mmon denominator $(2x+1)(x-2)^2$:\n\n$$6x^2-23x+12 \\equiv A(x-2)^2 + B(2x+1)(x-2) + C(2x+1)$$"},{"type":"text","order":1,"title":"Expand the right-hand side","content":"To equate coefficients, we must expand and simplify the right-hand side of our identity:\n\n1. Expand the first term:\n $$A(x-2)^2 = A(x^2 - 4x + 4) = Ax^2 - 4Ax + 4A$$\n\n2. Expand the second term:\n $$B(2x+1)(x-2) = B(2x^2 - 3x - 2) = 2Bx^2 - 3Bx - 2B$$\n\n3. Expand the third term:\n $$C(2x+1) = 2Cx + C$$\n\nNow, group these expanded terms together by powers of $x$:\n\n$$\\begin{aligned} R.H.S. &= (Ax^2 - 4Ax + 4A) + (2Bx^2 - 3Bx - 2B) + (2Cx + C) \\\\ &= (A+2B)x^2 + (-4A-3B+2C)x + (4A-2B+C) \\end{aligned}$$"},{"type":"text","order":2,"title":"Equate coefficients","content":"By comparing the coefficients of the expanded right-hand side to the left-hand side ($6x^2 - 23x + 12$), we set up a system of linear equations:\n\n* For $x^2$:\n $$A + 2B = 6 \\quad \\text{--- (Equation 1)}$$\n\n* For $x$:\n $$-4A - 3B + 2C = -23 \\quad \\text{--- (Equation 2)}$$\n\n* For the constant term:\n $$4A - 2B + C = 12 \\quad \\text{--- (Equation 3)}$$"},{"type":"text","order":3,"title":"Solve the system","content":"Let's solve the system of equations step by step:\n\nFrom Equation 1, express $A$ in terms of $B$:\n$$A = 6 - 2B$$\n\nNow substitute this expression for $A$ into Equations 2 and 3:\n\n1. Substitute into Equation 2:\n $$\\begin{aligned} -4(6 - 2B) - 3B + 2C &= -23 \\\\ -24 + 8B - 3B + 2C &= -23 \\\\ 5B + 2C &= 1 \\quad \\text{--- (Equation 4)} \\end{aligned}$$\n\n2. Substitute into Equation 3:\n $$\\begin{aligned} 4(6 - 2B) - 2B + C &= 12 \\\\ 24 - 8B - 2B + C &= 12 \\\\ -10B + C &= -12 \\quad \\Rightarrow \\quad C = 10B - 12 \\quad \\text{--- (Equation 5)} \\end{aligned}$$\n\nNow, substitute Equation 5 into Equation 4:\n$$\\begin{aligned} 5B + 2(10B - 12) &= 1 \\\\ 5B + 20B - 24 &= 1 \\\\ 25B &= 25 \\\\ B &= 1 \\end{aligned}$$\n\nUsing $B = 1$, we find $A$ and $C$:\n* $$A = 6 - 2(1) = 4$$\n* $$C = 10(1) - 12 = -2$$","calculator":[{"gdc":{"instructions":"Go to the Equation menu, select 'Simultaneous', and choose 3 unknowns. Input the coefficients matrix to solve the system."},"ti84":{"instructions":"Press [APPS] and select 'PlySmlt2'. Choose '2: SYSTEM SOLVER'. Set the number of equations to 3 and variables to 3. Enter the coefficients as a matrix:\n[[1, 2, 0, 6],\n [-4, -3, 2, -23],\n [4, -2, 1, 12]]\nPress [GRAPH] (SOLVE) to find the values: x1 (A) = 4, x2 (B) = 1, x3 (C) = -2."},"desmos":{"expression":"A=\\begin{bmatrix}1&2&0\\\\-4&-3&2\\\\4&-2&1\\end{bmatrix}, B=\\begin{bmatrix}6\\\\-23\\\\12\\end{bmatrix}","instructions":"Open the matrix calculator. Define matrix A as the 3x3 coefficient matrix and matrix B as the 3x1 constant matrix, then compute A^-1 * B to get the solutions."},"description":"You can solve this system of linear equations easily using a graphing calculator's system solver."}]},{"type":"text","order":4,"title":"State final answer","content":"Now we substitute our values of $A = 4$, $B = 1$, and $C = -2$ back into our initial partial fraction form:\n\n$$\\frac{6x^2-23x+12}{(2x+1)(x-2)^2} \\equiv \\frac{4}{2x+1} + \\frac{1}{x-2} - \\frac{2}{(x-2)^2}$$"}]}],"generatedAt":"2026-06-09T04:56:15.731Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null},{"id":"acf362fa-77c0-4622-bfbb-e81ffd30f439","revisionId":"10b0d85f-46d7-4fc5-ae11-b32b004f1755","questionNumber":null,"questionType":"LA","specification":"","difficulty":1,"difficultyBand":"medium","totalMarks":5,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"ea8c203a-4740-4713-aa39-c3913a1e72f5","label":"a","content":"Express $\\frac{4x^2 + 13x + 8}{(x-2)(x+3)^2}$ in partial fractions","markscheme":"- Sets up the identity $4x^2 + 13x + 8 = A(x+3)^2 + B(x-2)(x+3) + C(x-2)$ M1\n- Finds $A = 2$ and $C = -1$ A1\n- Finds $B = 2$ A1\n- Final answer: $\\frac{2}{x-2} + \\frac{2}{x+3} - \\frac{1}{(x+3)^2}$ A1\n- Correct use of partial fraction form A1","marks":5,"order":0,"

1.5.10 Partial fractions (A-level only)

Question 13

1.5.10 Partial fractions (A-level only) Questions

1.5.10 Partial fractions (A-level only)

Question 13

1.5.10 Partial fractions (A-level only) Questions