic division the coefficients of the polynomial are written in a row and the root a is used to bring down, multiply and add step by step. The last number is the remainder, and the others are the coefficients of the quotient. This is a short form of long division by x − a.

Factorization is used to solve quadratic equations, to simplify fractions and to find HCF and LCM. Always arrange terms in descending powers of the variable before starting.

Study tip: Follow a fixed order: take out the common factor, count the terms, then pick a method (two terms: squares or cubes; three terms: perfect square or splitting; four terms: grouping). Always multiply back to check. For factor theorem questions, test only the divisors of the constant term.

Definitions and flashcards

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Short questions with model answers

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Numericals with worked steps

Common mistakes

• Not taking out the highest common factor first, which leaves the expression only partly factorized.
• Writing a² + b² = (a + b)(a − b). A sum of squares has no real factors.
• Using the wrong signs in the cube formulas, especially the middle term of the quadratic factor.
• Using x + a instead of x − a in the remainder theorem. For divisor x + 2, substitute x = −2.
• Stopping early, for example leaving x⁴ − 81 as (x² − 9)(x² + 9) when x² − 9 can still be factorized.

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