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Solving Quadratic Equations by Factorisation

Class 10Quadratic Equations

Factorisation is the most fundamental and widely used method for solving quadratic equations at the CBSE Class 10 level. It is based on the simple but powerful principle that if the product of two expressions equals zero, then at least one of them must be zero. This principle, known as the zero product property, allows us to break a quadratic equation into two linear factors and solve each one separately. The factorisation method is preferred whenever the quadratic expression can be neatly split into factors, as it avoids the complexity of the quadratic formula and gives exact rational answers. In CBSE board examinations, factorisation problems are among the most frequently tested, and the skill of quickly factorising a quadratic expression is invaluable throughout mathematics. This topic will cover the theory behind the method, the systematic splitting-the-middle-term technique, and a comprehensive set of worked examples that progress from straightforward to challenging problems. By the end, students will be confident in their ability to recognise when factorisation is applicable and to carry it out efficiently.

What is Solving by Factorisation?

Factorisation of a quadratic equation means expressing the quadratic polynomial ax2 + bx + c as a product of two linear factors. If we can write:

ax2 + bx + c = (px + q)(rx + s)

then the equation ax2 + bx + c = 0 becomes (px + q)(rx + s) = 0. By the zero product property (if the product of two factors is zero, at least one factor must be zero), we get:

px + q = 0 → x = -q/p
rx + s = 0 → x = -s/r

These are the two roots of the quadratic equation.

The zero product property states: if A × B = 0, then A = 0 or B = 0 (or both). This property holds for real numbers and is the cornerstone of the factorisation method. It is crucial to understand that this property works ONLY when the product equals zero — if A × B = 6, we cannot conclude that A = 6 or B = 6.

Factorisation is most efficient when the roots are rational numbers, which happens when the discriminant D = b2 - 4ac is a perfect square. If D is not a perfect square, the roots are irrational and factorisation into linear factors with rational coefficients is not possible — in such cases, the quadratic formula or completing the square should be used.

Solving Quadratic Equations by Factorisation Formula

The Splitting the Middle Term Method:

For ax2 + bx + c = 0, find two numbers p and q such that:

p + q = b (sum equals the middle coefficient)
p × q = a × c (product equals the product of first and last coefficients)

Then rewrite bx as px + qx and factor by grouping:

ax2 + px + qx + c = 0
Group and factor: (common factor from first two) + (common factor from last two) = 0

Steps Summary:

  1. Write in standard form: ax2 + bx + c = 0
  2. Compute the product ac
  3. Find two numbers p and q such that p + q = b and p × q = ac
  4. Rewrite bx as px + qx
  5. Factor by grouping (take common factors)
  6. Apply zero product property to find the roots

Methods

Detailed Step-by-Step Process:

Step 1 — Standard Form: Ensure the equation is in the form ax2 + bx + c = 0. If not, rearrange.

Step 2 — Compute ac: Multiply the coefficient of x2 by the constant term. Note the sign.

Step 3 — Find the Split: Look for two numbers whose product is ac and whose sum is b. This is the most important step. Consider all factor pairs of ac (including negative ones) and check which pair sums to b.

Step 4 — Rewrite: Replace bx with the sum of the two terms found in Step 3.

Step 5 — Group and Factor: Group the four terms into two pairs. Take out the common factor from each pair. You should get two brackets that are identical.

Step 6 — Write as Product: Express as a product of two linear factors.

Step 7 — Solve: Set each factor equal to zero and solve for x.

How to Find the Right Pair in Step 3:

  • List all factor pairs of |ac|.
  • If ac is positive, both numbers have the same sign (both positive if b > 0, both negative if b < 0).
  • If ac is negative, the numbers have opposite signs. The larger magnitude number takes the sign of b.

Special Cases:

  • When c = 0: ax2 + bx = 0 → x(ax + b) = 0. One root is always x = 0.
  • When b = 0: ax2 + c = 0. This is a difference of squares if c < 0: ax2 = |c| → x2 = |c|/a → x = ± √(|c|/a).
  • Perfect square trinomial: ax2 + bx + c = (√a · x + √c)2 when b = 2√(ac).

Solved Examples

Example 1: Simple Factorisation with a = 1

Problem: Solve x2 - 7x + 12 = 0 by factorisation.

Solution:

Step 1: a = 1, b = -7, c = 12. Product ac = 12.

Step 2: Find two numbers with product 12 and sum -7.
Factor pairs of 12: (1,12), (2,6), (3,4), (-1,-12), (-2,-6), (-3,-4)
The pair (-3, -4) has sum -7 and product 12. ✓

Step 3: Split the middle term:
x2 - 3x - 4x + 12 = 0

Step 4: Group and factor:
x(x - 3) - 4(x - 3) = 0
(x - 3)(x - 4) = 0

Step 5: x - 3 = 0 or x - 4 = 0
x = 3 or x = 4

Answer: x = 3 or x = 4.

Example 2: Factorisation with a > 1

Problem: Solve 6x2 + x - 15 = 0 by factorisation.

Solution:

Step 1: a = 6, b = 1, c = -15. Product ac = 6 × (-15) = -90.

Step 2: Find two numbers with product -90 and sum 1.
Factor pairs of 90: (1,90), (2,45), (3,30), (5,18), (6,15), (9,10)
We need opposite signs summing to 1: 10 and -9 → 10 + (-9) = 1 and 10 × (-9) = -90. ✓

Step 3: Split: 6x2 + 10x - 9x - 15 = 0

Step 4: Group: 2x(3x + 5) - 3(3x + 5) = 0
(3x + 5)(2x - 3) = 0

Step 5: 3x + 5 = 0 → x = -5/3
2x - 3 = 0 → x = 3/2

Answer: x = -5/3 or x = 3/2.

Example 3: Factorisation When c = 0

Problem: Solve 3x2 - 12x = 0.

Solution:

Step 1: Take x as a common factor:
3x(x - 4) = 0

Step 2: Apply zero product property:
3x = 0 → x = 0
x - 4 = 0 → x = 4

Note: Do not divide both sides by x (you would lose the root x = 0).

Answer: x = 0 or x = 4.

Example 4: Factorisation with All Negative Middle Term

Problem: Solve 2x2 - 5x - 3 = 0 by factorisation.

Solution:

Step 1: a = 2, b = -5, c = -3. Product ac = 2 × (-3) = -6.

Step 2: Find two numbers with product -6 and sum -5.
Pairs: (1,-6): sum = -5. ✓

Step 3: Split: 2x2 + x - 6x - 3 = 0

Step 4: Group: x(2x + 1) - 3(2x + 1) = 0
(2x + 1)(x - 3) = 0

Step 5: 2x + 1 = 0 → x = -1/2
x - 3 = 0 → x = 3

Answer: x = -1/2 or x = 3.

Example 5: Perfect Square Trinomial

Problem: Solve 9x2 - 12x + 4 = 0 by factorisation.

Solution:

Step 1: a = 9, b = -12, c = 4. Product ac = 36.

Step 2: Find two numbers with product 36 and sum -12: -6 and -6.

Step 3: Split: 9x2 - 6x - 6x + 4 = 0

Step 4: Group: 3x(3x - 2) - 2(3x - 2) = 0
(3x - 2)(3x - 2) = 0
(3x - 2)2 = 0

Step 5: 3x - 2 = 0 → x = 2/3

This is a repeated root (equal roots). D = (-12)2 - 4(9)(4) = 144 - 144 = 0 confirms this.

Answer: x = 2/3 (a repeated root).

Example 6: Factorisation After Rearranging

Problem: Solve 5x2 = 4x + 1.

Solution:

Step 1: Rearrange: 5x2 - 4x - 1 = 0

Step 2: a = 5, b = -4, c = -1. Product ac = -5.

Step 3: Find two numbers with product -5 and sum -4: -5 and 1.

Step 4: Split: 5x2 - 5x + x - 1 = 0

Step 5: Group: 5x(x - 1) + 1(x - 1) = 0
(x - 1)(5x + 1) = 0

Step 6: x = 1 or x = -1/5

Answer: x = 1 or x = -1/5.

Example 7: Factorisation with Large Coefficients

Problem: Solve 12x2 - 7x - 10 = 0 by factorisation.

Solution:

Step 1: a = 12, b = -7, c = -10. Product ac = -120.

Step 2: Find two numbers with product -120 and sum -7.
We need factors of 120 with opposite signs summing to -7.
Try: 8 and -15: sum = -7, product = -120. ✓

Step 3: Split: 12x2 + 8x - 15x - 10 = 0

Step 4: Group: 4x(3x + 2) - 5(3x + 2) = 0
(3x + 2)(4x - 5) = 0

Step 5: x = -2/3 or x = 5/4

Answer: x = -2/3 or x = 5/4.

Example 8: Difference of Squares Pattern

Problem: Solve 4x2 - 49 = 0.

Solution:

Step 1: Recognise as a difference of squares:
4x2 - 49 = (2x)2 - 72 = (2x - 7)(2x + 7)

Step 2: Set each factor to zero:
2x - 7 = 0 → x = 7/2
2x + 7 = 0 → x = -7/2

Answer: x = 7/2 or x = -7/2.

Example 9: Word Problem Leading to Factorisation

Problem: The product of two consecutive positive integers is 306. Find them.

Solution:

Step 1: Let the integers be x and x + 1.
x(x + 1) = 306
x2 + x - 306 = 0

Step 2: a = 1, b = 1, c = -306. Product ac = -306.

Step 3: Find two numbers with product -306 and sum 1.
306 = 2 × 153 = 2 × 9 × 17. Try 18 and -17: sum = 1, product = -306. ✓

Step 4: Split: x2 + 18x - 17x - 306 = 0

Step 5: Group: x(x + 18) - 17(x + 18) = 0
(x + 18)(x - 17) = 0

Step 6: x = -18 or x = 17. Since x must be positive, x = 17.

The two integers are 17 and 18. Check: 17 × 18 = 306 ✓

Answer: The consecutive positive integers are 17 and 18.

Example 10: Equation Requiring Expansion Before Factorisation

Problem: Solve (2x + 3)(3x - 7) = 0.

Solution:

Step 1: The equation is already in factored form. Apply zero product property directly:

2x + 3 = 0 → x = -3/2
3x - 7 = 0 → x = 7/3

Note: There is no need to expand first when the equation is already a product equal to zero. Expanding would give 6x2 - 5x - 21 = 0, which would then need to be re-factored — an unnecessary detour.

Answer: x = -3/2 or x = 7/3.

Real-World Applications

The factorisation method for solving quadratic equations has widespread applications:

Number Theory: Finding numbers that satisfy given conditions about products, sums, and differences often leads to quadratic equations that are best solved by factorisation. For example, finding two consecutive integers whose product is 132 leads to x(x+1) = 132, or x2 + x - 132 = 0, which factors as (x - 11)(x + 12) = 0.

Geometry: Problems involving areas and dimensions of rectangles, triangles, and other shapes frequently produce quadratic equations with integer or rational roots that factorise neatly. For instance, finding the dimensions of a rectangle with area 60 cm2 and perimeter 34 cm leads to a factorisable quadratic.

Physics: Kinematic equations (involving distance, velocity, and acceleration) sometimes yield quadratic equations. When a ball is thrown upward with initial velocity 20 m/s, the equation h = 20t - 5t2 = 0 factors as 5t(4 - t) = 0, giving t = 0 or t = 4 seconds. The factorisation reveals both the starting and landing times.

Finance: Compound interest and profit/loss problems occasionally lead to quadratic equations. Factorisation gives exact answers without rounding errors. For example, if a sum doubles in n years at a certain rate, the resulting quadratic often has rational roots.

Computer Science: Algorithm analysis sometimes involves solving recurrence relations that reduce to quadratic equations. Factorisation provides clean closed-form solutions. The Fibonacci recurrence, when solved via its characteristic equation x2 - x - 1 = 0, illustrates this connection.

Daily Life: Planning dimensions (garden beds, tiles, rooms) subject to area constraints leads to quadratic equations. The factorisation method gives practical, exact dimensions. For example, tiling a floor where the number of tiles along one side is 3 more than along the other, with a total of 108 tiles, gives a quadratic equation that factors neatly.

Age Problems: Problems where the product of ages or the relationship between current and future ages leads to quadratic equations are commonly solved by factorisation in school-level mathematics.

Key Points to Remember

  • Factorisation uses the zero product property: if (px + q)(rx + s) = 0, then px + q = 0 or rx + s = 0.
  • The splitting the middle term method: find two numbers whose product is ac and sum is b.
  • After splitting, group the four terms into pairs and extract common factors.
  • When c = 0, always take x as a common factor first. Do not divide both sides by x — dividing by x loses the root x = 0.
  • Factorisation works best when D = b2 - 4ac is a perfect square. In this case, the roots are rational.
  • If D is not a perfect square, the quadratic cannot be factored over rationals — use the quadratic formula or completing the square instead.
  • Always write the equation in standard form ax2 + bx + c = 0 before attempting to factorise. The product-sum method requires this form.
  • A perfect square trinomial gives a repeated (double) root: for example, x2 + 6x + 9 = (x + 3)2 gives x = -3 as a double root.
  • The difference of squares pattern a2 - b2 = (a - b)(a + b) is a special factorisation case useful when b = 0.
  • Verify roots by substituting back into the original equation. This catches sign errors and arithmetic mistakes.
  • The product ac can be positive or negative. If ac > 0, both factors have the same sign. If ac < 0, the factors have opposite signs.
  • For exam efficiency, try small factor pairs of |ac| first, as most textbook problems use manageable numbers.

Practice Problems

  1. Solve by factorisation: x² + 5x + 6 = 0.
  2. Solve by factorisation: 2x² + 7x + 3 = 0.
  3. Solve: 3x² - 11x + 10 = 0.
  4. Solve: x² - 25 = 0.
  5. Solve: 4x² + 12x + 9 = 0.
  6. The sum of a number and its reciprocal is 10/3. Find the number using factorisation.

Frequently Asked Questions

Q1. What is the factorisation method for quadratic equations?

The factorisation method involves expressing the quadratic polynomial as a product of two linear factors, then using the zero product property to find the roots. The key technique is splitting the middle term.

Q2. What is the zero product property?

The zero product property states that if the product of two factors equals zero, then at least one of the factors must be zero. So if (x - 3)(x + 5) = 0, then x - 3 = 0 or x + 5 = 0.

Q3. How do I find the two numbers for splitting the middle term?

For ax² + bx + c = 0, find two numbers whose product is a×c and whose sum is b. List factor pairs of |a×c| and check which pair sums to b, considering signs.

Q4. When does factorisation not work?

Factorisation into rational linear factors does not work when the discriminant D = b² - 4ac is not a perfect square. In such cases, the roots are irrational, and you should use the quadratic formula or completing the square.

Q5. Can I divide both sides by x to simplify x² = 5x?

No! Dividing by x loses the root x = 0. Instead, write x² - 5x = 0, then factor as x(x - 5) = 0, giving x = 0 or x = 5.

Q6. What if the equation is not in standard form?

Always rearrange to standard form ax² + bx + c = 0 before factorising. Move all terms to one side, expand any products, and combine like terms.

Q7. How do I check my factors are correct?

Expand your two factors back and verify you get the original quadratic. Also substitute your roots into the original equation to verify each gives zero.

Q8. Is factorisation important for CBSE Class 10 exams?

Yes, factorisation is one of the most commonly tested topics. Nearly every quadratic equations question in board exams can involve factorisation, either as the primary method or as a step in a larger problem.

Q9. What is a perfect square trinomial?

A perfect square trinomial is a quadratic that factors as (px + q)², such as x² + 6x + 9 = (x + 3)² or 4x² - 12x + 9 = (2x - 3)². It occurs when D = 0, giving a repeated root.

Q10. What is the difference of squares and how is it used?

The difference of squares is the identity a² - b² = (a - b)(a + b). Quadratics of the form ax² - c = 0 (with b = 0) can often be written as a difference of squares and factored directly. For example, 9x² - 16 = (3x - 4)(3x + 4).

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