--- 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: 2026-11-12 --- > **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`. # 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, check proposed factors by expansion or substitution, and recover a missing bracket constant - Choose a useful factorising method by checking for a common factor or a difference-of-squares pattern ## Lesson: Factorising quadratics with a leading coefficient First check for a common factor or a special pattern. A difference of squares uses A² - B² = (A + B)(A - B); the cross terms cancel. For other quadratics, splitting the middle term is useful. Brackets (x + p)(x + q) have leading term x² because x times x is x². In any two linear brackets, multiply their x coefficients to find the leading coefficient. *(drawing: After removing a common factor, find the split pair, group the terms and check the result.)* For ax² + bx + c with integer coefficients and a not zero, find integers m and n with mn = ac and m + n = b. Replace bx with mx + nx. The sum condition keeps the expression equal to the original. The product condition lets the two groups reveal a shared bracket. If no signed integer pair meets both targets, there is no factorisation into two linear brackets with integer coefficients. **Example.** Factorise 2x² + 9x + 4. The product target is 2 times 4 = 8. The pair 8 and 1 adds to 9. Split and group: 2x² + 8x + x + 4 = 2x(x + 4) + 1(x + 4). Take out the shared bracket: (2x + 1)(x + 4). To check, multiply all four pairs: 2x², 8x, x and 4. These collect to 2x² + 9x + 4. **Example.** Remove a common factor from every term first. For 8x² + 24x + 16, take out 8 to get 8(x² + 3x + 2) = 8(x + 1)(x + 2). A common factor can contain x: 3x² + 6x = 3x(x + 2). Use the largest shared factor. With a negative leading term, taking out a negative factor can help: -8x² - 24x - 16 = -8(x + 1)(x + 2). Keep that outside factor in the answer. **Example.** Factorise 2x² + x - 3. The product target is -6. The pair 3 and -2 has product -6 and sum 1. Split: 2x² + 3x - 2x - 3. Group: x(2x + 3) - 1(2x + 3). The result is (x - 1)(2x + 3). Check the four products: 2x² + 3x - 2x - 3 = 2x² + x - 3. A rectangle's area is the product of its side lengths. If its area is 3x² + 13x + 4 square units, factorising gives (3x + 1)(x + 4). If one side is 3x + 1 units, the other is x + 4 units. Use values of x that make both lengths positive. Check the side expressions by multiplying them back to the area. Why use ac? Expanding (Ax + B)(Cx + D) gives a = AC, b = AD + BC and c = BD. The cross-term coefficients are AD and BC. Their sum is b, and their product is (AD)(BC) = (AC)(BD) = ac. This also works backward: a supplied split pair tells you b by addition. **Example.** Sometimes two brackets are opposites. Since 3 - x = -(x - 3), the expression 2x(x - 3) - 5(3 - x) becomes 2x(x - 3) + 5(x - 3). Now take out x - 3 to get (2x + 5)(x - 3). Change the outside sign when you reverse every sign inside a bracket. A full check compares all coefficients after expansion. Matching only at x = 0 checks the constant, not the whole quadratic. A mismatch at any input disproves a proposed factorisation. You can also find a missing bracket constant: (3x + 2)(x + t) has x coefficient 3t + 2. If that coefficient is 14, solve 3t + 2 = 14 to get t = 4. **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. *This lesson and its practice material were written for this dataset.* ## Practice 34 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)