cheme":"- Divide $g(x)$ by $(x+5)$ to get $2x^2 - 9x + 7$. M1 A1\n- Factorise quadratic: $(2x-7)(x-1)$. M1\n- State final form: $(x+5)(2x-7)(x-1)$. A1","marks":4,"order":1,"markingRubric":{"totalMarks":4,"points":[{"id":"p1","marks":2,"description":"Finds the quadratic factor $2x^2 - 9x + 7$.","code":"M"},{"id":"p2","marks":1,"description":"Factorises the quadratic into $(2x-7)(x-1)$.","code":"M"},{"id":"p3","marks":1,"description":"States the product of three linear factors.","code":"A"}],"finalAnswers":["(x+5)(2x-7)(x-1)"],"generatedAt":"2026-05-12T02:14:06.612Z","modelVersion":"gemini-3-flash-preview"},"explanation":{"methods":[{"id":"polynomial_division","label":"Divide then factorise","steps":[{"type":"text","order":0,"title":"Show the division","content":"Using the result from the previous part, $x+5$ is a factor. Divide by $x+5$, using the root $-5$:\n\n$$\\begin{array}{r|rrrr}-5&2&1&-38&35\\\\&&-10&45&-35\\\\\\hline&2&-9&7&0\\end{array}$$\n\nHence the quotient is $2x^2-9x+7$ and the remainder is $0$. Showing this division earns M1; obtaining the correct quotient earns A1.","highlights":[{"text":"Hence, show that $g(x)$ can be written as the product of three linear factors.","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\htmlData{anchor=split}{2x^2-9x+7}=2x^2-2x-7x+7"},{"latex":"2x^2-2x-7x+7=\\htmlData{anchor=grouped}{2x(x-1)-7(x-1)}"},{"latex":"2x(x-1)-7(x-1)=\\htmlData{anchor=factored}{(2x-7)(x-1)}"}],"highlights":[{"color":"orange","style":"text-color","anchor":"split"},{"color":"orange","style":"text-color","anchor":"grouped"},{"color":"orange","style":"text-color","anchor":"factored"}]},"order":2,"title":"Factorise the quadratic","description":"The cross terms must add to -9x, so split the middle term as -2x-7x. This factorisation earns the M1 mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\color{@green}{g(x)=(x+5)(2x-7)(x-1)}$"},{"text":"Therefore, this is a product of three linear factors."}]},"order":3,"title":"State the final form","description":"Write the complete equality. This earns the final accuracy mark A1."}]}],"generatedAt":"2026-07-21T04:15:14.468Z","modelVersion":"gpt-5.6-terra","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null},{"id":"df131400-df1a-41b3-9284-6b5359b1d75d","revisionId":"c9ef0e04-8304-4229-bf5a-1cb36fcd88cd","questionNumber":"30","questionType":"LA","specification":"","difficulty":1,"difficultyBand":"medium","totalMarks":5,"sourceTitle":"Question Set 7: Algebraic Methods","calculatorUse":null,"parts":[{"id":"6131cb28-b016-46ff-ab19-36ef278b1b70","label":"0","content":"$$N$ is an integer that is not divisible by 3. Prove that $N^2$ is of the form $3p + 1$, where $p$ is an integer.","markscheme":"- Let $N = 3n \\pm 1$. M1\n- Case 1: $N = 3n+1 \\Rightarrow N^2 = 9n^2 + 6n + 1 = 3(3n^2 + 2n) + 1$. M1 A1\n- Case 2: $N = 3n-1 \\Rightarrow N^2 = 9n^2 - 6n + 1 = 3(3n^2 - 2n) + 1$. M1\n- Conclude both are in form $3p+1$. A1","marks":5,"order":0,"markingRubric":{"totalMarks":5,"points":[{"id":"p1","marks":1,"description":"Identifies $N$ can be written as $3n+1$ or $3n-1$.","code":"M"},{"id":"p2","marks":2,"description":"Correctly expands and factors the case for $3n+1$.","code":"M"},{"id":"p3","marks":2,"description":"Correctly expands and factors the case for $3n-1$.","code":"M"}],"finalAnswers":[],"generatedAt":"2026-05-12T02:14:06.612Z","modelVersion":"gemini-3-flash-preview"},"explanation":{"methods":[{"id":"proof_by_cases","label":"Proof by cases","steps":[{"type":"text","order":0,"title":"Classify the integer","content":"Because $N$ is not divisible by $3$, division by $3$ leaves remainder $1$ or $2$. A remainder of $1$ gives $N=3n+1$; a remainder of $2$ is one less than the next multiple of $3$, so it gives $N=3n-1$, for an integer $n$. Therefore $N=3n+1$ or $N=3n-1$. Establishing these two forms, with $n$ an integer, earns the first M1 mark.","highl

Algebraic Methods

Question 27

Algebraic Methods Questions

Algebraic Methods

Question 27

Algebraic Methods Questions