Factorisation of Quadratic Expressions

Definition: Factorisation of a quadratic expression means expressing ax² + bx + c as a product of two linear factors.

ax² + bx + c = a(x − p)(x − q)

where p and q are the roots of the quadratic equation ax² + bx + c = 0.

Key relationships:

• Sum of roots: p + q = −b/a
• Product of roots: p × q = c/a

When can a quadratic be factorised over integers?

• The discriminant D = b² − 4ac must be a perfect square (including 0).
• If D is a perfect square, the roots are rational and the expression factorises over rationals.
• If D is not a perfect square, factorisation over integers is not possible — use the quadratic formula instead.

Step-by-step procedure — Splitting the Middle Term:

1. Write the expression in standard form: ax² + bx + c.
2. Calculate the product a × c.
3. Find two numbers m and n such that m + n = b and m × n = a × c.
4. Rewrite the middle term bx as mx + nx.
5. Group the four terms into two pairs.
6. Factor out the GCD from each pair.
7. Factor out the common binomial factor.

Example walkthrough:

• Factorise: 6x² + 11x + 3
• a × c = 6 × 3 = 18
• Find m, n such that m + n = 11 and m × n = 18.
• m = 9, n = 2 (since 9 + 2 = 11 and 9 × 2 = 18)
• Rewrite: 6x² + 9x + 2x + 3
• Group: (6x² + 9x) + (2x + 3) = 3x(2x + 3) + 1(2x + 3)
• Factor: (3x + 1)(2x + 3)

How to find m and n systematically:

• List the factor pairs of |a × c|.
• Check which pair sums to b (with correct signs).
• If a × c is positive, m and n have the same sign (both positive if b > 0, both negative if b < 0).
• If a × c is negative, m and n have opposite signs.

Types of quadratic expressions and their factorisation:

Type 1: a = 1 (Monic quadratic)

• Expression: x² + bx + c
• Find two numbers whose sum = b and product = c.
• Example: x² + 7x + 12 = (x + 3)(x + 4)

Type 2: a > 1 (Non-monic quadratic)

• Expression: ax² + bx + c where a ≠ 1
• Find m, n with m + n = b and m × n = ac.
• Example: 2x² + 7x + 3 = (2x + 1)(x + 3)

Type 3: Middle term is negative

• Example: x² − 5x + 6 = (x − 2)(x − 3)
• Both m and n are negative.

Type 4: Constant term is negative

• Example: x² + x − 12 = (x + 4)(x − 3)
• m and n have opposite signs.

Type 5: Difference of squares

• Example: 4x² − 9 = (2x + 3)(2x − 3)
• Use identity: a² − b² = (a + b)(a − b)

Type 6: Perfect square trinomial

• Example: x² + 6x + 9 = (x + 3)²
• Both factors are identical.

Method 1: Splitting the Middle Term

• The standard method for Class 10.
• Works when the discriminant is a perfect square.

Method 2: Using Algebraic Identities

• a² − b² = (a − b)(a + b)
• a² ± 2ab + b² = (a ± b)²
• Useful for special forms like 25x² − 16 or 9x² + 12x + 4.

Method 3: Trial and Error (Cross Method)

• For ax² + bx + c, try factor pairs of a and c.
• Check which combination gives the correct middle term.
• Faster for experienced solvers but less systematic.

When factorisation is not possible over integers:

• Calculate D = b² − 4ac.
• If D is not a perfect square, the expression cannot be factorised over integers.
• Use the quadratic formula: x = (−b ± √D) / 2a.

Common mistakes:

• Forgetting to check signs of m and n.
• Incorrect grouping — always verify by expanding the result.
• Missing a common factor of the entire expression (always extract GCD first).

Applications of factorisation of quadratic expressions:

• Solving quadratic equations — set each factor to zero to find roots. If (x − 3)(x + 5) = 0, then x = 3 or x = −5.
• Finding zeros of polynomials — the zeros (roots) of ax² + bx + c are the values of x that make the expression zero. These are found by factorisation.
• Simplifying algebraic fractions — factorise numerator and denominator separately, then cancel common factors. For example: (x² − 9)/(x² − 5x + 6) = (x+3)(x−3)/[(x−2)(x−3)] = (x+3)/(x−2) for x ≠ 3.
• Area and perimeter problems — geometric conditions lead to quadratic equations. A rectangle with area 180 and length = breadth + 3 gives x(x+3) = 180, solved by factorisation.
• Number problems — finding two numbers given their sum and product. If sum = 13 and product = 40, then x² − 13x + 40 = 0 → (x−5)(x−8) = 0 → numbers are 5 and 8.
• Projectile motion — height h = ut − (1/2)gt². Setting h = 0 gives t(u − gt/2) = 0 → t = 0 or t = 2u/g. Factorisation identifies the two times when the projectile is at ground level.
• Revenue and profit — revenue R = px − ax² (p = price, a = demand factor, x = quantity). Setting R = 0 gives the break-even points by factorisation.
• Graphing quadratic functions — the x-intercepts of y = ax² + bx + c are the roots found by factorisation. These determine where the parabola crosses the x-axis.

Connection to higher mathematics:

• Factorisation is a prerequisite for partial fractions (Class 11–12).
• Factor theorem and remainder theorem extend factorisation to higher-degree polynomials.
• The discriminant D = b² − 4ac determines whether factorisation over rationals is possible.