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bg-background","children":["$","$L3a",null,{"backHref":"/a-level/maths/edexcel/questions","basePath":"/a-level/maths/edexcel/a-level-pure-mathematics-1-3-partial-fractions/questions","formulaUrl":"https://assets.mathsgenie.co.uk/board-resources/a-level-maths/edexcel/formula-book.pdf","initialIndex":6,"initialQuestionWindow":[{"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,"isCalculation":true,"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"},"tableImages":[],"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":"amber","style":"underline","anchor":"remainder"}],"connections":[]},"order":2,"title":"Rewrite the improper fraction","description":"Using our division results, we rewrite the original expression as the sum of the constant quotient $2$ and a proper fraction with the remainder $4x-5$ in the numerator."},{"type":"text","order":3,"title":"Set up partial fractions","content":"Now we focus on decomposing the proper fraction part, $\\frac{4x-5}{(x-1)(x-2)}$, into partial fractions with linear denominators:\n\n$$\\frac{4x-5}{(x-1)(x-2)} \\equiv \\frac{B}{x-1} + \\frac{C}{x-2}$$\n\nMultiply both sides by $(x-1)(x-2)$ to clear the fractions:\n\n$$4x - 5 \\equiv B(x-2) + C(x-1)$$"},{"type":"text","order":4,"title":"Solve for B and C","content":"To find the constants $B$ and $C$, we can substitute convenient values of $x$ into the identity:\n\n**To find $B$, let $x = 1$:**\n\n$$\\begin{aligned} 4(1) - 5 &= B(1-2) + C(1-1) \\\\ -1 &= -B \\\\ B &= 1 \\end{aligned}$$\n\n**To find $C$, let $x = 2$:**\n\n$$\\begin{aligned} 4(2) - 5 &= B(2-2) + C(2-1) \\\\ 3 &= C \\\\ C &= 3 \\end{aligned}$$\n\nThis gives us the proper fraction decomposition:\n\n$$\\frac{4x-5}{(x-1)(x-2)} = \\frac{1}{x-1} + \\frac{3}{x-2}$$\n\n*(Note: You can also find $A=2$ by setting up the equivalent unified identity $2x^2-2x-1 = A(x-1)(x-2) + B(x-2) + C(x-1)$ as indicated in the markscheme, which yields the exact same constants.)*"},{"type":"text","order":5,"title":"Write the final answer","content":"Combining our integer quotient $2$ and the partial fraction decomposition of the proper fraction, we obtain our final answer:\n\n$$\\frac{2x^2-2x-1}{(x-1)(x-2)} = 2 + \\frac{1}{x-1} + \\frac{3}{x-2}$$"}]},{"id":"direct_identity","label":"Direct Identity Method","steps":[{"type":"text","order":0,"title":"Analyze the fraction type","content":"Before setting up our partial fractions, we first check the degrees of the numerator and denominator.\n\n* **Numerator:** $2x^2-2x-1$ is a polynomial of degree $2$.\n* **Denominator:** $(x-1)(x-2) = x^2-3x+2$ is also a polynomial of degree $2$.\n\nSince the degree of the numerator is equal to the degree of the denominator, this is an **improper algebraic fraction**. \n\nTherefore, our partial fraction expansion must include a constant term $A$ (the quotient) alongside the usual terms for linear factors: \n\n$$\\frac{2x^2-2x-1}{(x-1)(x-2)} \\equiv A + \\frac{B}{x-1} + \\frac{C}{x-2}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\frac{2x^2-2x-1}{(x-1)(x-2)} \\equiv \\htmlData{anchor=const-A}{\\color{@sky}{A}} + \\frac{\\htmlData{anchor=const-B}{\\color{@amber}{B}}}{x-1} + \\frac{\\htmlData{anchor=const-C}{\\color{@purple}{C}}}{x-2}$$"},{"text":"Multiplying both sides of the identity by the common denominator $(x-1)(x-2)$ gives:"},{"latex":"$$$2x^2-2x-1 \\equiv \\htmlData{anchor=term-A}{\\color{@sky}{A(x-1)(x-2)}} + \\htmlData{anchor=term-B}{\\color{@amber}{B(x-2)}} + \\htmlData{anchor=term-C}{\\color{@purple}{C(x-1)}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"const-A"},{"color":"amber","style":"text-color","anchor":"const-B"},{"color":"purple","style":"text-color","anchor":"const-C"}],"connections":[{"to":"term-A","from":"const-A","color":"sky","label":"multiply by (x-1)(x-2)","style":"solid"},{"to":"term-B","from":"const-B","color":"amber","label":"multiply by (x-2)","style":"solid"},{"to":"term-C","from":"const-C","color":"purple","label":"multiply by (x-1)","style":"solid"}]},"order":1,"title":"Set up the direct identity","description":"We establish the general identity directly with the quotient constant and partial fraction terms."},{"type":"text","order":2,"title":"Solve for B","content":"To solve for the constants, we can substitute convenient values of $x$ into our identity:\n\n$$2x^2-2x-1 \\equiv A(x-1)(x-2) + B(x-2) + C(x-1)$$\n\nTo eliminate $A$ and $C$, we choose $x = 1$:\n\n$$\\begin{aligned} 2(1)^2 - 2(1) - 1 &= A(1-1)(1-2) + B(1-2) + C(1-1) \\\\ 2 - 2 - 1 &= A(0) + B(-1) + C(0) \\\\ -1 &= -B \\\\ B &= 1 \\end{aligned}$$\n\nThis awards us the second mark A1."},{"type":"text","order":3,"title":"Solve for C","content":"Now we substitute $x = 2$ to eliminate the terms containing $A$ and $B$:\n\n$$\\begin{aligned} 2(2)^2 - 2(2) - 1 &= A(2-1)(2-2) + B(2-2) + C(2-1) \\\\ 2(4) - 4 - 1 &= A(0) + B(0) + C(1) \\\\ 8 - 5 &= C \\\\ C &= 3 \\end{aligned}$$\n\nThis gives us $C = 3$ and awards another mark A1."},{"type":"text","order":4,"title":"Solve for A","content":"To find $A$, we can equate the coefficients of the $x^2$ terms on both sides of our identity:\n\n* **Left-hand side ($x^2$ coefficient):** $2$\n* **Right-hand side ($x^2$ coefficient):** Expanding $A(x-1)(x-2)$ gives $A(x^2 - 3x + 2)$, so the coefficient is $A$. The terms $B(x-2)$ and $C(x-1)$ have no $x^2$ terms.\n\nEquating these directly yields:\n\n$$A = 2$$\n\nThis awards another mark A1."},{"type":"text","order":5,"title":"Write final answer","content":"Substitute our discovered constants back into our original expression to get the final form:\n\n$$\\frac{2x^2-2x-1}{(x-1)(x-2)} \\equiv 2 + \\frac{1}{x-1} + \\frac{3}{x-2}$$\n\nThis secures the final accuracy mark A1."}]}],"generatedAt":"2026-06-09T09:00:26.628Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"e32a7ea2-3837-41f8-9ce1-1bb18a4f8097: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":"amber","style":"underline","anchor":"remainder"}],"connections":[]},"order":2,"title":"Rewrite the improper fraction","description":"Using our division results, we rewrite the original expression as the sum of the constant quotient $2$ and a proper fraction with the remainder $4x-5$ in the numerator."},{"type":"text","order":3,"title":"Set up partial fractions","content":"Now we focus on decomposing the proper fraction part, $\\frac{4x-5}{(x-1)(x-2)}$, into partial fractions with linear denominators:\n\n$$\\frac{4x-5}{(x-1)(x-2)} \\equiv \\frac{B}{x-1} + \\frac{C}{x-2}$$\n\nMultiply both sides by $(x-1)(x-2)$ to clear the fractions:\n\n$$4x - 5 \\equiv B(x-2) + C(x-1)$$"},{"type":"text","order":4,"title":"Solve for B and C","content":"To find the constants $B$ and $C$, we can substitute convenient values of $x$ into the identity:\n\n**To find $B$, let $x = 1$:**\n\n$$\\begin{aligned} 4(1) - 5 &= B(1-2) + C(1-1) \\\\ -1 &= -B \\\\ B &= 1 \\end{aligned}$$\n\n**To find $C$, let $x = 2$:**\n\n$$\\begin{aligned} 4(2) - 5 &= B(2-2) + C(2-1) \\\\ 3 &= C \\\\ C &= 3 \\end{aligned}$$\n\nThis gives us the proper fraction decomposition:\n\n$$\\frac{4x-5}{(x-1)(x-2)} = \\frac{1}{x-1} + \\frac{3}{x-2}$$\n\n*(Note: You can also find $A=2$ by setting up the equivalent unified identity $2x^2-2x-1 = A(x-1)(x-2) + B(x-2) + C(x-1)$ as indicated in the markscheme, which yields the exact same constants.)*"},{"type":"text","order":5,"title":"Write the final answer","content":"Combining our integer quotient $2$ and the partial fraction decomposition of the proper fraction, we obtain our final answer:\n\n$$\\frac{2x^2-2x-1}{(x-1)(x-2)} = 2 + \\frac{1}{x-1} + \\frac{3}{x-2}$$"}]},{"id":"e32a7ea2-3837-41f8-9ce1-1bb18a4f8097:direct_identity","label":"Direct Identity Method","steps":[{"type":"text","order":0,"title":"Analyze the fraction type","content":"Before setting up our partial fractions, we first check the degrees of the numerator and denominator.\n\n* **Numerator:** $2x^2-2x-1$ is a polynomial of degree $2$.\n* **Denominator:** $(x-1)(x-2) = x^2-3x+2$ is also a polynomial of degree $2$.\n\nSince the degree of the numerator is equal to the degree of the denominator, this is an **improper algebraic fraction**. \n\nTherefore, our partial fraction expansion must include a constant term $A$ (the quotient) alongside the usual terms for linear factors: \n\n$$\\frac{2x^2-2x-1}{(x-1)(x-2)} \\equiv A + \\frac{B}{x-1} + \\frac{C}{x-2}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\frac{2x^2-2x-1}{(x-1)(x-2)} \\equiv \\htmlData{anchor=const-A}{\\color{@sky}{A}} + \\frac{\\htmlData{anchor=const-B}{\\color{@amber}{B}}}{x-1} + \\frac{\\htmlData{anchor=const-C}{\\color{@purple}{C}}}{x-2}$$"},{"text":"Multiplying both sides of the identity by the common denominator $(x-1)(x-2)$ gives:"},{"latex":"$$$2x^2-2x-1 \\equiv \\htmlData{anchor=term-A}{\\color{@sky}{A(x-1)(x-2)}} + \\htmlData{anchor=term-B}{\\color{@amber}{B(x-2)}} + \\htmlData{anchor=term-C}{\\color{@purple}{C(x-1)}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"const-A"},{"color":"amber","style":"text-color","anchor":"const-B"},{"color":"purple","style":"text-color","anchor":"const-C"}],"connections":[{"to":"term-A","from":"const-A","color":"sky","label":"multiply by (x-1)(x-2)","style":"solid"},{"to":"term-B","from":"const-B","color":"amber","label":"multiply by (x-2)","style":"solid"},{"to":"term-C","from":"const-C","color":"purple","label":"multiply by (x-1)","style":"solid"}]},"order":1,"title":"Set up the direct identity","description":"We establish the general identity directly with the quotient constant and partial fraction terms."},{"type":"text","order":2,"title":"Solve for B","content":"To solve for the constants, we can substitute convenient values of $x$ into our identity:\n\n$$2x^2-2x-1 \\equiv A(x-1)(x-2) + B(x-2) + C(x-1)$$\n\nTo eliminate $A$ and $C$, we choose $x = 1$:\n\n$$\\begin{aligned} 2(1)^2 - 2(1) - 1 &= A(1-1)(1-2) + B(1-2) + C(1-1) \\\\ 2 - 2 - 1 &= A(0) + B(-1) + C(0) \\\\ -1 &= -B \\\\ B &= 1 \\end{aligned}$$\n\nThis awards us the second mark A1."},{"type":"text","order":3,"title":"Solve for C","content":"Now we substitute $x = 2$ to eliminate the terms containing $A$ and $B$:\n\n$$\\begin{aligned} 2(2)^2 - 2(2) - 1 &= A(2-1)(2-2) + B(2-2) + C(2-1) \\\\ 2(4) - 4 - 1 &= A(0) + B(0) + C(1) \\\\ 8 - 5 &= C \\\\ C &= 3 \\end{aligned}$$\n\nThis gives us $C = 3$ and awards another mark A1."},{"type":"text","order":4,"title":"Solve for A","content":"To find $A$, we can equate the coefficients of the $x^2$ terms on both sides of our identity:\n\n* **Left-hand side ($x^2$ coefficient):** $2$\n* **Right-hand side ($x^2$ coefficient):** Expanding $A(x-1)(x-2)$ gives $A(x^2 - 3x + 2)$, so the coefficient is $A$. The terms $B(x-2)$ and $C(x-1)$ have no $x^2$ terms.\n\nEquating these directly yields:\n\n$$A = 2$$\n\nThis awards another mark A1."},{"type":"text","order":5,"title":"Write final answer","content":"Substitute our discovered constants back into our original expression to get the final form:\n\n$$\\frac{2x^2-2x-1}{(x-1)(x-2)} \\equiv 2 + \\frac{1}{x-1} + \\frac{3}{x-2}$$\n\nThis secures the final accuracy mark A1."}]}],"generatedAt":"2026-06-09T09:00:39.994Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":47},"walkthroughVideo":null},{"id":"edbf14ba-6206-41a7-aaa9-7fc7f936d530","revisionId":"57e48674-0e1a-40fb-a843-c2fd0c5ad24f","questionNumber":null,"questionType":"LA","specification":"","difficulty":1,"difficultyBand":"medium","totalMarks":3,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"a0f822bb-933b-4e3d-9c3e-ad34f9b4a2c0","label":"a","content":"Express $\\frac{13x+3}{2(3x+1)(x+1)}$ in partial fractions","markscheme":"- Sets up the identity $\\frac{13x+3}{2} = A(x+1) + B(3x+1)$ M1\n- Finds $A = -1$ and $B = \\frac{5}{2}$ A1\n- Final answer: $-\\frac{1}{3x+1} + \\frac{5}{2(x+1)}$ A1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Sets up the identity $\\frac{13x+3}{2} = A(x+1) + B(3x+1)$ or equivalent.","code":"M"},{"id":"p2","marks":1,"description":"Obtains $A = -1$ and $B = 2.5$ through substitution or coefficients comparison.","code":"A"},{"id":"p3","marks":1,"description":"Correct final partial fraction form $-\\frac{1}{3x+1} + \\frac{5}{2(x+1)}$.","code":"A"}],"finalAnswers":["-\\frac{1}{3x+1} + \\frac{5}{2(x+1)}","\\frac{2.5}{x+1} - \\frac{1}{3x+1}"],"generatedAt":"2026-05-06T14:03:46.654Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"substitution","label":"Substitution Method","steps":[{"type":"text","order":0,"title":"Set up the identity","content":"To express the algebraic fraction in partial fractions, we first write it in the standard partial fraction form for distinct linear factors in the denominator:\n\n$$\\frac{13x+3}{2(3x+1)(x+1)} = \\frac{A}{3x+1} + \\frac{B}{x+1}$$\n\nNext, we multiply both sides by the denominator of the left-hand side, which is $2(3x+1)(x+1)$, or we can keep the constant $2$ on the left side to match the markscheme's formulation. \n\nMultiplying both sides of the equation by $(3x+1)(x+1)$ gives the identity:\n\n$$\\frac{13x+3}{2} = A(x+1) + B(3x+1)$$\n\nThis is a crucial first step that earns you **1 mark** (M1)."},{"type":"text","order":1,"title":"Substitute $x = -1$","content":"To find the constant $B$, we can eliminate the term containing $A$ by substituting $x = -1$ (the root of $x+1 = 0$) into the identity:\n\n$$\\frac{13(-1)+3}{2} = A(-1+1) + B(3(-1)+1)$$\n\nNow, let's simplify both sides:\n\n$$\\frac{-13+3}{2} = A(0) + B(-3+1)$$\n\n$$\\frac{-10}{2} = -2B$$\n\n$$-5 = -2B \\implies B = \\frac{5}{2}$$"},{"type":"text","order":2,"title":"Substitute $x = -\\frac{1}{3}$","content":"To find the constant $A$, we can eliminate the term containing $B$ by substituting $x = -\\frac{1}{3}$ (the root of $3x+1 = 0$) into the identity:\n\n$$\\frac{13\\left(-\\frac{1}{3}\\right)+3}{2} = A\\left(-\\frac{1}{3}+1\\right) + B\\left(3\\left(-\\frac{1}{3}\\right)+1\\right)$$\n\nSimplifying both sides:\n\n$$\\frac{-\\frac{13}{3}+3}{2} = A\\left(\\frac{2}{3}\\right) + B(0)$$\n\n$$\\frac{-\\frac{4}{3}}{2} = \\frac{2}{3}A$$\n\n$$-\\frac{2}{3} = \\frac{2}{3}A \\implies A = -1$$\n\nFinding these correct values of $A = -1$ and $B = \\frac{5}{2}$ earns you **1 mark** (A1)."},{"type":"text","order":3,"title":"Write the final answer","content":"Now we substitute the values of $A = -1$ and $B = \\frac{5}{2}$ back into our original partial fraction expression:\n\n$$\\frac{-1}{3x+1} + \\frac{\\frac{5}{2}}{x+1}$$\n\nWe simplify this to get the final form:\n\n$$-\\frac{1}{3x+1} + \\frac{5}{2(x+1)}$$\n\nWriting this final correct expression earns you the final **1 mark** (A1)."}]},{"id":"equating_coefficients","label":"Equating Coefficients","steps":[{"type":"text","order":0,"title":"Set up the identity","content":"To express the algebraic fraction in partial fractions, we first write it in the form:\n\n$$\\frac{13x+3}{2(3x+1)(x+1)} \\equiv \\frac{A}{3x+1} + \\frac{B}{x+1}$$\n\nTo make it easier to work with, we can group the factor of $2$ in the denominator with the numerator:\n\n$$\\frac{\\frac{13x+3}{2}}{(3x+1)(x+1)} \\equiv \\frac{A(x+1) + B(3x+1)}{(3x+1)(x+1)}$$\n\nMultiplying both sides by the common denominator $(3x+1)(x+1)$, we obtain the identity:\n\n$$\\frac{13x+3}{2} \\equiv A(x+1) + B(3x+1)$$\n\nThis is our starting point to solve for the constants $A$ and $B$."},{"type":"text","order":1,"title":"Expand and group terms","content":"Using the **Equating Coefficients** approach, we expand the right-hand side of the identity and then group the terms by the powers of $x$.\n\nExpanding the right-hand side:\n\n$$A(x+1) + B(3x+1) = Ax + A + 3Bx + B$$\n\nGrouping the $x$ terms and the constant terms together:\n\n$$(A + 3B)x + (A + B)$$\n\nNow, our identity is written as:\n\n$$\\frac{13}{2}x + \\frac{3}{2} \\equiv (A + 3B)x + (A + B)$$\n\nWe can now equate the coefficients on both sides of the identity."},{"type":"text","order":2,"title":"Create simultaneous equations","content":"By equating the coefficients of $x$ and the constant terms from both sides, we get a system of two linear equations:\n\n1. For the coefficients of $x$:\n $$A + 3B = \\frac{13}{2}$$\n\n2. For the constant terms:\n $$A + B = \\frac{3}{2}$$\n\nWe can solve this system to find the values of $A$ and $B$."},{"type":"text","order":3,"title":"Solve the system","content":"Let's subtract the second equation from the first to eliminate $A$:\n\n$$\\begin{aligned} (A + 3B) - (A + B) &= \\frac{13}{2} - \\frac{3}{2} \\\\ 2B &= \\frac{10}{2} \\\\ 2B &= 5 \\\\ B &= \\frac{5}{2} \\end{aligned}$$\n\nNow, substitute $B = \\frac{5}{2}$ back into the second equation to find $A$:\n\n$$\\begin{aligned} A + \\frac{5}{2} &= \\frac{3}{2} \\\\ A &= \\frac{3}{2} - \\frac{5}{2} \\\\ A &= -1 \\end{aligned}$$\n\nThis gives us $A = -1$ and $B = \\frac{5}{2}$."},{"type":"text","order":4,"title":"Write the final answer","content":"Substitute the values of $A = -1$ and $B = \\frac{5}{2}$ back into our original partial fractions form:\n\n$$\\frac{-1}{3x+1} + \\frac{\\frac{5}{2}}{x+1}$$\n\nSimplifying the second term, we get our final expression:\n\n$$-\\frac{1}{3x+1} + \\frac{5}{2(x+1)}$$"}]}],"generatedAt":"2026-06-09T09:38:22.213Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"a0f822bb-933b-4e3d-9c3e-ad34f9b4a2c0:substitution","label":"Substitution Method","steps":[{"type":"text","order":0,"title":"Set up the identity","content":"To express the algebraic fraction in partial fractions, we first write it in the standard partial fraction form for distinct linear factors in the denominator:\n\n$$\\frac{13x+3}{2(3x+1)(x+1)} = \\frac{A}{3x+1} + \\frac{B}{x+1}$$\n\nNext, we multiply both sides by the denominator of the left-hand side, which is $2(3x+1)(x+1)$, or we can keep the constant $2$ on the left side to match the markscheme's formulation. \n\nMultiplying both sides of the equation by $(3x+1)(x+1)$ gives the identity:\n\n$$\\frac{13x+3}{2} = A(x+1) + B(3x+1)$$\n\nThis is a crucial first step that earns you **1 mark** (M1)."},{"type":"text","order":1,"title":"Substitute $x = -1$","content":"To find the constant $B$, we can eliminate the term containing $A$ by substituting $x = -1$ (the root of $x+1 = 0$) into the identity:\n\n$$\\frac{13(-1)+3}{2} = A(-1+1) + B(3(-1)+1)$$\n\nNow, let's simplify both sides:\n\n$$\\frac{-13+3}{2} = A(0) + B(-3+1)$$\n\n$$\\frac{-10}{2} = -2B$$\n\n$$-5 = -2B \\implies B = \\frac{5}{2}$$"},{"type":"text","order":2,"title":"Substitute $x = -\\frac{1}{3}$","content":"To find the constant $A$, we can eliminate the term containing $B$ by substituting $x = -\\frac{1}{3}$ (the root of $3x+1 = 0$) into the identity:\n\n$$\\frac{13\\left(-\\frac{1}{3}\\right)+3}{2} = A\\left(-\\frac{1}{3}+1\\right) + B\\left(3\\left(-\\frac{1}{3}\\right)+1\\right)$$\n\nSimplifying both sides:\n\n$$\\frac{-\\frac{13}{3}+3}{2} = A\\left(\\frac{2}{3}\\right) + B(0)$$\n\n$$\\frac{-\\frac{4}{3}}{2} = \\frac{2}{3}A$$\n\n$$-\\frac{2}{3} = \\frac{2}{3}A \\implies A = -1$$\n\nFinding these correct values of $A = -1$ and $B = \\frac{5}{2}$ earns you **1 mark** (A1)."},{"type":"text","order":3,"title":"Write the final answer","content":"Now we substitute the values of $A = -1$ and $B = \\frac{5}{2}$ back into our original partial fraction expression:\n\n$$\\frac{-1}{3x+1} + \\frac{\\frac{5}{2}}{x+1}$$\n\nWe simplify this to get the final form:\n\n$$-\\frac{1}{3x+1} + \\frac{5}{2(x+1)}$$\n\nWriting this final correct expression earns you the final **1 mark** (A1)."}]},{"id":"a0f822bb-933b-4e3d-9c3e-ad34f9b4a2c0:equating_coefficients","label":"Equating Coefficients","steps":[{"type":"text","order":0,"title":"Set up the identity","content":"To express the algebraic fraction in partial fractions, we first write it in the form:\n\n$$\\frac{13x+3}{2(3x+1)(x+1)} \\equiv \\frac{A}{3x+1} + \\frac{B}{x+1}$$\n\nTo make it easier to work with, we can group the factor of $2$ in the denominator with the numerator:\n\n$$\\frac{\\frac{13x+3}{2}}{(3x+1)(x+1)} \\equiv \\frac{A(x+1) + B(3x+1)}{(3x+1)(x+1)}$$\n\nMultiplying both sides by the common denominator $(3x+1)(x+1)$, we obtain the identity:\n\n$$\\frac{13x+3}{2} \\equiv A(x+1) + B(3x+1)$$\n\nThis is our starting point to solve for the constants $A$ and $B$."},{"type":"text","order":1,"title":"Expand and group terms","content":"Using the **Equating Coefficients** approach, we expand the right-hand side of the identity and then group the terms by the powers of $x$.\n\nExpanding the right-hand side:\n\n$$A(x+1) + B(3x+1) = Ax + A + 3Bx + B$$\n\nGrouping the $x$ terms and the constant terms together:\n\n$$(A + 3B)x + (A + B)$$\n\nNow, our identity is written as:\n\n$$\\frac{13}{2}x + \\frac{3}{2} \\equiv (A + 3B)x + (A + B)$$\n\nWe can now equate the coefficients on both sides of the identity."},{"type":"text","order":2,"title":"Create simultaneous equations","content":"By equating the coefficients of $x$ and the constant terms from both sides, we get a system of two linear equations:\n\n1. For the coefficients of $x$:\n $$A + 3B = \\frac{13}{2}$$\n\n2. For the constant terms:\n $$A + B = \\frac{3}{2}$$\n\nWe can solve this system to find the values of $A$ and $B$."},{"type":"text","order":3,"title":"Solve the system","content":"Let's subtract the second equation from the first to eliminate $A$:\n\n$$\\begin{aligned} (A + 3B) - (A + B) &= \\frac{13}{2} - \\frac{3}{2} \\\\ 2B &= \\frac{10}{2} \\\\ 2B &= 5 \\\\ B &= \\frac{5}{2} \\end{aligned}$$\n\nNow, substitute $B = \\frac{5}{2}$ back into the second equation to find $A$:\n\n$$\\begin{aligned} A + \\frac{5}{2} &= \\frac{3}{2} \\\\ A &= \\frac{3}{2} - \\frac{5}{2} \\\\ A &= -1 \\end{aligned}$$\n\nThis gives us $A = -1$ and $B = \\frac{5}{2}$."},{"type":"text","order":4,"title":"Write the final answer","content":"Substitute the values of $A = -1$ and $B = \\frac{5}{2}$ back into our original partial fractions form:\n\n$$\\frac{-1}{3x+1} + \\frac{\\frac{5}{2}}{x+1}$$\n\nSimplifying the second term, we get our final expression:\n\n$$-\\frac{1}{3x+1} + \\frac{5}{2(x+1)}$$"}]}],"generatedAt":"2026-06-09T09:38:59.934Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":67},"walkthroughVideo":null},{"id":"d877c2ca-f576-439d-9a69-de51fbdd6d47","revisionId":"155f838b-750f-44aa-9740-a0d859ea2012","questionNumber":null,"questionType":"LA","specification":"","difficulty":1,"difficultyBand":"medium","totalMarks":3,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"9f307832-ffe3-4a13-bc46-654bd2c89dfe","label":"a","content":"Express $\\frac{5x+2}{(x-2)(x+2)}$ in the form $\\frac{A}{(x-2)} + \\frac{B}{(x+2)}$","markscheme":"- Sets up the identity $5x+2 = A(x+2) + B(x-2)$ M1\n- Finds $A = 3$ and $B = 2$ by substituting $x = 2$ and $x = -2$ A1\n- Final answer: $\\frac{3}{(x-2)} + \\frac{2}{(x+2)}$ A1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Establishes the identity $5x+2 = A(x+2) + B(x-2)$.","code":"M"},{"id":"p2","marks":1,"description":"Finds the correct constants $A = 3$ and $B = 2$.","code":"A"},{"id":"p3","marks":1,"description":"Writes the final expression in the form $\\frac{3}{(x-2)} + \\frac{2}{(x+2)}$.","code":"A"}],"finalAnswers":["\\frac{3}{(x-2)} + \\frac{2}{(x+2)}"],"generatedAt":"2026-05-06T14:03:32.967Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"substitution","label":"Substitution Method","steps":[{"type":"text","order":0,"title":"Set up the identity","content":"To express the algebraic fraction as partial fractions, we begin by setting up an identity. We equate the original fraction to the sum of the partial fractions:\n\n$$\\frac{5x+2}{(x-2)(x+2)} \\equiv \\frac{A}{x-2} + \\frac{B}{x+2}$$\n\nTo clear the denominators, we multiply both sides of the identity by the common denominator $(x-2)(x+2)$:\n\n$$5x+2 \\equiv A(x+2) + B(x-2)$$\n\nSetting up this correct identity earns the first method mark (M1)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$5\\htmlData{anchor=x-val-2}{\\color{@sky}{x}} + 2 \\equiv A(\\htmlData{anchor=x-sub-2-a}{\\color{@sky}{x}} + 2) + B(\\htmlData{anchor=x-sub-2-b}{\\color{@sky}{x}} - 2)$"},{"latex":"$$\\text{Substitute } \\htmlData{anchor=sub-val-2}{\\color{@sky}{x = 2}}:$"},{"latex":"$$5(\\color{@sky}{2}) + 2 = A(\\color{@sky}{2} + 2) + B(\\color{@sky}{2} - 2)$"},{"latex":"$$12 = 4A + 0 \\implies \\htmlData{anchor=val-a}{\\color{@sky}{A = 3}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sub-val-2"},{"color":"sky","style":"box","anchor":"val-a"}],"connections":[{"to":"x-val-2","from":"sub-val-2","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Find A using substitution","description":"We can choose any value of $x$ to substitute into our identity. Choosing $x = 2$ is strategic because it makes the $(x - 2)$ term zero, allowing us to isolate and solve for $A$ directly."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$5\\htmlData{anchor=x-val-neg2}{\\color{@amber}{x}} + 2 \\equiv A(\\htmlData{anchor=x-sub-neg2-a}{\\color{@amber}{x}} + 2) + B(\\htmlData{anchor=x-sub-neg2-b}{\\color{@amber}{x}} - 2)$"},{"latex":"$$\\text{Substitute } \\htmlData{anchor=sub-val-neg2}{\\color{@amber}{x = -2}}:$"},{"latex":"$$5(\\color{@amber}{-2}) + 2 = A(\\color{@amber}{-2} + 2) + B(\\color{@amber}{-2} - 2)$"},{"latex":"$$-8 = 0 - 4B \\implies \\htmlData{anchor=val-b}{\\color{@amber}{B = 2}}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"sub-val-neg2"},{"color":"amber","style":"box","anchor":"val-b"}],"connections":[{"to":"x-val-neg2","from":"sub-val-neg2","color":"amber","label":null,"style":"solid"}]},"order":2,"title":"Find B using substitution","description":"Next, we substitute $x = -2$ to make the $(x + 2)$ term zero, which eliminates the $A$ term and lets us solve for $B$ directly. Correctly finding both values of $A$ and $B$ earns the second mark (A1)."},{"type":"text","order":3,"title":"State the final expression","content":"Now we substitute our values $A = 3$ and $B = 2$ back into the partial fraction template:\n\n$$\\frac{3}{x-2} + \\frac{2}{x+2}$$\n\nThis final answer completes the expression and earns the third mark (A1)."}]},{"id":"equating_coefficients","label":"Equating Coefficients","steps":[{"type":"text","order":0,"title":"Set up the identity","content":"To express the algebraic fraction as partial fractions, we start by setting up an identity with the unknown constants $A$ and $B$:\n\n$$\\frac{5x+2}{(x-2)(x+2)} \\equiv \\frac{A}{x-2} + \\frac{B}{x+2}$$\n\nTo clear the denominators, we multiply both sides of the identity by the common denominator, $(x-2)(x+2)$:\n\n$$5x+2 \\equiv A(x+2) + B(x-2)$$\n\nThis gives us our starting equation for finding the values of $A$ and $B$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$5x+2 \\equiv \\htmlData{anchor=expand1}{\\color{@sky}{A(x+2)}} + \\htmlData{anchor=expand2}{\\color{@amber}{B(x-2)}}$"},{"latex":"$$5x+2 \\equiv \\color{@sky}{Ax} + \\color{@sky}{2A} + \\color{@amber}{Bx} - \\color{@amber}{2B}$"},{"latex":"$$5x+2 \\equiv \\htmlData{anchor=grouped}{\\color{@purple}{(A+B)x + (2A-2B)}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"expand1"},{"color":"amber","style":"text-color","anchor":"expand2"},{"color":"purple","style":"box","anchor":"grouped"}],"connections":[{"to":"grouped","from":"expand1","color":"purple","label":"Group coefficients","style":"solid"}]},"order":1,"title":"Expand and group terms","description":"Now, we expand the brackets on the right-hand side and collect like terms to group the coefficients of $x$ and the constant terms."},{"type":"text","order":2,"title":"Equate the coefficients","content":"Since this is an identity, the coefficients on both sides of the equation must be equal. We can equate the coefficients of $x$ and the constant terms:\n\n* **For the coefficients of $x$:**\n $$A+B = 5$$ (Equation 1)\n\n* **For the constant terms:**\n $$2A-2B = 2$$ \n\nWe can simplify the constant term equation by dividing both sides by $2$:\n $$A-B = 1$$ (Equation 2)"},{"type":"text","order":3,"title":"Solve the system","content":"Now we can solve this system of linear equations.\n\nFirst, we add Equation 1 and Equation 2 together:\n$$(A+B) + (A-B) = 5 + 1$$\n\n$$2A = 6$$\n\n$$A = 3$$\n\nNext, we substitute $A = 3$ back into Equation 1 to find $B$:\n\n$$3+B = 5$$\n\n$$B = 2$$\n\nSo, our constants are $A = 3$ and $B = 2$."},{"type":"text","order":4,"title":"State final answer","content":"Substitute the values of $A$ and $B$ back into our original partial fractions structure to write down our final answer:\n\n$$\\frac{5x+2}{(x-2)(x+2)} \\equiv \\frac{3}{x-2} + \\frac{2}{x+2}$$"}]}],"generatedAt":"2026-06-09T09:18:51.568Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"9f307832-ffe3-4a13-bc46-654bd2c89dfe:substitution","label":"Substitution Method","steps":[{"type":"text","order":0,"title":"Set up the identity","content":"To express the algebraic fraction as partial fractions, we begin by setting up an identity. We equate the original fraction to the sum of the partial fractions:\n\n$$\\frac{5x+2}{(x-2)(x+2)} \\equiv \\frac{A}{x-2} + \\frac{B}{x+2}$$\n\nTo clear the denominators, we multiply both sides of the identity by the common denominator $(x-2)(x+2)$:\n\n$$5x+2 \\equiv A(x+2) + B(x-2)$$\n\nSetting up this correct identity earns the first method mark (M1)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$5\\htmlData{anchor=x-val-2}{\\color{@sky}{x}} + 2 \\equiv A(\\htmlData{anchor=x-sub-2-a}{\\color{@sky}{x}} + 2) + B(\\htmlData{anchor=x-sub-2-b}{\\color{@sky}{x}} - 2)$"},{"latex":"$$\\text{Substitute } \\htmlData{anchor=sub-val-2}{\\color{@sky}{x = 2}}:$"},{"latex":"$$5(\\color{@sky}{2}) + 2 = A(\\color{@sky}{2} + 2) + B(\\color{@sky}{2} - 2)$"},{"latex":"$$12 = 4A + 0 \\implies \\htmlData{anchor=val-a}{\\color{@sky}{A = 3}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sub-val-2"},{"color":"sky","style":"box","anchor":"val-a"}],"connections":[{"to":"x-val-2","from":"sub-val-2","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Find A using substitution","description":"We can choose any value of $x$ to substitute into our identity. Choosing $x = 2$ is strategic because it makes the $(x - 2)$ term zero, allowing us to isolate and solve for $A$ directly."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$5\\htmlData{anchor=x-val-neg2}{\\color{@amber}{x}} + 2 \\equiv A(\\htmlData{anchor=x-sub-neg2-a}{\\color{@amber}{x}} + 2) + B(\\htmlData{anchor=x-sub-neg2-b}{\\color{@amber}{x}} - 2)$"},{"latex":"$$\\text{Substitute } \\htmlData{anchor=sub-val-neg2}{\\color{@amber}{x = -2}}:$"},{"latex":"$$5(\\color{@amber}{-2}) + 2 = A(\\color{@amber}{-2} + 2) + B(\\color{@amber}{-2} - 2)$"},{"latex":"$$-8 = 0 - 4B \\implies \\htmlData{anchor=val-b}{\\color{@amber}{B = 2}}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"sub-val-neg2"},{"color":"amber","style":"box","anchor":"val-b"}],"connections":[{"to":"x-val-neg2","from":"sub-val-neg2","color":"amber","label":null,"style":"solid"}]},"order":2,"title":"Find B using substitution","description":"Next, we substitute $x = -2$ to make the $(x + 2)$ term zero, which eliminates the $A$ term and lets us solve for $B$ directly. Correctly finding both values of $A$ and $B$ earns the second mark (A1)."},{"type":"text","order":3,"title":"State the final expression","content":"Now we substitute our values $A = 3$ and $B = 2$ back into the partial fraction template:\n\n$$\\frac{3}{x-2} + \\frac{2}{x+2}$$\n\nThis final answer completes the expression and earns the third mark (A1)."}]},{"id":"9f307832-ffe3-4a13-bc46-654bd2c89dfe:equating_coefficients","label":"Equating Coefficients","steps":[{"type":"text","order":0,"title":"Set up the identity","content":"To express the algebraic fraction as partial fractions, we start by setting up an identity with the unknown constants $A$ and $B$:\n\n$$\\frac{5x+2}{(x-2)(x+2)} \\equiv \\frac{A}{x-2} + \\frac{B}{x+2}$$\n\nTo clear the denominators, we multiply both sides of the identity by the common denominator, $(x-2)(x+2)$:\n\n$$5x+2 \\equiv A(x+2) + B(x-2)$$\n\nThis gives us our starting equation for finding the values of $A$ and $B$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$5x+2 \\equiv \\htmlData{anchor=expand1}{\\color{@sky}{A(x+2)}} + \\htmlData{anchor=expand2}{\\color{@amber}{B(x-2)}}$"},{"latex":"$$5x+2 \\equiv \\color{@sky}{Ax} + \\color{@sky}{2A} + \\color{@amber}{Bx} - \\color{@amber}{2B}$"},{"latex":"$$5x+2 \\equiv \\htmlData{anchor=grouped}{\\color{@purple}{(A+B)x + (2A-2B)}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"expand1"},{"color":"amber","style":"text-color","anchor":"expand2"},{"color":"purple","style":"box","anchor":"grouped"}],"connections":[{"to":"grouped","from":"expand1","color":"purple","label":"Group coefficients","style":"solid"}]},"order":1,"title":"Expand and group terms","description":"Now, we expand the brackets on the right-hand side and collect like terms to group the coefficients of $x$ and the constant terms."},{"type":"text","order":2,"title":"Equate the coefficients","content":"Since this is an identity, the coefficients on both sides of the equation must be equal. We can equate the coefficients of $x$ and the constant terms:\n\n* **For the coefficients of $x$:**\n $$A+B = 5$$ (Equation 1)\n\n* **For the constant terms:**\n $$2A-2B = 2$$ \n\nWe can simplify the constant term equation by dividing both sides by $2$:\n $$A-B = 1$$ (Equation 2)"},{"type":"text","order":3,"title":"Solve the system","content":"Now we can solve this system of linear equations.\n\nFirst, we add Equation 1 and Equation 2 together:\n$$(A+B) + (A-B) = 5 + 1$$\n\n$$2A = 6$$\n\n$$A = 3$$\n\nNext, we substitute $A = 3$ back into Equation 1 to find $B$:\n\n$$3+B = 5$$\n\n$$B = 2$$\n\nSo, our constants are $A = 3$ and $B = 2$."},{"type":"text","order":4,"title":"State final answer","content":"Substitute the values of $A$ and $B$ back into our original partial fractions structure to write down our final answer:\n\n$$\\frac{5x+2}{(x-2)(x+2)} \\equiv \\frac{3}{x-2} + \\frac{2}{x+2}$$"}]}],"generatedAt":"2026-06-09T09:19:04.651Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths","summaries":[{"id":"f209f2a2-5723-43ba-82a1-fc17edfc8c67","difficultyBand":"easy","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"9d3c6e50-5295-4a06-b680-b33bd067334d","difficultyBand":"medium","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"ccd01f36-c50a-4e45-ad97-44a918590c37","difficultyBand":"medium","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"61bf9309-865b-4108-a293-c8fc670013b0","difficultyBand":"medium","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"ba7e7e9f-cffd-4764-bae7-369bfddc13e7","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"bf5c00ea-3f06-43fb-bd0d-ec2689b1e986","difficultyBand":"easy","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"edbf14ba-6206-41a7-aaa9-7fc7f936d530","difficultyBand":"medium","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"d877c2ca-f576-439d-9a69-de51fbdd6d47","difficultyBand":"medium","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"f9c1abed-3a80-46e0-87b8-dba96c072800","difficultyBand":"medium","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"1fa2f970-a7dc-494e-8df0-cb6f72019df4","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"5e736fe7-d52f-4446-9999-56c60bf286a7","difficultyBand":"hard","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"60f1d250-7fc0-4450-bfae-f94c3355b1db","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"faa915aa-44fb-41cd-9000-18defed8f4cf","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true}],"title":"1.3 Partial Fractions","topicId":"5d96c6a6-9bd0-49bd-9a4a-f7ce0af48c3d","topicSidebar":"$L3b","topicSlug":"a-level-pure-mathematics-1-3-partial-fractions","topicTitle":"1.3 Partial Fractions"}]}],"$L3c","$L3d"]}]

1.3 Partial Fractions

Question 5

1.3 Partial Fractions Questions

1.3 Partial Fractions

Question 5

1.3 Partial Fractions Questions