--- title: "Factorising Quadratics with a Leading Coefficient" description: "Factorise quadratics such as 6x² + 11x + 3, where the x² coefficient is not 1, by splitting the middle term and grouping." canonical: https://lightmysky.com/learn/mathematics/factorising-quadratics-with-a-leading-coefficient-mt_IdKGCzbH07 source: https://lightmysky.com/learn/mathematics/factorising-quadratics-with-a-leading-coefficient-mt_IdKGCzbH07.md retrieved: 2026-09-02 --- > **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`. # Factorising Quadratics with a Leading Coefficient Factorise quadratics such as 6x² + 11x + 3, where the x² coefficient is not 1, by splitting the middle term and grouping. Subject: Mathematics · Area: Algebra · Ages 14 to 15 Page: https://lightmysky.com/learn/mathematics/factorising-quadratics-with-a-leading-coefficient-mt_IdKGCzbH07 ## Ready when they can - Split the middle term of 6x² + 11x + 3 using the pair that multiplies to 18 and adds to 11 - Factorise by grouping once the middle term has been split - Take out a common factor first when the whole quadratic shares one, as in 2x² + 10x + 12 ## Lesson: Factorising Quadratics with a Leading Coefficient In the last two stops you factorised x squared + bx + c with the pair that multiplies to c and adds to b, and you learned to spot the special patterns on sight. Now the x squared term can carry a coefficient, like 2x squared + 7x + 3. The simple pair method stops working, because (x + a)(x + b) always gives an x squared with coefficient 1. Instead, you split the middle term into two pieces, then factor by grouping. *(drawing: The method in five moves. The pair does two jobs at once: it multiplies to a times c and adds to b, then it splits the middle term.)* Here is the trick. For ax squared + bx + c, multiply a and c together. Find the pair that multiplies to that product and adds to b, then rewrite the middle term bx as the sum of the pair. If no pair of integers does both jobs, the quadratic cannot be factorised over the integers, and saying so is the correct answer. **Example.** Factorise 2x squared + 7x + 3. Here a is 2 and c is 3, so the product is 6. The pair that multiplies to 6 and adds to 7 is 6 and 1. Split the middle term: 2x squared + 6x + x + 3. Group the pairs: 2x(x + 3) + 1(x + 3). Now the bracket (x + 3) is common, so take it out: (2x + 1)(x + 3). Check by expanding: 2x times 3 gives 6x, and 1 times x gives x, which collect to 7x, and 1 times 3 gives 3. **Example.** Sometimes the whole quadratic shares a factor first. Factorise 4x squared + 12x + 8. Every term divides by 4, so take out 4: 4(x squared + 3x + 2). The inside is the simple case from the first stop: the pair 1 and 2 multiplies to 2 and adds to 3, so the inside is (x + 1)(x + 2). The full answer is 4(x + 1)(x + 2). Check: expand the inside back to x squared + 3x + 2, then multiply by 4. **Example.** Now a negative constant. Factorise 2x squared + x - 3. The product a times c is -6, so the pair multiplies to -6 and adds to 1: that is 3 and -2. Split the middle term: 2x squared + 3x - 2x - 3. Group: x(2x + 3) - 1(2x + 3). Take out the common bracket: (x - 1)(2x + 3). Check by expanding: x times 3 gives 3x, and -1 times 2x gives -2x, which collect to x, and -1 times 3 gives -3. **Tip.** The order of the split pair never matters. 2x squared + 6x + x + 3 and 2x squared + x + 6x + 3 group into the same brackets. Use whichever order makes the common factor easier to see. **Recap.** Multiply a and c, find the pair that multiplies to that product and adds to b, split the middle term, group, take out the shared bracket, then check by expanding. ## Practice 24 questions on this page, each with its working shown. ## Needs first - [Difference of Two Squares and Perfect Square Trinomials](https://lightmysky.com/learn/mathematics/difference-of-two-squares-and-perfect-square-trinomials-mt_GSpsHuserT) ## Opens up - [Solving Quadratic Equations by Factorising](https://lightmysky.com/learn/mathematics/solving-quadratic-equations-by-factorising-mt_tg_sqRzDjU)