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factors of 20 that sum to -9.","code":"M"},{"id":"p2","marks":1,"description":"Final answer $(x - 2)(2x - 5)$.","code":"A"}],"finalAnswers":["(x - 2)(2x - 5)"],"generatedAt":"2026-05-07T02:06:38.085Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"ac-grouping","label":"Split and group","steps":[{"type":"text","order":0,"title":"Find the two numbers","content":"For $2x^2-9x+10$, multiply the coefficient of $x^2$ by the constant term:\n\n$$2\\times10=20$$\n\nWe need two numbers that multiply to $20$ and add to $-9$. They are $-4$ and $-5$.\n\nThis identifies the required pair, earning the **M1** mark.","highlights":[{"text":"Factorise $2x^2 - 9x + 10$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"2x^2+\\htmlData{anchor=midterm}{(-9x)}+10"},{"latex":"2x^2-4x-5x+10"},{"latex":"\\left(2x^2-4x\\right)+\\left(-5x+10\\right)"}],"highlights":[{"color":"sky","style":"underline","anchor":"midterm"}]},"order":1,"title":"Split the middle term","description":"Replace $-9x$ with $-4x-5x$, then group the four terms into two pairs."},{"type":"hyper_latex","block":{"nodes":[{"latex":"2x(x-2)-5\\htmlData{anchor=commonbracket}{(x-2)}"},{"latex":"(2x-5)(x-2)"},{"latex":"\\boxed{(x-2)(2x-5)}"}],"highlights":[{"color":"green","style":"box","anchor":"commonbracket"}]},"order":2,"title":"Factor each group","description":"Take out the highest common factor from each bracket. Both groups now contain the same bracket, $x-2$."}]},{"id":"binomial-inspection","label":"Inspect binomial factors","steps":[{"type":"text","order":0,"title":"Choose possible binomials","content":"The first terms must multiply to $2x^2$, so try $x$ and $2x$. The constant terms must multiply to $10$; because the middle term is negative but the final constant is positive, use two negative constants.\n\nUse $-2$ and $-5$: their products with the opposite $x$-terms give $-4x$ and $-5x$. These add to $-9x$, exactly the required middle term.\n\nFinding the factor pair of $20$ that totals $-9$ — namely $-4$ and $-5$ — earns the **M1**.\n\nTherefore,\n$$\n2x^2-9x+10=(x-2)(2x-5).\n$$\nThis completed factorisation earns the **A1**.","highlights":[{"text":"Factorise $2x^2 - 9x + 10$","location":"specification"}]}]}],"generatedAt":"2026-09-01T09:03:53.283Z","modelVersion":"gpt-5.6-terra+diagrams-2","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"14a8841a-a335-4b3e-b116-765bc83b8d98:ac-grouping","label":"Split and group","steps":[{"type":"text","order":0,"title":"Find the two numbers","content":"For $2x^2-9x+10$, multiply the coefficient of $x^2$ by the constant term:\n\n$$2\\times10=20$$\n\nWe need two numbers that multiply to $20$ and add to $-9$. They are $-4$ and $-5$.\n\nThis identifies the required pair, earning the **M1** mark.","highlights":[{"text":"Factorise $2x^2 - 9x + 10$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"2x^2+\\htmlData{anchor=midterm}{(-9x)}+10"},{"latex":"2x^2-4x-5x+10"},{"latex":"\\left(2x^2-4x\\right)+\\left(-5x+10\\right)"}],"highlights":[{"color":"sky","style":"underline","anchor":"midterm"}]},"order":1,"title":"Split the middle term","description":"Replace $-9x$ with $-4x-5x$, then group the four terms into two pairs."},{"type":"hyper_latex","block":{"nodes":[{"latex":"2x(x-2)-5\\htmlData{anchor=commonbracket}{(x-2)}"},{"latex":"(2x-5)(x-2)"},{"latex":"\\boxed{(x-2)(2x-5)}"}],"highlights":[{"color":"green","style":"box","anchor":"commonbracket"}]},"order":2,"title":"Factor each group","description":"Take out the highest common factor from each bracket. Both groups now contain the same bracket, $x-2$."}]},{"id":"14a8841a-a335-4b3e-b116-765bc83b8d98:binomial-inspection","label":"Inspect binomial factors","steps":[{"type":"text","order":0,"title":"Choose possible binomials","content":"The first terms must multiply to $2x^2$, so try $x$ and $2x$. The constant terms must multiply to $10$; because the middle term is negative but the final constant is positive, use two negative constants.\n\nUse $-2$ and $-5$: their products with the opposite $x$-terms give $-4x$ and $-5x$. These add to $-9x$, exactly the required middle term.\n\nFinding the factor pair of $20$ that totals $-9$ — namely $-4$ and $-5$ — earns the **M1**.\n\nTherefore,\n$$\n2x^2-9x+10=(x-2)(2x-5).\n$$\nThis completed factorisation earns the **A1**.","highlights":[{"text":"Factorise $2x^2 - 9x + 10$","location":"specification"}]}]}],"generatedAt":"2026-09-01T09:03:53.548Z","modelVersion":"gpt-5.6-terra+diagrams-2","explanationVersion":"v3"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"b221100c-3a06-45f6-9e9a-c56bd50f6b6c","revisionId":"aac59874-63d0-433b-ba3a-242a9e13e1cb","questionNumber":"10","questionType":"LA","specification":"","difficulty":1,"difficultyBand":"easy","totalMarks":2,"sourceTitle":"Factorising Harder Quadratics","calculatorUse":null,"isCalculation":true,"parts":[{"id":"d8331684-500d-4fe0-b7f2-beecda243dcd","label":"a","content":"Factorise $3x^2 + 16x + 21$","markscheme":"- Identifies factors of $ac = 63$ that sum to 16 (7 and 9) M1\n- Final answer $(3x + 7)(x + 3)$ A1","marks":2,"order":0,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Identifies factors of 63 that sum to 16.","code":"M"},{"id":"p2","marks":1,"description":"Final answer $(3x + 7)(x + 3)$.","code":"A"}],"finalAnswers":["(3x + 7)(x + 3)"],"generatedAt":"2026-05-07T02:06:38.085Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"split_middle_term","label":"Split the middle term","steps":[{"type":"text","order":0,"title":"Find the factor pair","content":"Multiply the coefficient of $x^2$ by the constant: $3\\times21=63$. Find two factors of $63$ that add to $16$: they are $7$ and $9$.\n\nThis earns **M1**: identifying the factors of $63$ that sum to $16$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"3x^2+\\htmlData{anchor=MID}{\\color{@sky}{16x}}+21"},{"latex":"3x^2+7x+9x+21"}],"highlights":[{"color":"sky","style":"text-color-with-background","anchor":"MID"}]},"order":1,"title":"Split the middle term","description":"Replace 16x with 7x + 9x, using the factor pair that sums to 16."},{"type":"hyper_latex","block":{"nodes":[{"latex":"3x^2+7x+9x+21"},{"latex":"(3x^2+7x)+(9x+21)"}]},"order":2,"title":"Group the terms","description":"Put the first two terms in one bracket and the last two terms in another."},{"type":"hyper_latex","block":{"nodes":[{"latex":"(3x^2+7x)+(9x+21)"},{"latex":"x(3x+7)+3(3x+7)"}]},"order":3,"title":"Factor each group","description":"Take out the common factor from each bracket. Both groups now contain the same binomial, 3x + 7."},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\htmlData{anchor=REPEAT}{x\\left(3x+7\\right)+3\\left(3x+7\\right)}"},{"latex":"\\boxed{\\htmlData{anchor=ANSWER}{\\left(3x+7\\right)\\left(x+3\\right)}}"}],"connections":[{"to":"ANSWER","from":"REPEAT","color":"orange","label":"factor out 3x + 7","style":"solid"}]},"order":4,"title":"Write the factors","description":"Now factor out the repeated bracket. This earns A1: (3x + 7)(x + 3)."}]}],"generatedAt":"2026-07-21T04:15:37.067Z","modelVersion":"gpt-5.6-terra","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"644f82b0-5ae3-4077-934a-fc2d73b02100","revisionId":"87b02f4b-7e03-4ce4-946c-ce5105b67fd3","questionNumber":"11","questionType":"LA","specification":"","difficulty":1,"difficultyBand":"easy","totalMarks":2,"sourceTitle":"Factorising Harder Quadratics","calculatorUse":null,"isCalculation":true,"parts":[{"id":"085cec46-d96a-4055-809c-197de1a51755","label":"a","content":"Factorise $2x^2 - 21x + 54$","markscheme":"- Identifies factors of $ac = 108$ that sum to -21 (-9 and -12) M1\n- Final answer $(2x - 9)(x - 6)$ A1","marks":2,"order":0,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Identifies factors of 108 that sum to -21.","code":"M"},{"id":"p2","marks":1,"description":"Final answer $(2x - 9)(x - 6)$.","code":"A"}],"finalAnswers":["(2x - 9)(x - 6)"],"generatedAt":"2026-05-07T02:06:38.085Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"split_middle_term","label":"Split the middle term","steps":[{"type":"text","order":0,"title":"Find the factor pair","content":"For $2x^2-21x+54$, multiply the coefficient of $x^2$ by the constant: $2\\times54=108$.\n\nFind two factors of $108$ which add to $-21$. They are $-9$ and $-12$, because $(-9)\\times(-12)=108$ and $-9+(-12)=-21$.\n\nThis identifies the required pair and earns the **M1** mark.","highlights":[{"text":"Factorise $2x^2 - 21x + 54$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\htmlData{anchor=original}{\\color{@sky}{2x^2-21x+54}}"},{"latex":"\\htmlData{anchor=split}{\\color{@sky}{2x^2-9x-12x+54}}"},{"latex":"\\htmlData{anchor=grouped}{\\color{@amber}{x(2x-9)-6(2x-9)}}"},{"latex":"\\htmlData{anchor=answer}{\\color{@green}{(2x-9)(x-6)}}"}],"highlights":[{"color":"green","style":"text-color-with-background","anchor":"answer"}],"connections":[{"to":"split","from":"original","color":"sky","label":"split using -9, -12","style":"dashed"},{"to":"grouped","from":"split","color":"amber","label":"group terms","style":"dashed"},{"to":"answer","from":"grouped","color":"green","label":"factor common bracket","style":"dashed"}]},"order":1,"title":"Factor by grouping","description":"Split the middle term using the factor pair, then take out the common bracket. The final line is the factorisation, earning A1."}]}],"generatedAt":"2026-07-21T04:19:33.638Z","modelVersion":"gpt-5.6-terra","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"igcse_maths__oxford_aqa","summaries":[{"id":"3bb21d01-9a45-4be2-a065-a4a9579601f5","difficultyBand":"medium","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"ee9f9894-2d44-4dd3-ae6b-057f2e161fee","difficultyBand":"medium","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"4dc9e8d0-c31f-4c54-9ac0-0e607bffb5bd","difficultyBand":"medium","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"0b0feb0e-aae1-41a6-b12f-9caf0922c76e","difficultyBand":"easy","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"a51a9195-19cc-4ffd-8f21-4b478da954e7","difficultyBand":"medium","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"9ca38bdc-aae3-4446-afff-bca6e8fbff63","difficultyBand":"medium","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"39e0a108-d7d2-4bcd-b0ee-46742e62458f","difficultyBand":"medium","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"65e58598-1de1-4e0c-a13b-07edefa8c4f8","difficultyBand":"easy","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"6bf8e53a-f5aa-4e47-bc80-b75668c474a3","difficultyBand":"medium","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"b221100c-3a06-45f6-9e9a-c56bd50f6b6c","difficultyBand":"easy","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"644f82b0-5ae3-4077-934a-fc2d73b02100","difficultyBand":"easy","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"eb6956bd-2510-4959-a5ce-c557de9458b0","difficultyBand":"medium","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"23334b17-ffaa-487c-be33-f6d1c6a211c3","difficultyBand":"medium","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"4157f0a2-a26d-41c7-879c-358678be4cbe","difficultyBand":"medium","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"19feebda-9403-457c-8fa3-e0792636902b","difficultyBand":"medium","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"c46305ea-f88d-487f-a44e-026584333267","difficultyBand":"medium","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"f65fd5fc-d81c-4081-a46f-d0a2b9bc59a7","difficultyBand":"medium","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"d5f39c2a-88a0-43c6-978c-57d476c1ba4f","difficultyBand":"easy","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"d7d8fdb6-7e25-4074-b845-414983bf3111","difficultyBand":"easy","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"7a0f3bd3-b52d-427e-8430-042e02488a5b","difficultyBand":"medium","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"154ea538-be4c-45df-bd45-0eecd7491ea1","difficultyBand":"medium","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true}],"title":"Factorising Harder Quadratics","topicId":"33714bef-2511-43cd-8ea5-f98f12dd98fe","topicSidebar":"$L3c","topicSlug":"grade-7-7-5-factorising-harder-quadratics","topicTitle":"Factorising Harder Quadratics"}]}],"$L3d","$L3e"]}]

Factorising Harder Quadratics

Question 3

Factorising Harder Quadratics Questions

Factorising Harder Quadratics

Question 3

Factorising Harder Quadratics Questions