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bg-background","children":["$","$L3b",null,{"backHref":"/a-level/maths/wjec/questions","basePath":"/a-level/maths/wjec/algebra-and-functions-9/questions","formulaUrl":"https://www.wjec.co.uk/media/4jfdytcz/ss-w27-18-e.pdf","initialIndex":2,"initialQuestionWindow":[{"id":"b31d8fc6-27de-4952-ac70-23d1e4cd0e24","revisionId":"0a01fc3a-8695-4cd3-a0ed-e1e7c672e7f6","questionNumber":null,"questionType":"LA","specification":"","difficulty":1,"difficultyBand":"easy","tier":null,"totalMarks":3,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"2b753903-63bc-4ba6-b093-5cb084855b5d","label":"a","content":"Express $\\frac{2x^2 + 7x - 10}{x+4}$ in the form $Ax + B + \\frac{C}{x+4}$","markscheme":"- Sets up the identity $2x^2 + 7x - 10 = Ax(x+4) + B(x+4) + C$ or equivalent long division M1\n- Finds $A = 2$ and $B = -1$ A1\n- Finds $C = -6$ and gives the expression $2x - 1 - \\frac{6}{x+4}$ A1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Sets up the identity $2x^2 + 7x - 10 = Ax(x+4) + B(x+4) + C$ or performs an equivalent division step.","code":"M"},{"id":"p2","marks":1,"description":"Calculates correct values for two constants, e.g., $A = 2$ and $B = -1$.","code":"A"},{"id":"p3","marks":1,"description":"Finds all constants correctly ($A=2, B=-1, C=-6$) and states the full expression $2x - 1 - \\frac{6}{x+4}$.","code":"A","acceptable":["2x - 1 + \\frac{-6}{x+4}"]}],"finalAnswers":["2x - 1 - \\frac{6}{x+4}"],"generatedAt":"2026-05-06T14:03:45.486Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"longdivision","label":"Polynomial Long Division","steps":[{"type":"text","order":0,"title":"Set up the division","content":"To express the algebraic fraction $\\frac{2x^2 + 7x - 10}{x+4}$ in the required form, we will use **polynomial long division**.\n\nWe divide the numerator (the dividend) $2x^2 + 7x - 10$ by the denominator (the divisor) $x+4$.\n\n$$\\begin{aligned} \\text{Divisor:} & \\quad x + 4 \\\\ \\text{Dividend:} & \\quad 2x^2 + 7x - 10 \\end{aligned}$$"},{"type":"text","order":1,"title":"Divide the first term","content":"First, divide the leading term of the dividend, $2x^2$, by the leading term of the divisor, $x$:\n\n$$\\frac{2x^2}{x} = 2x$$\n\nMultiply $2x$ by the entire divisor $(x + 4)$:\n\n$$2x(x + 4) = 2x^2 + 8x$$\n\nSubtract this product from the dividend to find the first remainder:\n\n$$(2x^2 + 7x) - (2x^2 + 8x) = -x$$\n\nNow, bring down the next term, $-10$, to get the new expression:\n\n$$-x - 10$$"},{"type":"text","order":2,"title":"Divide the second term","content":"Next, divide the leading term of our new expression, $-x$, by the leading term of the divisor, $x$:\n\n$$\\frac{-x}{x} = -1$$\n\nMultiply $-1$ by the divisor $(x + 4)$:\n\n$$-1(x + 4) = -x - 4$$\n\nSubtract this from our expression to find the final remainder:\n\n$$(-x - 10) - (-x - 4) = -10 + 4 = -6$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\text{Quotient} = \\htmlData{anchor=quotient}{\\color{@sky}{2x - 1}}"},{"latex":"\\text{Remainder} = \\htmlData{anchor=remainder}{\\color{@red}{-6}}"},{"latex":"\\frac{2x^2 + 7x - 10}{x+4} = \\htmlData{anchor=target-q}{\\color{@sky}{2x - 1}} + \\frac{\\htmlData{anchor=target-r}{\\color{@red}{-6}}}{x+4}"}],"highlights":[{"color":"sky","style":"text-color","anchor":"quotient"},{"color":"red","style":"text-color","anchor":"remainder"}],"connections":[{"to":"target-q","from":"quotient","color":"sky","label":null,"style":"solid"},{"to":"target-r","from":"remainder","color":"red","label":null,"style":"solid"}]},"order":3,"title":"Write the final expression","description":"Combining our quotient and remainder, we can write the complete division result."},{"type":"text","order":4,"title":"Identify coefficients","content":"By comparing our result to the required form $Ax + B + \\frac{C}{x+4}$, we can state the values of the constants:\n\n* $A = 2$\n* $B = -1$\n* $C = -6$\n\nThis gives our final simplified expression:\n\n$$2x - 1 - \\frac{6}{x+4}$$"}]},{"id":"equatingcoefficients","label":"Equating Coefficients","steps":[{"type":"text","order":0,"title":"Set up the identity","content":"To express the algebraic fraction in the desired form, we start by setting up an equation:\n\n$$\\frac{2x^2 + 7x - 10}{x+4} = Ax + B + \\frac{C}{x+4}$$\n\nMultiply both sides of this equation by the denominator, $(x+4)$, to eliminate the fractions and form an algebraic identity:\n\n$$2x^2 + 7x - 10 = (Ax + B)(x + 4) + C$$\n\nThis setup earns the first method mark **(M1)**."},{"type":"hyper_latex","block":{"nodes":[{"latex":"2x^2 + 7x - 10 = \\htmlData{anchor=expanded-form}{\\color{@sky}{A}x^2 + (4\\color{@sky}{A} + \\color{@amber}{B})x + (4\\color{@amber}{B} + \\color{@purple}{C})}"}],"highlights":[{"color":"sky","style":"underline","anchor":"expanded-form"}],"connections":[]},"order":1,"title":"Expand the right-hand side","description":"Now, we expand and group the terms on the right-hand side by powers of $x$:\n\n$$\\begin{aligned} (Ax + B)(x + 4) + C &= Ax^2 + 4Ax + Bx + 4B + C \\\\ &= Ax^2 + (4A + B)x + (4B + C) \\end{aligned}$$"},{"type":"text","order":2,"title":"Equate the coefficients","content":"Since this is an identity, the coefficients of corresponding powers of $x$ on both sides must be equal.\n\nLet's equate the coefficients of $x^2$, $x$, and the constant terms:\n\n* **For $x^2$:**\n $$A = 2$$\n\n* **For $x$:**\n $$4A + B = 7$$\n Substituting $A = 2$:\n $$\\begin{aligned} 4(2) + B &= 7 \\\\ 8 + B &= 7 \\\\ B &= -1 \\end{aligned}$$\n\nFinding both $A = 2$ and $B = -1$ earns the second mark **(A1)**."},{"type":"text","order":3,"title":"Find the constant term","content":"Now we equate the constant terms ($x^0$):\n\n$$4B + C = -10$$\n\nSubstitute our value of $B = -1$ into this equation:\n\n$$\\begin{aligned} 4(-1) + C &= -10 \\\\ -4 + C &= -10 \\\\ C &= -6 \\end{aligned}$$\n\nThis gives us our third coefficient."},{"type":"text","order":4,"title":"Write the final expression","content":"Substituting $A = 2$, $B = -1$, and $C = -6$ back into our original expression gives:\n\n$$2x - 1 - \\frac{6}{x+4}$$\n\nThis completes our solution and earns the final mark **(A1)**."}]},{"id":"syntheticdivision","label":"Synthetic Division","steps":[{"type":"text","order":0,"title":"Identify the key components","content":"To express the algebraic fraction $\\frac{2x^2 + 7x - 10}{x+4}$ in the form $Ax + B + \\frac{C}{x+4}$, we can use **synthetic division**.\n\nFirst, we identify the coefficients of the numerator (dividend):\n- $x^2$-coefficient: $2$\n- $x$-coefficient: $7$\n- Constant term: $-10$\n\nNext, we find the root of the divisor by setting the denominator equal to zero:\n$$x + 4 = 0 \\implies x = -4$$"},{"type":"text","order":1,"title":"Set up synthetic division","content":"We set up our synthetic division table using the root $-4$ on the left, and the coefficients $2$, $7$, and $-10$ on the top row:\n\n$$\n\\begin{array}{r|rrr}\n-4 & 2 & 7 & -10 \\\\\n & & & \\\\\n\\hline\n & & & \\\\\n\\end{array}\n$$"},{"type":"text","order":2,"title":"Perform synthetic division","content":"Now, we carry out the synthetic division step-by-step:\n\n1. **Bring down** the first coefficient, $2$, to the bottom row:\n $$\n \\begin{array}{r|rrr}\n -4 & 2 & 7 & -10 \\\\\n & & & \\\\\n \\hline\n & 2 & & \\\\\n \\end{array}\n $$\n\n2. **Multiply** the $2$ by the root $-4$ to get $-8$, place this under the $7$, and **add** them:\n $$7 + (-8) = -1$$\n $$\n \\begin{array}{r|rrr}\n -4 & 2 & 7 & -10 \\\\\n & & -8 & \\\\\n \\hline\n & 2 & -1 & \\\\\n \\end{array}\n $$\n\n3. **Multiply** the $-1$ by the root $-4$ to get $4$, place this under $-10$, and **add** them:\n $$-10 + 4 = -6$$\n $$\n \\begin{array}{r|rrr}\n -4 & 2 & 7 & -10 \\\\\n & & -8 & 4 \\\\\n \\hline\n & 2 & -1 & -6 \\\\\n \\end{array}\n $$"},{"type":"hyper_latex","block":{"nodes":[{"text":"The bottom row gives the coefficients of our quotient and remainder:"},{"latex":"\\text{Quotient: } \\htmlData{anchor=quotient}{\\color{@sky}{2x - 1}}"},{"latex":"\\text{Remainder: } \\htmlData{anchor=remainder}{\\color{@amber}{-6}}"}],"highlights":[{"color":"sky","style":"text-color","anchor":"quotient"},{"color":"amber","style":"text-color","anchor":"remainder"}],"connections":[]},"order":3,"title":"Interpret the results","description":"The first two numbers on the bottom row, 2 and -1, represent the coefficients of the linear quotient $Ax + B$. The final number, -6, is the remainder $C$."},{"type":"text","order":4,"title":"Write the final answer","content":"Using our quotient $2x - 1$ and remainder $-6$, we write the expression in the required form:\n\n$$\\frac{2x^2 + 7x - 10}{x+4} = 2x - 1 - \\frac{6}{x+4}$$\n\nBy comparing this to $Ax + B + \\frac{C}{x+4}$, we find:\n- $A = 2$\n- $B = -1$\n- $C = -6$"}]}],"generatedAt":"2026-06-09T07:48:57.877Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"2b753903-63bc-4ba6-b093-5cb084855b5d:longdivision","label":"Polynomial Long Division","steps":[{"type":"text","order":0,"title":"Set up the division","content":"To express the algebraic fraction $\\frac{2x^2 + 7x - 10}{x+4}$ in the required form, we will use **polynomial long division**.\n\nWe divide the numerator (the dividend) $2x^2 + 7x - 10$ by the denominator (the divisor) $x+4$.\n\n$$\\begin{aligned} \\text{Divisor:} & \\quad x + 4 \\\\ \\text{Dividend:} & \\quad 2x^2 + 7x - 10 \\end{aligned}$$"},{"type":"text","order":1,"title":"Divide the first term","content":"First, divide the leading term of the dividend, $2x^2$, by the leading term of the divisor, $x$:\n\n$$\\frac{2x^2}{x} = 2x$$\n\nMultiply $2x$ by the entire divisor $(x + 4)$:\n\n$$2x(x + 4) = 2x^2 + 8x$$\n\nSubtract this product from the dividend to find the first remainder:\n\n$$(2x^2 + 7x) - (2x^2 + 8x) = -x$$\n\nNow, bring down the next term, $-10$, to get the new expression:\n\n$$-x - 10$$"},{"type":"text","order":2,"title":"Divide the second term","content":"Next, divide the leading term of our new expression, $-x$, by the leading term of the divisor, $x$:\n\n$$\\frac{-x}{x} = -1$$\n\nMultiply $-1$ by the divisor $(x + 4)$:\n\n$$-1(x + 4) = -x - 4$$\n\nSubtract this from our expression to find the final remainder:\n\n$$(-x - 10) - (-x - 4) = -10 + 4 = -6$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\text{Quotient} = \\htmlData{anchor=quotient}{\\color{@sky}{2x - 1}}"},{"latex":"\\text{Remainder} = \\htmlData{anchor=remainder}{\\color{@red}{-6}}"},{"latex":"\\frac{2x^2 + 7x - 10}{x+4} = \\htmlData{anchor=target-q}{\\color{@sky}{2x - 1}} + \\frac{\\htmlData{anchor=target-r}{\\color{@red}{-6}}}{x+4}"}],"highlights":[{"color":"sky","style":"text-color","anchor":"quotient"},{"color":"red","style":"text-color","anchor":"remainder"}],"connections":[{"to":"target-q","from":"quotient","color":"sky","label":null,"style":"solid"},{"to":"target-r","from":"remainder","color":"red","label":null,"style":"solid"}]},"order":3,"title":"Write the final expression","description":"Combining our quotient and remainder, we can write the complete division result."},{"type":"text","order":4,"title":"Identify coefficients","content":"By comparing our result to the required form $Ax + B + \\frac{C}{x+4}$, we can state the values of the constants:\n\n* $A = 2$\n* $B = -1$\n* $C = -6$\n\nThis gives our final simplified expression:\n\n$$2x - 1 - \\frac{6}{x+4}$$"}]},{"id":"2b753903-63bc-4ba6-b093-5cb084855b5d:equatingcoefficients","label":"Equating Coefficients","steps":[{"type":"text","order":0,"title":"Set up the identity","content":"To express the algebraic fraction in the desired form, we start by setting up an equation:\n\n$$\\frac{2x^2 + 7x - 10}{x+4} = Ax + B + \\frac{C}{x+4}$$\n\nMultiply both sides of this equation by the denominator, $(x+4)$, to eliminate the fractions and form an algebraic identity:\n\n$$2x^2 + 7x - 10 = (Ax + B)(x + 4) + C$$\n\nThis setup earns the first method mark **(M1)**."},{"type":"hyper_latex","block":{"nodes":[{"latex":"2x^2 + 7x - 10 = \\htmlData{anchor=expanded-form}{\\color{@sky}{A}x^2 + (4\\color{@sky}{A} + \\color{@amber}{B})x + (4\\color{@amber}{B} + \\color{@purple}{C})}"}],"highlights":[{"color":"sky","style":"underline","anchor":"expanded-form"}],"connections":[]},"order":1,"title":"Expand the right-hand side","description":"Now, we expand and group the terms on the right-hand side by powers of $x$:\n\n$$\\begin{aligned} (Ax + B)(x + 4) + C &= Ax^2 + 4Ax + Bx + 4B + C \\\\ &= Ax^2 + (4A + B)x + (4B + C) \\end{aligned}$$"},{"type":"text","order":2,"title":"Equate the coefficients","content":"Since this is an identity, the coefficients of corresponding powers of $x$ on both sides must be equal.\n\nLet's equate the coefficients of $x^2$, $x$, and the constant terms:\n\n* **For $x^2$:**\n $$A = 2$$\n\n* **For $x$:**\n $$4A + B = 7$$\n Substituting $A = 2$:\n $$\\begin{aligned} 4(2) + B &= 7 \\\\ 8 + B &= 7 \\\\ B &= -1 \\end{aligned}$$\n\nFinding both $A = 2$ and $B = -1$ earns the second mark **(A1)**."},{"type":"text","order":3,"title":"Find the constant term","content":"Now we equate the constant terms ($x^0$):\n\n$$4B + C = -10$$\n\nSubstitute our value of $B = -1$ into this equation:\n\n$$\\begin{aligned} 4(-1) + C &= -10 \\\\ -4 + C &= -10 \\\\ C &= -6 \\end{aligned}$$\n\nThis gives us our third coefficient."},{"type":"text","order":4,"title":"Write the final expression","content":"Substituting $A = 2$, $B = -1$, and $C = -6$ back into our original expression gives:\n\n$$2x - 1 - \\frac{6}{x+4}$$\n\nThis completes our solution and earns the final mark **(A1)**."}]},{"id":"2b753903-63bc-4ba6-b093-5cb084855b5d:syntheticdivision","label":"Synthetic Division","steps":[{"type":"text","order":0,"title":"Identify the key components","content":"To express the algebraic fraction $\\frac{2x^2 + 7x - 10}{x+4}$ in the form $Ax + B + \\frac{C}{x+4}$, we can use **synthetic division**.\n\nFirst, we identify the coefficients of the numerator (dividend):\n- $x^2$-coefficient: $2$\n- $x$-coefficient: $7$\n- Constant term: $-10$\n\nNext, we find the root of the divisor by setting the denominator equal to zero:\n$$x + 4 = 0 \\implies x = -4$$"},{"type":"text","order":1,"title":"Set up synthetic division","content":"We set up our synthetic division table using the root $-4$ on the left, and the coefficients $2$, $7$, and $-10$ on the top row:\n\n$$\n\\begin{array}{r|rrr}\n-4 & 2 & 7 & -10 \\\\\n & & & \\\\\n\\hline\n & & & \\\\\n\\end{array}\n$$"},{"type":"text","order":2,"title":"Perform synthetic division","content":"Now, we carry out the synthetic division step-by-step:\n\n1. **Bring down** the first coefficient, $2$, to the bottom row:\n $$\n \\begin{array}{r|rrr}\n -4 & 2 & 7 & -10 \\\\\n & & & \\\\\n \\hline\n & 2 & & \\\\\n \\end{array}\n $$\n\n2. **Multiply** the $2$ by the root $-4$ to get $-8$, place this under the $7$, and **add** them:\n $$7 + (-8) = -1$$\n $$\n \\begin{array}{r|rrr}\n -4 & 2 & 7 & -10 \\\\\n & & -8 & \\\\\n \\hline\n & 2 & -1 & \\\\\n \\end{array}\n $$\n\n3. **Multiply** the $-1$ by the root $-4$ to get $4$, place this under $-10$, and **add** them:\n $$-10 + 4 = -6$$\n $$\n \\begin{array}{r|rrr}\n -4 & 2 & 7 & -10 \\\\\n & & -8 & 4 \\\\\n \\hline\n & 2 & -1 & -6 \\\\\n \\end{array}\n $$"},{"type":"hyper_latex","block":{"nodes":[{"text":"The bottom row gives the coefficients of our quotient and remainder:"},{"latex":"\\text{Quotient: } \\htmlData{anchor=quotient}{\\color{@sky}{2x - 1}}"},{"latex":"\\text{Remainder: } \\htmlData{anchor=remainder}{\\color{@amber}{-6}}"}],"highlights":[{"color":"sky","style":"text-color","anchor":"quotient"},{"color":"amber","style":"text-color","anchor":"remainder"}],"connections":[]},"order":3,"title":"Interpret the results","description":"The first two numbers on the bottom row, 2 and -1, represent the coefficients of the linear quotient $Ax + B$. The final number, -6, is the remainder $C$."},{"type":"text","order":4,"title":"Write the final answer","content":"Using our quotient $2x - 1$ and remainder $-6$, we write the expression in the required form:\n\n$$\\frac{2x^2 + 7x - 10}{x+4} = 2x - 1 - \\frac{6}{x+4}$$\n\nBy comparing this to $Ax + B + \\frac{C}{x+4}$, we find:\n- $A = 2$\n- $B = -1$\n- $C = -6$"}]}],"generatedAt":"2026-06-09T07:49:15.067Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":75},"walkthroughVideo":null},{"id":"07b9dc6a-1b72-4311-9c5e-072f0a0b629d","revisionId":"2b0c1f5c-f83f-4a7c-a92c-e295ee6a5c2d","questionNumber":null,"questionType":"LA","specification":"","difficulty":2,"difficultyBand":"easy","tier":null,"totalMarks":2,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"0e48f0c9-767b-4d7b-a958-ac8579a13f4b","label":null,"content":"The expression \\[(15x^2 + 4x + 9) / (5x - 2)\\] can be written as \\[Ax + B + \\frac{C}{5x - 2}\\] State the value of \$A\$.","markscheme":"- Sets up the identity $15x^2 + 4x + 9 = (Ax + B)(5x - 2) + C$ or performs the first step of algebraic division $(15x^2) \\div (5x)$ M1\n- Equates coefficients of $x^2$ to find $15 = 5A$ or completes the division to find $A = 3$ A1","marks":2,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"polynomial_division","label":"Polynomial Long Division","steps":[{"type":"text","order":0,"title":"Set up long division","content":"To write the expression $\\frac{15x^2 + 4x + 9}{5x - 2}$ in the form $Ax + B + \\frac{C}{5x - 2}$, we can perform algebraic long division of the quadratic numerator by the linear denominator.\n\nWe start by setting up the division:\n\n$$5x - 2 \\ \\overline{) \\ 15x^2 + 4x + 9}$$"},{"type":"text","order":1,"title":"Divide leading terms","content":"Next, we divide the leading term of the dividend, $15x^2$, by the leading term of the divisor, $5x$.\n\n$$\\frac{15x^2}{5x} = 3x$$\n\nThis $3x$ is the first term of our quotient, and it represents the $Ax$ term in the required format.\n\nThis first division step earns the method mark M1."},{"type":"text","order":2,"title":"Identify coefficient A","content":"By comparing the first term of our quotient, $3x$, to the term $Ax$ in the target expression $Ax + B + \\frac{C}{5x - 2}$, we can directly identify the coefficient $A$:\n\n$$A = 3$$\n\nThis completes the division step required to find the value of $A$ and earns the accuracy mark A1."}]},{"id":"equating_coefficients","label":"Equating Coefficients","steps":[{"type":"text","order":0,"title":"Set up the identity","content":"To find the value of $A$ using the **equating coefficients** method, we start by setting up an algebraic identity.\n\nWe are given the expression:\n\n$$\\frac{15x^2 + 4x + 9}{5x - 2} = Ax + B + \\frac{C}{5x - 2}$$\n\nMultiply both sides of this identity by the denominator, $(5x - 2)$, to eliminate the fraction:\n\n$$15x^2 + 4x + 9 = (Ax + B)(5x - 2) + C$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$15x^2 + 4x + 9 = (\\htmlData{anchor=ax-term}{\\color{@sky}{Ax}} + B)(\\htmlData{anchor=fivex-term}{\\color{@sky}{5x}} - 2) + C$"},{"latex":"$$\\htmlData{anchor=lhs-x2}{\\color{@sky}{15}}x^2 + 4x + 9 = \\htmlData{anchor=rhs-x2}{\\color{@sky}{5A}}x^2 + (5B - 2A)x + (C - 2B)$"}],"highlights":[{"color":"sky","style":"box","anchor":"lhs-x2"},{"color":"sky","style":"box","anchor":"rhs-x2"}],"connections":[{"to":"rhs-x2","from":"ax-term","color":"sky","label":"multiply","style":"dashed"},{"to":"rhs-x2","from":"fivex-term","color":"sky","label":"multiply","style":"dashed"}]},"order":1,"title":"Focus on $x^2$ terms","description":"Expand the right-hand side, focusing specifically on how the $x^2$ term is formed. The only way to get an $x^2$ term on the right is by multiplying $Ax$ by $5x$."},{"type":"text","order":2,"title":"Solve for A","content":"Now, equate the coefficients of $x^2$ from both sides of the identity:\n\n$$15 = 5A$$\n\nDivide both sides by $5$ to find $A$:\n\n$$A = 3$$"}]}],"generatedAt":"2026-06-09T03:21:23.152Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"0e48f0c9-767b-4d7b-a958-ac8579a13f4b:polynomial_division","label":"Polynomial Long Division","steps":[{"type":"text","order":0,"title":"Set up long division","content":"To write the expression $\\frac{15x^2 + 4x + 9}{5x - 2}$ in the form $Ax + B + \\frac{C}{5x - 2}$, we can perform algebraic long division of the quadratic numerator by the linear denominator.\n\nWe start by setting up the division:\n\n$$5x - 2 \\ \\overline{) \\ 15x^2 + 4x + 9}$$"},{"type":"text","order":1,"title":"Divide leading terms","content":"Next, we divide the leading term of the dividend, $15x^2$, by the leading term of the divisor, $5x$.\n\n$$\\frac{15x^2}{5x} = 3x$$\n\nThis $3x$ is the first term of our quotient, and it represents the $Ax$ term in the required format.\n\nThis first division step earns the method mark M1."},{"type":"text","order":2,"title":"Identify coefficient A","content":"By comparing the first term of our quotient, $3x$, to the term $Ax$ in the target expression $Ax + B + \\frac{C}{5x - 2}$, we can directly identify the coefficient $A$:\n\n$$A = 3$$\n\nThis completes the division step required to find the value of $A$ and earns the accuracy mark A1."}]},{"id":"0e48f0c9-767b-4d7b-a958-ac8579a13f4b:equating_coefficients","label":"Equating Coefficients","steps":[{"type":"text","order":0,"title":"Set up the identity","content":"To find the value of $A$ using the **equating coefficients** method, we start by setting up an algebraic identity.\n\nWe are given the expression:\n\n$$\\frac{15x^2 + 4x + 9}{5x - 2} = Ax + B + \\frac{C}{5x - 2}$$\n\nMultiply both sides of this identity by the denominator, $(5x - 2)$, to eliminate the fraction:\n\n$$15x^2 + 4x + 9 = (Ax + B)(5x - 2) + C$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$15x^2 + 4x + 9 = (\\htmlData{anchor=ax-term}{\\color{@sky}{Ax}} + B)(\\htmlData{anchor=fivex-term}{\\color{@sky}{5x}} - 2) + C$"},{"latex":"$$\\htmlData{anchor=lhs-x2}{\\color{@sky}{15}}x^2 + 4x + 9 = \\htmlData{anchor=rhs-x2}{\\color{@sky}{5A}}x^2 + (5B - 2A)x + (C - 2B)$"}],"highlights":[{"color":"sky","style":"box","anchor":"lhs-x2"},{"color":"sky","style":"box","anchor":"rhs-x2"}],"connections":[{"to":"rhs-x2","from":"ax-term","color":"sky","label":"multiply","style":"dashed"},{"to":"rhs-x2","from":"fivex-term","color":"sky","label":"multiply","style":"dashed"}]},"order":1,"title":"Focus on $x^2$ terms","description":"Expand the right-hand side, focusing specifically on how the $x^2$ term is formed. The only way to get an $x^2$ term on the right is by multiplying $Ax$ by $5x$."},{"type":"text","order":2,"title":"Solve for A","content":"Now, equate the coefficients of $x^2$ from both sides of the identity:\n\n$$15 = 5A$$\n\nDivide both sides by $5$ to find $A$:\n\n$$A = 3$$"}]}],"generatedAt":"2026-06-09T03:21:45.167Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths__wjec","summaries":[{"id":"5c640287-30a0-46df-943c-b85d9d16211c","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"89261831-e48a-4f04-a4ec-635a56739676","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"b31d8fc6-27de-4952-ac70-23d1e4cd0e24","difficultyBand":"easy","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"c7323b8c-a165-4150-9bd0-5c15fbc2f6b7","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"12e59216-4520-4b11-9dde-f14159667ea7","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"82011ee7-32c7-4a1e-a167-5261849ebc29","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"b9c621f4-6f84-4298-bb60-1aa4e912acb6","difficultyBand":"medium","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"f94d84f8-bbb7-42a1-94af-b4a4fe3e9ce3","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"07c4ec0e-6e71-41fa-ace8-019ab033236a","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"5f873329-2af6-4952-bf3e-7c72af9f27de","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"788d3dde-8f8c-4011-b6af-41a1774ce4bd","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"a656bf81-a91c-4d24-b308-37afa8420a5d","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"f4e2426d-21b1-4aa8-b63d-a6658b2de000","difficultyBand":"medium","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"7e861995-4441-452e-a7d6-b8a5541db516","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"7a634fe3-8158-4388-b0f7-4c3181cff893","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"82388296-1670-4839-8b9a-640eb9ea4641","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"f2048a87-4092-4ac5-b99c-512271643d6f","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"007fb9f8-93ed-4e25-b579-6753fac2b779","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"043fd4ae-9a3f-4ae5-9475-2e9189a2baf6","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"06fb0f0d-1b2c-445f-a0db-d1f9b3106c9b","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"0730257a-447a-4bb2-a583-413fda8691f6","difficultyBand":"medium","calculatorUse":null,"totalMarks":12,"questionType":"LA","isCalculation":true},{"id":"07b9dc6a-1b72-4311-9c5e-072f0a0b629d","difficultyBand":"easy","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"0b570e33-d69c-4171-b80c-9ab8a251bdd5","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"0ec40464-8a56-4084-b1ef-394dd4f60ff0","difficultyBand":"hard","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"0f52e988-34c0-45a6-9c77-f282497d9dee","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"179fcd92-9fe8-4e27-aaa5-c90641138f26","difficultyBand":"medium","calculatorUse":null,"totalMarks":12,"questionType":"LA","isCalculation":true},{"id":"21113931-cbbe-45cf-ac23-9f59c64e6acf","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"2c5d421f-3ef9-4fba-8e1b-2226606fb77b","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"2fe221a4-c3f9-48d8-b71c-feaf4adde48c","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"312b4f89-7426-4320-8a0a-c9bd408bcdd9","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"3223b54e-69f3-42f1-b669-47dd81b89221","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"3fb39c35-3f84-41d7-9b3d-af21faccbd83","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"49387849-ae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Algebra and Functions","topicId":"c1005309-f8b4-4344-8c07-cd05be08b008","topicSidebar":"$L3c","topicSlug":"algebra-and-functions-9","topicTitle":"1.2.8 Algebra and Functions"}]}],"$L3d","$L3e"]}]

1.2.8 Algebra and Functions

Question 1

1.2.8 Algebra and Functions Questions

1.2.8 Algebra and Functions

Question 1

1.2.8 Algebra and Functions Questions