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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":3,"initialQuestionWindow":[{"id":"ccd01f36-c50a-4e45-ad97-44a918590c37","revisionId":"f439f861-1b8c-4996-bc35-74f23bb6c9e5","questionNumber":null,"questionType":"LA","specification":"","difficulty":1,"difficultyBand":"medium","totalMarks":3,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"bb84d143-6ec7-457d-afa0-df137f4aed9d","label":"a","content":"Express $\\frac{3x-1}{5(x+1)(x-3)}$ in partial fractions","markscheme":"- Sets up the identity $\\frac{3x-1}{5} = A(x-3) + B(x+1)$ M1\n- Finds $A = \\frac{1}{5}$ and $B = \\frac{2}{5}$ A1\n- Final answer: $\\frac{1}{5(x+1)} + \\frac{2}{5(x-3)}$ A1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Sets up the correct partial fraction identity by clearing the common denominator.","code":"M"},{"id":"p2","marks":1,"description":"Solves for the constants $A$ and $B$ correctly.","code":"A"},{"id":"p3","marks":1,"description":"States the final expression in partial fractions correctly.","code":"A"}],"finalAnswers":["\\frac{1}{5(x+1)} + \\frac{2}{5(x-3)}","\\frac{0.2}{x+1} + \\frac{0.4}{x-3}"],"generatedAt":"2026-05-06T14:03:44.609Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"substitution","label":"Substitution Method","steps":[{"type":"text","order":0,"title":"Set up partial fractions","content":"To express the algebraic fraction in partial fractions, we first set up the standard identity.\n\nWe can factor out the constant $\\frac{1}{5}$ or include it in the denominators. Let's write the decomposition as:\n\n$$\\frac{3x-1}{5(x+1)(x-3)} \\equiv \\frac{A}{5(x+1)} + \\frac{B}{5(x-3)}$$\n\nTo make the substitution method simpler, we can multiply both sides of the identity by the entire denominator $5(x+1)(x-3)$ to eliminate the fractions:\n\n$$3x - 1 \\equiv A(x-3) + B(x+1)$$"},{"type":"text","order":1,"title":"Substitute the first root","content":"To find $A$, we can substitute the root of the other linear factor, which is $x = -1$, into our identity to eliminate $B$:\n\n$$3(-1) - 1 = A(-1 - 3) + B(-1 + 1)$$\n\n$$-4 = -4A$$\n\n$$A = 1$$"},{"type":"text","order":2,"title":"Substitute the second root","content":"Next, to find $B$, we substitute the root $x = 3$ to eliminate $A$:\n\n$$3(3) - 1 = A(3 - 3) + B(3 + 1)$$\n\n$$8 = 4B$$\n\n$$B = 2$$"},{"type":"text","order":3,"title":"Form final partial fractions","content":"Now that we have $A = 1$ and $B = 2$, we can substitute these values back into our original setup:\n\n$$\\frac{3x-1}{5(x+1)(x-3)} = \\frac{1}{5(x+1)} + \\frac{2}{5(x-3)}$$\n\nThis matches the final answer from the markscheme."}]},{"id":"equating_coefficients","label":"Equating Coefficients Method","steps":[{"type":"text","order":0,"title":"Set up the decomposition","content":"To express the expression as partial fractions, we set up an identity where each linear factor in the denominator gets its own fraction:\n\n$$\\frac{3x-1}{5(x+1)(x-3)} \\equiv \\frac{A}{x+1} + \\frac{B}{x-3}$$\n\nTo find the constants $A$ and $B$, we multiply both sides of the identity by the entire denominator $5(x+1)(x-3)$:\n\n$$3x - 1 \\equiv 5A(x-3) + 5B(x+1)$$\n\nAlternatively, we can group the factor of $\\frac{1}{5}$ with the numerators and write:\n\n$$\\frac{3x-1}{5} \\equiv A(x-3) + B(x+1)$$\n\nLet's use this simplified version to find $A$ and $B$ by equating coefficients."},{"type":"text","order":1,"title":"Expand and group terms","content":"We expand the right-hand side of the identity and collect the terms with $x$ and the constant terms together:\n\n$$\\frac{3}{5}x - \\frac{1}{5} \\equiv A(x-3) + B(x+1)$$\n\n$$\\frac{3}{5}x - \\frac{1}{5} \\equiv (A+B)x + (-3A+B)$$"},{"type":"text","order":2,"title":"Equate the coefficients","content":"By equating the coefficients of $x$ and the constant terms from both sides of the identity, we get a system of simultaneous equations:\n\n**For $x$ coefficients:**\n$$A + B = \\frac{3}{5}$$\n\n**For constant terms:**\n$$-3A + B = -\\frac{1}{5}$$"},{"type":"text","order":3,"title":"Solve the equations","content":"We can solve this system by subtracting the second equation from the first equation to eliminate $B$:\n\n$$(A + B) - (-3A + B) = \\frac{3}{5} - \\left(-\\frac{1}{5}\\right)$$\n\n$$4A = \\frac{4}{5}$$\n\n$$A = \\frac{1}{5}$$\n\nNow substitute $A = \\frac{1}{5}$ back into the first equation to find $B$:\n\n$$\\frac{1}{5} + B = \\frac{3}{5}$$\n\n$$B = \\frac{2}{5}$$"},{"type":"text","order":4,"title":"State the final fraction","content":"Substitute our values for $A$ and $B$ back into the original partial fraction setup:\n\n$$\\frac{3x-1}{5(x+1)(x-3)} \\equiv \\frac{\\frac{1}{5}}{x+1} + \\frac{\\frac{2}{5}}{x-3}$$\n\nSimplifying this gives the final answer:\n\n$$\\frac{1}{5(x+1)} + \\frac{2}{5(x-3)}$$"}]}],"generatedAt":"2026-06-09T09:06:55.897Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"bb84d143-6ec7-457d-afa0-df137f4aed9d:substitution","label":"Substitution Method","steps":[{"type":"text","order":0,"title":"Set up partial fractions","content":"To express the algebraic fraction in partial fractions, we first set up the standard identity.\n\nWe can factor out the constant $\\frac{1}{5}$ or include it in the denominators. Let's write the decomposition as:\n\n$$\\frac{3x-1}{5(x+1)(x-3)} \\equiv \\frac{A}{5(x+1)} + \\frac{B}{5(x-3)}$$\n\nTo make the substitution method simpler, we can multiply both sides of the identity by the entire denominator $5(x+1)(x-3)$ to eliminate the fractions:\n\n$$3x - 1 \\equiv A(x-3) + B(x+1)$$"},{"type":"text","order":1,"title":"Substitute the first root","content":"To find $A$, we can substitute the root of the other linear factor, which is $x = -1$, into our identity to eliminate $B$:\n\n$$3(-1) - 1 = A(-1 - 3) + B(-1 + 1)$$\n\n$$-4 = -4A$$\n\n$$A = 1$$"},{"type":"text","order":2,"title":"Substitute the second root","content":"Next, to find $B$, we substitute the root $x = 3$ to eliminate $A$:\n\n$$3(3) - 1 = A(3 - 3) + B(3 + 1)$$\n\n$$8 = 4B$$\n\n$$B = 2$$"},{"type":"text","order":3,"title":"Form final partial fractions","content":"Now that we have $A = 1$ and $B = 2$, we can substitute these values back into our original setup:\n\n$$\\frac{3x-1}{5(x+1)(x-3)} = \\frac{1}{5(x+1)} + \\frac{2}{5(x-3)}$$\n\nThis matches the final answer from the markscheme."}]},{"id":"bb84d143-6ec7-457d-afa0-df137f4aed9d:equating_coefficients","label":"Equating Coefficients Method","steps":[{"type":"text","order":0,"title":"Set up the decomposition","content":"To express the expression as partial fractions, we set up an identity where each linear factor in the denominator gets its own fraction:\n\n$$\\frac{3x-1}{5(x+1)(x-3)} \\equiv \\frac{A}{x+1} + \\frac{B}{x-3}$$\n\nTo find the constants $A$ and $B$, we multiply both sides of the identity by the entire denominator $5(x+1)(x-3)$:\n\n$$3x - 1 \\equiv 5A(x-3) + 5B(x+1)$$\n\nAlternatively, we can group the factor of $\\frac{1}{5}$ with the numerators and write:\n\n$$\\frac{3x-1}{5} \\equiv A(x-3) + B(x+1)$$\n\nLet's use this simplified version to find $A$ and $B$ by equating coefficients."},{"type":"text","order":1,"title":"Expand and group terms","content":"We expand the right-hand side of the identity and collect the terms with $x$ and the constant terms together:\n\n$$\\frac{3}{5}x - \\frac{1}{5} \\equiv A(x-3) + B(x+1)$$\n\n$$\\frac{3}{5}x - \\frac{1}{5} \\equiv (A+B)x + (-3A+B)$$"},{"type":"text","order":2,"title":"Equate the coefficients","content":"By equating the coefficients of $x$ and the constant terms from both sides of the identity, we get a system of simultaneous equations:\n\n**For $x$ coefficients:**\n$$A + B = \\frac{3}{5}$$\n\n**For constant terms:**\n$$-3A + B = -\\frac{1}{5}$$"},{"type":"text","order":3,"title":"Solve the equations","content":"We can solve this system by subtracting the second equation from the first equation to eliminate $B$:\n\n$$(A + B) - (-3A + B) = \\frac{3}{5} - \\left(-\\frac{1}{5}\\right)$$\n\n$$4A = \\frac{4}{5}$$\n\n$$A = \\frac{1}{5}$$\n\nNow substitute $A = \\frac{1}{5}$ back into the first equation to find $B$:\n\n$$\\frac{1}{5} + B = \\frac{3}{5}$$\n\n$$B = \\frac{2}{5}$$"},{"type":"text","order":4,"title":"State the final fraction","content":"Substitute our values for $A$ and $B$ back into the original partial fraction setup:\n\n$$\\frac{3x-1}{5(x+1)(x-3)} \\equiv \\frac{\\frac{1}{5}}{x+1} + \\frac{\\frac{2}{5}}{x-3}$$\n\nSimplifying this gives the final answer:\n\n$$\\frac{1}{5(x+1)} + \\frac{2}{5(x-3)}$$"}]}],"generatedAt":"2026-06-09T09:07:06.616Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":59},"walkthroughVideo":null},{"id":"61bf9309-865b-4108-a293-c8fc670013b0","revisionId":"e5ce1501-2cdd-4aff-88cb-b40183b2651f","questionNumber":"2","questionType":"LA","specification":"","difficulty":1,"difficultyBand":"medium","totalMarks":3,"sourceTitle":"Question Set 1: Algebraic Methods","calculatorUse":null,"isCalculation":true,"parts":[{"id":"64510fdf-2642-4f2a-817e-37f3a371759b","label":"","content":"Express $\\frac{9x-5}{2(2x-1)(x-1)}$ in partial fractions","markscheme":"- Sets up the identity $\\frac{9x-5}{2} = A(x-1) + B(2x-1)$ M1\n- Finds $A = \\frac{1}{2}$ and $B = 2$ A1\n- Final answer: $\\frac{1}{2(2x-1)} + \\frac{2}{x-1}$ A1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Sets up the correct identity $\\frac{9x-5}{2} = A(x-1) + B(2x-1)$.","code":"M"},{"id":"p2","marks":1,"description":"Correctly calculates the values of $A$ and $B$.","code":"A"},{"id":"p3","marks":1,"description":"States the final expression correctly.","code":"A"}],"finalAnswers":["\\frac{1}{2(2x-1)} + \\frac{2}{x-1}"],"generatedAt":"2026-05-06T02:36:06.047Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"substitution","label":"Substitution Method","steps":[{"type":"text","order":0,"title":"Set up partial fractions","content":"To express the algebraic fraction in partial fractions, we first write it as a sum of two simpler fractions with unknown constant numerators $A$ and $B$:\n\n$$\\frac{9x-5}{2(2x-1)(x-1)} = \\frac{A}{2x-1} + \\frac{B}{x-1}$$\n\nThis setup allows us to keep the constant $2$ on the left-hand side, leading directly to the form required by the markscheme."},{"type":"text","order":1,"title":"Clear the denominators","content":"Next, we multiply both sides of the identity by the common denominator of the right-hand side, which is $(2x-1)(x-1)$:\n\n$$\\frac{9x-5}{2} = A(x-1) + B(2x-1)$$\n\nThis gives us a linear identity where we can easily apply the **Substitution Method** to find the values of $A$ and $B$."},{"type":"text","order":2,"title":"Substitute to find B","content":"To isolate $B$, we can substitute a value of $x$ that makes the term containing $A$ equal to zero. Let's substitute $x = 1$:\n\n$$\\begin{aligned} \\frac{9(1)-5}{2} &= A(1-1) + B(2(1)-1) \\\\ \\frac{4}{2} &= A(0) + B(1) \\\\ 2 &= B \\end{aligned}$$\n\nThus, $B = 2$."},{"type":"text","order":3,"title":"Substitute to find A","content":"To isolate $A$, we choose a value of $x$ that makes the term containing $B$ equal to zero. Substituting $x = \\frac{1}{2}$ gives:\n\n$$\\begin{aligned} \\frac{9\\left(\\frac{1}{2}\\right)-5}{2} &= A\\left(\\frac{1}{2}-1\\right) + B\\left(2\\left(\\frac{1}{2}\\right)-1\\right) \\\\ \\frac{4.5-5}{2} &= A\\left(-\\frac{1}{2}\\right) + B(0) \\\\ -0.25 &= -0.5A \\\\ A &= \\frac{1}{2} \\end{aligned}$$\n\nThus, $A = \\frac{1}{2}$."},{"type":"text","order":4,"title":"Write final expression","content":"Now we substitute our values of $A = \\frac{1}{2}$ and $B = 2$ back into our initial partial fraction setup:\n\n$$\\frac{9x-5}{2(2x-1)(x-1)} = \\frac{\\frac{1}{2}}{2x-1} + \\frac{2}{x-1}$$\n\nWe simplify the first term by bringing the $2$ in the numerator down to the denominator:\n\n$$\\frac{1}{2(2x-1)} + \\frac{2}{x-1}$$"}]},{"id":"equating-coefficients","label":"Equating Coefficients","steps":[{"type":"text","order":0,"title":"Set up partial fractions","content":"To express the algebraic fraction in partial fractions, we first write it as a sum of simpler fractions with constant numerators $A$ and $B$:\n\n$$\\frac{9x-5}{2(2x-1)(x-1)} \\equiv \\frac{A}{2x-1} + \\frac{B}{x-1}$$\n\nThis is our starting identity, where we need to find the values of $A$ and $B$. This step helps earn the first method mark M1."},{"type":"text","order":1,"title":"Clear the denominators","content":"Next, we clear the denominators by multiplying both sides of our identity by $(2x-1)(x-1)$:\n\n$$\\frac{9x-5}{2} \\equiv A(x-1) + B(2x-1)$$\n\nTo make this linear equation easier to work with, we can multiply the entire equation by $2$ to clear the fraction on the left-hand side:\n\n$$9x-5 \\equiv 2A(x-1) + 2B(2x-1)$$"},{"type":"text","order":2,"title":"Expand and group terms","content":"Using the **Equating Coefficients** approach, we expand the terms on the right-hand side and group like powers of $x$:\n\n$$\\begin{aligned} 9x - 5 &\\equiv 2A(x - 1) + 2B(2x - 1) \\\\ 9x - 5 &\\equiv 2Ax - 2A + 4Bx - 2B \\\\ 9x - 5 &\\equiv (2A + 4B)x - (2A + 2B) \\end{aligned}$$"},{"type":"text","order":3,"title":"Equate the coefficients","content":"Since this is an identity, the coefficients of like powers of $x$ on both sides must be equal.\n\nBy comparing the coefficients of $x$:\n$$2A + 4B = 9 \\quad \\text{--- (Equation 1)}$$\n\nBy comparing the constant terms:\n$$-2A - 2B = -5 \\implies 2A + 2B = 5 \\quad \\text{--- (Equation 2)}$$"},{"type":"text","order":4,"title":"Solve for the constants","content":"We can solve this system of linear equations by subtracting Equation 2 from Equation 1:\n\n$$\\begin{aligned} (2A + 4B) - (2A + 2B) &= 9 - 5 \\\\ 2B &= 4 \\\\ B &= 2 \\end{aligned}$$\n\nNow, substitute $B = 2$ back into Equation 2 to find $A$:\n\n$$\\begin{aligned} 2A + 2(2) &= 5 \\\\ 2A + 4 &= 5 \\\\ 2A &= 1 \\\\ A &= \\frac{1}{2} \\end{aligned}$$\n\nThis gives us $A = \\frac{1}{2}$ and $B = 2$, which awards the second mark A1."},{"type":"text","order":5,"title":"Write the final answer","content":"Finally, we substitute the values of $A$ and $B$ back into our initial partial fraction expression:\n\n$$\\frac{\\frac{1}{2}}{2x-1} + \\frac{2}{x-1} = \\frac{1}{2(2x-1)} + \\frac{2}{x-1}$$\n\nThis is our final simplified answer, earning the final mark A1."}]}],"generatedAt":"2026-06-09T05:30:23.545Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"64510fdf-2642-4f2a-817e-37f3a371759b:substitution","label":"Substitution Method","steps":[{"type":"text","order":0,"title":"Set up partial fractions","content":"To express the algebraic fraction in partial fractions, we first write it as a sum of two simpler fractions with unknown constant numerators $A$ and $B$:\n\n$$\\frac{9x-5}{2(2x-1)(x-1)} = \\frac{A}{2x-1} + \\frac{B}{x-1}$$\n\nThis setup allows us to keep the constant $2$ on the left-hand side, leading directly to the form required by the markscheme."},{"type":"text","order":1,"title":"Clear the denominators","content":"Next, we multiply both sides of the identity by the common denominator of the right-hand side, which is $(2x-1)(x-1)$:\n\n$$\\frac{9x-5}{2} = A(x-1) + B(2x-1)$$\n\nThis gives us a linear identity where we can easily apply the **Substitution Method** to find the values of $A$ and $B$."},{"type":"text","order":2,"title":"Substitute to find B","content":"To isolate $B$, we can substitute a value of $x$ that makes the term containing $A$ equal to zero. Let's substitute $x = 1$:\n\n$$\\begin{aligned} \\frac{9(1)-5}{2} &= A(1-1) + B(2(1)-1) \\\\ \\frac{4}{2} &= A(0) + B(1) \\\\ 2 &= B \\end{aligned}$$\n\nThus, $B = 2$."},{"type":"text","order":3,"title":"Substitute to find A","content":"To isolate $A$, we choose a value of $x$ that makes the term containing $B$ equal to zero. Substituting $x = \\frac{1}{2}$ gives:\n\n$$\\begin{aligned} \\frac{9\\left(\\frac{1}{2}\\right)-5}{2} &= A\\left(\\frac{1}{2}-1\\right) + B\\left(2\\left(\\frac{1}{2}\\right)-1\\right) \\\\ \\frac{4.5-5}{2} &= A\\left(-\\frac{1}{2}\\right) + B(0) \\\\ -0.25 &= -0.5A \\\\ A &= \\frac{1}{2} \\end{aligned}$$\n\nThus, $A = \\frac{1}{2}$."},{"type":"text","order":4,"title":"Write final expression","content":"Now we substitute our values of $A = \\frac{1}{2}$ and $B = 2$ back into our initial partial fraction setup:\n\n$$\\frac{9x-5}{2(2x-1)(x-1)} = \\frac{\\frac{1}{2}}{2x-1} + \\frac{2}{x-1}$$\n\nWe simplify the first term by bringing the $2$ in the numerator down to the denominator:\n\n$$\\frac{1}{2(2x-1)} + \\frac{2}{x-1}$$"}]},{"id":"64510fdf-2642-4f2a-817e-37f3a371759b:equating-coefficients","label":"Equating Coefficients","steps":[{"type":"text","order":0,"title":"Set up partial fractions","content":"To express the algebraic fraction in partial fractions, we first write it as a sum of simpler fractions with constant numerators $A$ and $B$:\n\n$$\\frac{9x-5}{2(2x-1)(x-1)} \\equiv \\frac{A}{2x-1} + \\frac{B}{x-1}$$\n\nThis is our starting identity, where we need to find the values of $A$ and $B$. This step helps earn the first method mark M1."},{"type":"text","order":1,"title":"Clear the denominators","content":"Next, we clear the denominators by multiplying both sides of our identity by $(2x-1)(x-1)$:\n\n$$\\frac{9x-5}{2} \\equiv A(x-1) + B(2x-1)$$\n\nTo make this linear equation easier to work with, we can multiply the entire equation by $2$ to clear the fraction on the left-hand side:\n\n$$9x-5 \\equiv 2A(x-1) + 2B(2x-1)$$"},{"type":"text","order":2,"title":"Expand and group terms","content":"Using the **Equating Coefficients** approach, we expand the terms on the right-hand side and group like powers of $x$:\n\n$$\\begin{aligned} 9x - 5 &\\equiv 2A(x - 1) + 2B(2x - 1) \\\\ 9x - 5 &\\equiv 2Ax - 2A + 4Bx - 2B \\\\ 9x - 5 &\\equiv (2A + 4B)x - (2A + 2B) \\end{aligned}$$"},{"type":"text","order":3,"title":"Equate the coefficients","content":"Since this is an identity, the coefficients of like powers of $x$ on both sides must be equal.\n\nBy comparing the coefficients of $x$:\n$$2A + 4B = 9 \\quad \\text{--- (Equation 1)}$$\n\nBy comparing the constant terms:\n$$-2A - 2B = -5 \\implies 2A + 2B = 5 \\quad \\text{--- (Equation 2)}$$"},{"type":"text","order":4,"title":"Solve for the constants","content":"We can solve this system of linear equations by subtracting Equation 2 from Equation 1:\n\n$$\\begin{aligned} (2A + 4B) - (2A + 2B) &= 9 - 5 \\\\ 2B &= 4 \\\\ B &= 2 \\end{aligned}$$\n\nNow, substitute $B = 2$ back into Equation 2 to find $A$:\n\n$$\\begin{aligned} 2A + 2(2) &= 5 \\\\ 2A + 4 &= 5 \\\\ 2A &= 1 \\\\ A &= \\frac{1}{2} \\end{aligned}$$\n\nThis gives us $A = \\frac{1}{2}$ and $B = 2$, which awards the second mark A1."},{"type":"text","order":5,"title":"Write the final answer","content":"Finally, we substitute the values of $A$ and $B$ back into our initial partial fraction expression:\n\n$$\\frac{\\frac{1}{2}}{2x-1} + \\frac{2}{x-1} = \\frac{1}{2(2x-1)} + \\frac{2}{x-1}$$\n\nThis is our final simplified answer, earning the final mark A1."}]}],"generatedAt":"2026-06-09T05:30:48.609Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":57},"walkthroughVideo":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,"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}],"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 3

1.3 Partial Fractions Questions

1.3 Partial Fractions

Question 3

1.3 Partial Fractions Questions