Solving Quadratic Equations by Factorisation

Solving Quadratic Equations by Factorisation means rewriting a quadratic expression as a product of two factors and using the zero-product rule. This method applies when the equation can be expressed as ax² + bx + c = 0. The roots are obtained by setting each factor equal to zero.

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What Is Solving Quadratic Equations by Factorisation?

Solving a quadratic equation by factorisation involves expressing ax² + bx + c as two linear factors and equating each factor to zero. If AB = 0, then A = 0 or B = 0.

First, write the equation in standard form ax² + bx + c = 0. Factorise the left-hand side, then apply the zero-product property. For example, x² + 5x + 6 = 0 becomes (x + 2)(x + 3) = 0, giving x = −2 or x = −3.

Solving Quadratic Equations by Factorisation Video Lessons

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Quadratic Equations by Factorisation

Learn how to solve quadratic equations using the factorisation method, including identifying factors and applying them to find the equation’s roots.

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Solving Quadratic Equations by Factorisation Formula & Tricks

Important Formulas

Standard form
ax² + bx + c = 0, where a ≠ 0

A quadratic equation has degree 2, so the coefficient of x² must be non-zero.

Zero-product property
If (px + q)(rx + s) = 0, then px + q = 0 or rx + s = 0

Set each linear factor equal to zero and solve the resulting equations.

Factorisation condition
For ax² + bx + c, find m and n such that mn = ac and m + n = b

The middle term bx is split into mx + nx before grouping.

Difference of two squares
u² − v² = (u − v)(u + v)

This identity directly factorises quadratic expressions containing two perfect squares with a minus sign.

Quick Tricks

Use the signs of the constant term

If c is positive, the two constant factors have the same sign. If c is negative, they have opposite signs. Their signs must also produce the sign of b when added.

Example: For x² − 7x + 12, use −3 and −4 because (−3)(−4) = 12 and −3 + (−4) = −7. Thus x² − 7x + 12 = (x − 3)(x − 4).
Use the ac method for a ≠ 1

Multiply the coefficient of x² by the constant term. Find two numbers whose product is ac and whose sum is b, split the middle term, and factor by grouping.

Example: For 2x² + 7x + 3, ac = 6. The numbers 6 and 1 give 2x² + 6x + x + 3 = (2x + 1)(x + 3).
Check the roots in the original equation

Substitution confirms both the factorisation and the signs of the roots.

Example: For (x + 2)(x + 3) = 0, the roots are −2 and −3. Substituting either value into x² + 5x + 6 gives 0.

Solving Quadratic Equations by Factorisation Concepts

Factorising a monic quadratic

When the coefficient of x² is 1, find two numbers whose product is c and whose sum is b.

For x² + bx + c, write the expression as (x + p)(x + q), where pq = c and p + q = b. Then solve x + p = 0 and x + q = 0.

Example: x² − x − 12 = 0. The numbers 3 and −4 have product −12 and sum −1. Therefore, (x + 3)(x − 4) = 0, so x = −3 or x = 4.

Factorising when the leading coefficient is not 1

For ax² + bx + c with a ≠ 1, use two numbers whose product is ac and whose sum is b.

Split bx into mx + nx, where mn = ac and m + n = b. Group the four terms and take out the common binomial factor.

Example: 3x² + 10x + 8 = 0. Since ac = 24, use 6 and 4: 3x² + 6x + 4x + 8 = 0. Hence (3x + 4)(x + 2) = 0, giving x = −4/3 or x = −2.

Using the zero-product property

If a product of two factors is zero, at least one factor must be zero.

After factorisation, do not set the product equal to one or solve both factors simultaneously. Set each factor separately equal to zero. For (2x − 5)(x + 1) = 0, either 2x − 5 = 0 or x + 1 = 0.

Example: From (2x − 5)(x + 1) = 0, the solutions are x = 5/2 and x = −1.

Special quadratic factorisations

Some quadratics can be factorised directly using identities such as the difference of squares or a perfect-square trinomial.

Use u² − v² = (u − v)(u + v), u² + 2uv + v² = (u + v)², and u² − 2uv + v² = (u − v)². A repeated factor gives one repeated root.

Example: x² − 10x + 25 = (x − 5)², so the repeated solution is x = 5. Also, x² − 16 = (x − 4)(x + 4), giving x = 4 or x = −4.

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Practice Solving Quadratic Equations by Factorisation Questions

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1A shopkeeper sells goods to a customer at a profit of p% more than the cost price. Also, he cheats the customer by giving 850 g instead of 1 kg. His total profit percentage is 30%. Find the value of p.→ 2The value of the expression 3{- 4 × 0.125 + 1/4 × 23 + 2 × (- 2) ÷ (- 4)} is ______.→ 3The price of an article was ₹3,600 in 2024. In the year 2025, the article's price becomes ₹3,960. Find the percentage increase in the price of the article.→ 4The given pie chart shows the expenditure of a family in percentage. Study the chart and answer the question that follows. The total monthly income is ₹60,000.What is the expenditure on House Rent per month?→ 5Mohan, Sohan and Rohan started a business. Mohan invests ₹12000 for 18 months , Sohan invests ₹18000 for 15 months and Rohan invests ₹24000 for 12 months. If their profit after 18 months is ₹43000 , then the share of profit of Rohan is :→ 6The expression 5^(2n) - 2^(3n) has a factor:→ 7A record entry gives √32 . Simplify.→ 8A trader lost 10% by selling a shirt for ₹900. If he gains x% by selling it for ₹1,440, the value of x is:→ 9An AC was bought for ₹31,000. The value of the AC was depreciated by 5% per annum. Find the value of the AC after 2 years.→ 10A shopkeeper marks an article at ₹x and offers a discount of 63% on the marked price. He sells it for ₹481 after charging VAT of 25% on the discounted price. What is the value of x?→

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Quick Revision Notes

Solving Quadratic Equations by Factorisation: Revision Points

Use these rules to factorise and solve quadratic equations.

  • Write the equation in the form ax² + bx + c = 0 before factorising.
  • For x² + bx + c, find factors of c that add to b.
  • For ax² + bx + c, find m and n such that mn = ac and m + n = b.
  • Apply the zero-product rule after factorisation.
  • A factor x − r gives the root x = r; a factor x + r gives the root x = −r.
  • Always check the obtained roots in the original equation.

Related Topic Explainers

Substitution Method for Simultaneous Equations → Cross Multiplication Method for Linear Equations →

Solving Quadratic Equations by Factorisation FAQs

What is the factorisation method for a quadratic equation?

Write the equation as ax² + bx + c = 0, factorise the quadratic into two linear factors, and set each factor equal to zero. For example, (x + 2)(x + 3) = 0 gives x = −2 or x = −3.

How do you solve 2x² + 7x + 3 = 0 by factorisation?

Split 7x as 6x + x: 2x² + 6x + x + 3 = (2x + 1)(x + 3). Therefore, x = −1/2 or x = −3.

How do you solve x² − 9 = 0 by factorisation?

Use the difference of squares: x² − 9 = (x − 3)(x + 3). Hence, x = 3 or x = −3.

What happens when a quadratic has a repeated factor?

A repeated factor gives one repeated root. For example, x² − 6x + 9 = (x − 3)², so the repeated root is x = 3.

Why must a quadratic equation be written equal to zero before factorisation?

The zero-product property applies when a product equals zero. Therefore, rearrange all terms to one side so the equation has the form ax² + bx + c = 0.

How are the signs of factors selected in x² + bx + c?

The product of the two constants must equal c, and their sum must equal b. For x² − 5x + 6, use −2 and −3 because their product is 6 and their sum is −5.

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