SSC CGL Preparation: Concepts and Practice

Learning outcome

Factor expressions using common factors, grouping and identities, and simplify algebraic fractions without losing their original restrictions.

Concepts and assumptions

Variables are real unless stated otherwise. Factorisation rewrites an expression as a product with the same value. It reverses expansion: uv + uw = u(v + w). This equality holds even when u = 0 because factoring is not division by u.

First look for a common numerical factor and variable powers present in every term. Taking the smallest shared power leaves nonnegative integer powers inside the bracket.

Grouping creates a repeated bracket that can become another common factor. For x² + px + q, seek numbers r and s with r + s = p and rs = q, because (x + r)(x + s) expands to that polynomial.

For Ax² + Bx + C with A ≠ 0, a useful grouping method splits Bx using two numbers whose sum is B and product is AC. Suitable integer numbers are not guaranteed for every polynomial.

The identity a² − b² = (a − b)(a + b) factors a difference of squares. A sum of squares does not follow this rule.

Factoring alone introduces no exclusions. However, before simplifying a fraction, record where its original denominator is zero. Cancellation divides numerator and denominator by the same nonzero factor; it cannot restore excluded inputs.

Factoring an expression is not solving an equation. The zero-product rule applies only when a product is set equal to zero: then at least one factor must be zero.

Worked examples

Example 1 — Common factors. Factor 12x²y − 18xy².

The greatest common numerical factor is 6; both terms contain xy. Therefore 12x²y − 18xy² = 6xy(2x − 3y). Expanding back gives 12x²y − 18xy². This factorisation remains valid when x = 0 or y = 0.

Example 2 — Splitting a middle term. Factor x² + x − 12.

The required numbers have sum 1 and product −12: they are 4 and −3. Rewrite and group: x² + 4x − 3x − 12 = x(x + 4) − 3(x + 4) = (x + 4)(x − 3).

Example 3 — Preserving exclusions. Simplify (x² − 25)/(x² − 5x).

Denominator = x(x − 5), so x ≠ 0, 5. Numerator = (x − 5)(x + 5). Cancel the nonzero common factor x − 5: result = (x + 5)/x, with x ≠ 0, 5. Although the final formula can be evaluated at 5, the original fraction cannot.

Common mistakes

Dropping a common factor instead of extracting it; choosing numbers with the wrong sum or product; factoring a sum as a difference of squares; forgetting cancelled denominator zeros.

Practice questions

1. Factor 15a²b + 10ab².
2. Factor 9y² − 16.
3. Factor 6t² + 7t − 3.
4. State the domain and simplify (z² − 9)/(z² + 3z).

Worked solutions

1. The common factor is 5ab; the remaining factors are 3a and 2b. Thus 5ab(3a + 2b) expands to the original expression for all real a and b, including zero.
2. Write (3y)² − 4². The factors are (3y − 4)(3y + 4), valid for every real y.
3. The product is 6(−3) = −18 and required sum is 7. Choose 9 and −2: 6t² + 9t − 2t − 3 = 3t(2t + 3) − (2t + 3) = (3t − 1)(2t + 3).
4. Denominator = z(z + 3), requiring z ≠ 0, −3. Numerator = (z − 3)(z + 3). Cancel z + 3 to obtain (z − 3)/z, retaining both exclusions.