g another value for $x$, we can use the **Hybrid Method** to find $B$ more quickly by equating the coefficients of the $x^2$ terms from both sides of the identity.\n\nLet's look at the terms that produce $x^2$ on the right-hand side:\n- $A(x+2)^2$ expands to $A(x^2 + 4x + 4)$, which contributes $Ax^2$.\n- $B(x-4)(x+2)$ expands to $B(x^2 - 2x - 8)$, which contributes $Bx^2$.\n- $C(x-4)$ has no $x^2$ term.\n\nOn the left-hand side, there is no $x^2$ term, which means its coefficient is $0$. Therefore, we can write:\n\n$$0 = A + B$$\n\nSince we already know $A = 1$:\n\n$$\\begin{aligned} 0 &= 1 + B \\\\ B &= -1 \\end{aligned}$$\n\nFinding $B = -1$ earns the third mark (**A1**)."},{"type":"text","order":4,"title":"Write the final answer","content":"Now we substitute our values of $A = 1$, $B = -1$, and $C = 2$ back into our original partial fraction template:\n\n$$\\frac{8x+4}{(x-4)(x+2)^2} = \\frac{1}{x-4} - \\frac{1}{x+2} + \\frac{2}{(x+2)^2}$$\n\nThis final form earns the fourth and final mark (**A1**)."}]}],"generatedAt":"2026-06-09T06:02:58.680Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null},{"id":"ba7e7e9f-cffd-4764-bae7-369bfddc13e7","revisionId":"8a7e1262-6ae1-493c-b7df-dffa0faa6536","questionNumber":null,"questionType":"LA","specification":"","difficulty":1,"difficultyBand":"medium","totalMarks":5,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"e32a7ea2-3837-41f8-9ce1-1bb18a4f8097","label":"a","content":"Express $\\frac{2x^2-2x-1}{(x-1)(x-2)}$ in partial fractions","markscheme":"- Sets up the identity $2x^2-2x-1 = A(x-1)(x-2) + B(x-2) + C(x-1)$ M1\n- Finds $B = 1$ A1\n- Finds $C = 3$ A1\n- Finds $A = 2$ A1\n- Final answer: $2 + \\frac{1}{x-1} + \\frac{3}{x-2}$ A1","marks":5,"order":0,"markingRubric":{"totalMarks":5,"points":[{"id":"p1","marks":1,"description":"Sets up the correct form of the identity $2x^2 - 2x - 1 = A(x-1)(x-2) + B(x-2) + C(x-1)$. ","code":"M"},{"id":"p2","marks":1,"description":"Correctly finds the numerator $B = 1$.","code":"A"},{"id":"p3","marks":1,"description":"Correctly finds the numerator $C = 3$.","code":"A"},{"id":"p4","marks":1,"description":"Correctly finds the constant term $A = 2$.","code":"A"},{"id":"p5","marks":1,"description":"Expresses the final answer correctly as $2 + \\frac{1}{x-1} + \\frac{3}{x-2}$.","code":"A"}],"finalAnswers":[],"generatedAt":"2026-05-06T14:03:58.492Z","modelVersion":"gemini-3-flash-preview"},"explanation":{"methods":[{"id":"division_then_partial_fractions","label":"Polynomial Division and Partial Fractions","steps":[{"type":"text","order":0,"title":"Analyze the fraction type","content":"To express \n\n$$\\frac{2x^2-2x-1}{(x-1)(x-2)}$$\n\nin partial fractions, we first examine the degrees of the numerator and the denominator.\n\n- The degree of the numerator $2x^2 - 2x - 1$ is $2$.\n- Expanding the denominator $(x-1)(x-2) = x^2 - 3x + 2$ shows its degree is also $2$.\n\nSince the degree of the numerator is equal to the degree of the denominator, this is an **improper fraction**. We must perform algebraic division first to write it as an integer quotient plus a proper fraction."},{"type":"text","order":1,"title":"Perform algebraic division","content":"Let's divide the numerator $2x^2 - 2x - 1$ by the expanded denominator $x^2 - 3x + 2$ using polynomial long division:\n\n$$\\frac{2x^2 - 2x - 1}{x^2 - 3x + 2}$$\n\nHow many times does $x^2$ go into $2x^2$? It goes in $2$ times.\n\nMultiply $2$ by the divisor:\n\n$$2(x^2 - 3x + 2) = 2x^2 - 6x + 4$$\n\nSubtract this from our original numerator:\n\n$$\\begin{aligned} (2x^2 - 2x - 1) - (2x^2 - 6x + 4) &= -2x - (-6x) - 1 - 4 \\\\ &= 4x - 5 \\end{aligned}$$\n\nSo, the quotient is $2$ and the remainder is $4x - 5$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\frac{2x^2-2x-1}{(x-1)(x-2)} = \\htmlData{anchor=quotient}{\\color{@sky}{2}} + \\frac{\\htmlData{anchor=remainder}{\\color{@amber}{4x-5}}}{(x-1)(x-2)}$$"}],"highlights":[{"color":"sky","style":"box","anchor":"quotient"},{"color"

1.2.25 Partial fractions (A-level only)

Question 10

1.2.25 Partial fractions (A-level only) Questions

1.2.25 Partial fractions (A-level only)

Question 10

1.2.25 Partial fractions (A-level only) Questions