--- title: "Solving quadratic equations by factorising" description: "Rearrange a quadratic equation so one side is zero, factorise it, then use the fact that a product is zero only when one factor is zero to read off both solutions." canonical: https://lightmysky.com/learn/mathematics/solving-quadratic-equations-by-factorising-mt_tg_sqRzDjU source: https://lightmysky.com/learn/mathematics/solving-quadratic-equations-by-factorising-mt_tg_sqRzDjU.md retrieved: 2026-11-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # Solving quadratic equations by factorising Rearrange a quadratic equation so one side is zero, factorise it, then use the fact that a product is zero only when one factor is zero to read off both solutions. Subject: Mathematics · Area: Algebra · Ages 14 to 15 Page: https://lightmysky.com/learn/mathematics/solving-quadratic-equations-by-factorising-mt_tg_sqRzDjU ## Ready when they can - Rearrange x² + 5x = 14 into x² + 5x - 14 = 0 before factorising - Apply the zero product rule only when a product equals zero, and find all distinct solutions - Reject a solution that cannot fit the context, such as a negative length ## Lesson: Solving quadratic equations by factorising Factorise quadratics with the pair method, special patterns or splitting the middle term to solve equations. A product is zero only when a factor is zero, so solve a zero case from each bracket. For 2x²+7x+3=0, the pair's product is the leading coefficient times the constant: 2 times 3=6. Its sum is 7, so use 6 and 1. Split: 2x²+6x+x+3=0. Factor each pair: 2x(x+3)+1(x+3)=0. Take out x+3: (2x+1)(x+3)=0. Solve 2x+1=0 or x+3=0: x=-0.5 or x=-3. Add: -0.5+(-3)=-3.5. *(drawing: The moves of solving by factorising, in the order they happen.)* If AB = 0, then A = 0 or B = 0. Solve these alternative cases separately. From (x - 3)(x + 4) = 0, solve x - 3 = 0 or x + 4 = 0, giving x = 3 or x = -4. Do not set the factors equal to each other. If a product equals a nonzero number, neither factor can be zero, and the zero product rule does not apply. **Example.** Solve x² + 3x - 10 = 0. The pair 5 and -2 has product -10 and sum 3, so (x + 5)(x - 2) = 0. Set each factor to zero separately: x + 5 = 0 or x - 2 = 0. Thus x = -5 or x = 2. Check both in the original equation: 25 - 15 - 10 = 0 and 4 + 6 - 10 = 0. **Example.** The equation 3x squared = 12x is not yet set to zero. Move every term to one side: 3x squared - 12x = 0. Take out the common factor 3x: 3x(x - 4) = 0. The rule gives x = 0 or x = 4. Do not divide both sides by x here: that assumes x is not zero and drops the solution x = 0. **Example.** A garden has width x metres, length x + 3 metres and area 40 square metres. Area = width times length, so x(x + 3) = 40. Rearrange: x² + 3x - 40 = 0. The pair 8 and -5 gives (x + 8)(x - 5) = 0, so x = -8 or x = 5. A length cannot be negative. Keep x = 5: the garden is 5 by 8 metres, and 5 times 8 = 40. Two factors may give the same solution. For (x - 9)² = 0, both factors are x - 9, so both zero cases give x = 9. Count distinct values, not brackets: this equation has one distinct solution. A nonzero constant factor does not add a solution, since that constant cannot equal zero. You can build an equation from its roots. A root r gives a factor x - r. Roots 4 and -8 give (x - 4)(x + 8) = 0, which expands to x² + 4x - 32 = 0. For roots r and s, expanding (x - r)(x - s) gives x² - (r + s)x + rs. Thus the middle coefficient is the negative of the roots' sum. Translate a number story before solving. If a number is 6 less than twice its square, write x = 2x² - 6. Move all terms to one side: 2x² - x - 6 = 0. Factorise: (2x + 3)(x - 2) = 0, so x = -3/2 or x = 2. If the story asks for the positive number, keep 2. Read 'less than' carefully: subtract 6 from twice the square. A difference of squares uses A² - B² = (A - B)(A + B). Expanding gives A² + AB - AB - B², so the cross terms cancel. For x² - 121 = 0, 121 = 11², giving (x - 11)(x + 11) = 0. Solve x - 11 = 0 or x + 11 = 0: x = 11 or x = -11. Both check because 11² = (-11)² = 121. To solve x² = 11x - 24, subtract 11x and add 24 on both sides: x² - 11x + 24 = 0. For a quadratic with x² coefficient 1, find two numbers whose product is the constant and whose sum is the x coefficient. Here -3 and -8 have product 24 and sum -11, so (x - 3)(x - 8) = 0. Solve each zero case: x = 3 or x = 8. Check in the original: 9 = 33 - 24 and 64 = 88 - 24. A squared bracket means two identical brackets. For (x + 3)² = 25, expand (x + 3)(x + 3): x² + 3x + 3x + 9 = 25. Combine like terms and subtract 25 from both sides: x² + 6x - 16 = 0. The pair 8 and -2 has product -16 and sum 6, so (x + 8)(x - 2) = 0. Solve x + 8 = 0 or x - 2 = 0: x = -8 or x = 2. Their sum is -8 + 2 = -6. Check: (-8 + 3)² = 25 and (2 + 3)² = 25. **Recap.** Move every term to one side so the other is zero, factorise, set each factor to zero, and keep the solutions that fit the context. *This lesson and its practice questions were written for this dataset.* ## Practice 29 questions on this page, each with its working shown. ## Needs first - [Factorising quadratics with a leading coefficient](https://lightmysky.com/learn/mathematics/factorising-quadratics-with-a-leading-coefficient-mt_IdKGCzbH07) ## Opens up - [Simultaneous Equations with One Quadratic](https://lightmysky.com/learn/mathematics/simultaneous-equations-with-one-quadratic-mt_D89ql2sJvc) - [Solving Quadratic Inequalities](https://lightmysky.com/learn/mathematics/solving-quadratic-inequalities-mt_dukJtB_0i8) - [Polynomial Division and the Factor Theorem](https://lightmysky.com/learn/mathematics/polynomial-division-and-the-factor-theorem-mt_haL2UkGa5C) - [Simplifying algebraic fractions](https://lightmysky.com/learn/mathematics/simplifying-algebraic-fractions-mt_scDNSoz3eW) - [Function notation and evaluating functions](https://lightmysky.com/learn/mathematics/function-notation-and-evaluating-functions-mt_YWKWzHK_6Z)