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A1","marks":2,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"binomial_term_formula","label":"Direct General Term Formula","steps":[{"type":"text","order":0,"title":"Identify general term","content":"To find the constant term in the expansion of $(3x - 2)^4$, we can use the general term formula from the Binomial Theorem.\n\nFor an expansion of the form $(a + b)^n$, the general term $T_{r+1}$ is given by:\n\n$$T_{r+1} = \\binom{n}{r} a^{n-r} b^r$$\n\nIn this problem, we have:\n- $a = 3x$\n- $b = -2$\n- $n = 4$\n\nSubstituting these values into the formula gives:\n\n$$T_{r+1} = \\binom{4}{r} (3x)^{4-r} (-2)^r$$","highlights":[{"text":"$$(3x - 2)^4$","location":"specification"}]},{"type":"text","order":1,"title":"Set power to zero","content":"A constant term is independent of $x$, meaning the exponent of $x$ must be $0$.\n\nLet's isolate the $x$ part in our general term:\n\n$$T_{r+1} = \\binom{4}{r} 3^{4-r} (-2)^r x^{4-r}$$\n\nTo find the constant term, we set the exponent of $x$ to $0$:\n\n$$4 - r = 0 \\implies r = 4$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$T_{4+1} = \\binom{4}{\\htmlData{anchor=r-val}{\\color{@sky}{4}}} (3x)^{4-\\htmlData{anchor=r-sub1}{\\color{@sky}{4}}} (\\htmlData{anchor=b-val}{\\color{@amber}{-2}})^{\\htmlData{anchor=r-sub2}{\\color{@sky}{4}}}$"},{"latex":"$$T_5 = \\binom{4}{4} (3x)^0 (-2)^4$"},{"latex":"$$T_5 = 1 \\times 1 \\times 16 = 16$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"r-val"},{"color":"amber","style":"text-color","anchor":"b-val"}],"connections":[{"to":"r-sub1","from":"r-val","color":"sky","label":"r = 4","style":"solid"},{"to":"r-sub2","from":"r-val","color":"sky","label":null,"style":"solid"}]},"order":2,"title":"Evaluate constant term","description":"Substitute $r = 4$ back into our general term expression to compute the constant term."},{"type":"text","order":3,"title":"Confirm final answer","content":"The constant term in the expansion is **$16$**.\n\n### Markscheme Guidance\n- **1 mark (M1)**: Correctly identifying the term independent of $x$ as $\\binom{4}{4} (3x)^0 (-2)^4$.\n- **1 mark (A1)**: Correctly evaluating this term to $16$."}]},{"id":"full_expansion","label":"Full Binomial Expansion","steps":[{"type":"text","order":0,"title":"Set up the expansion","content":"To find the constant term by fully expanding the expression, we use the Binomial Theorem:\n\n$$(a + b)^n = \\sum_{r=0}^{n} \\binom{n}{r} a^{n-r} b^r$$\n\nIn our case, we have:\n- $a = 3x$\n- $b = -2$\n- $n = 4$\n\nLet's write down the five terms of this expansion, corresponding to $r = 0, 1, 2, 3, 4$:\n\n$$\\begin{aligned} (3x - 2)^4 = &\\binom{4}{0}(3x)^4(-2)^0 + \\binom{4}{1}(3x)^3(-2)^1 + \\binom{4}{2}(3x)^2(-2)^2 \\\\ &+ \\binom{4}{3}(3x)^1(-2)^3 + \\binom{4}{4}(3x)^0(-2)^4 \\end{aligned}$$"},{"type":"text","order":1,"title":"Calculate each term","content":"Now we evaluate each part of the sum step by step:\n\n- **For $r=0$:** \n $$1 \\times 81x^4 \\times 1 = 81x^4$$\n- **For $r=1$:** \n $$4 \\times 27x^3 \\times (-2) = -216x^3$$\n- **For $r=2$:** \n $$6 \\times 9x^2 \\times 4 = 216x^2$$\n- **For $r=3$:** \n $$4 \\times 3x \\times (-8) = -96x$$\n- **For $r=4$:** \n $$1 \\times 1 \\times 16 = 16$$"},{"type":"text","order":2,"title":"Write out full polynomial","content":"Combining all the calculated terms gives us the complete expanded polynomial:\n\n$$(3x - 2)^4 = 81x^4 - 216x^3 + 216x^2 - 96x + 16$$"},{"type":"text","order":3,"title":"Identify the constant term","content":"The constant term is the term independent of $x$ (the term where the exponent of $x$ is zero).\n\nFrom our expanded polynomial, this is the final term:\n\n$$\\text{Constant term} = 16$$"}]},{"id":"manual_multiplication","label":"Manual Algebraic Expansion","steps":[{"type":"text","order":0,"title":"Rewrite the expression","content":"To expand $(3x - 2)^4$ manually, we can rewrite it as the product of two quadratic expressions:\n\n$$(3x - 2)^4 = (3x - 2)^2(3x - 2)^2$$"},{"type":"text","order":1,"title":"Expand the quadratic term","content":"Let's first expand the quadratic term $(3x - 2)^2$:\n\n$$\\begin{aligned} (3x - 2)^2 &= (3x - 2)(3x - 2) \\\\ &= (3x)(3x) + (3x)(-2) + (-2)(3x) + (-2)(-2) \\\\ &= 9x^2 - 6x - 6x + 4 \\\\ &= 9x^2 - 12x + 4 \\end{aligned}$$"},{"type":"text","order":2,"title":"Set up the full product","content":"Now substitute this result back into our original expression:\n\n$$(3x - 2)^4 = (9x^2 - 12x + 4)(9x^2 - 12x + 4)$$"},{"type":"text","order":3,"title":"Identify the constant term","content":"The question asks for the **constant term** (the term independent of $x$, containing no $x$ variables).\n\nWhen multiplying the two trinomials, the only product that results in a constant term is the multiplication of the constant term from the first bracket by the constant term from the second bracket:\n\n$$\\text{Constant Term} = 4 \\times 4 = 16$$"},{"type":"text","order":4,"title":"State the final answer","content":"The constant term in the expansion of $(3x - 2)^4$ is **$16$**."}]}],"generatedAt":"2026-06-11T03:21:26.524Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"5223e7bd-a957-4fe3-b334-bc66a9604836:binomial_term_formula","label":"Direct General Term Formula","steps":[{"type":"text","order":0,"title":"Identify general term","content":"To find the constant term in the expansion of $(3x - 2)^4$, we can use the general term formula from the Binomial Theorem.\n\nFor an expansion of the form $(a + b)^n$, the general term $T_{r+1}$ is given by:\n\n$$T_{r+1} = \\binom{n}{r} a^{n-r} b^r$$\n\nIn this problem, we have:\n- $a = 3x$\n- $b = -2$\n- $n = 4$\n\nSubstituting these values into the formula gives:\n\n$$T_{r+1} = \\binom{4}{r} (3x)^{4-r} (-2)^r$$","highlights":[{"text":"$$(3x - 2)^4$","location":"specification"}]},{"type":"text","order":1,"title":"Set power to zero","content":"A constant term is independent of $x$, meaning the exponent of $x$ must be $0$.\n\nLet's isolate the $x$ part in our general term:\n\n$$T_{r+1} = \\binom{4}{r} 3^{4-r} (-2)^r x^{4-r}$$\n\nTo find the constant term, we set the exponent of $x$ to $0$:\n\n$$4 - r = 0 \\implies r = 4$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$T_{4+1} = \\binom{4}{\\htmlData{anchor=r-val}{\\color{@sky}{4}}} (3x)^{4-\\htmlData{anchor=r-sub1}{\\color{@sky}{4}}} (\\htmlData{anchor=b-val}{\\color{@amber}{-2}})^{\\htmlData{anchor=r-sub2}{\\color{@sky}{4}}}$"},{"latex":"$$T_5 = \\binom{4}{4} (3x)^0 (-2)^4$"},{"latex":"$$T_5 = 1 \\times 1 \\times 16 = 16$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"r-val"},{"color":"amber","style":"text-color","anchor":"b-val"}],"connections":[{"to":"r-sub1","from":"r-val","color":"sky","label":"r = 4","style":"solid"},{"to":"r-sub2","from":"r-val","color":"sky","label":null,"style":"solid"}]},"order":2,"title":"Evaluate constant term","description":"Substitute $r = 4$ back into our general term expression to compute the constant term."},{"type":"text","order":3,"title":"Confirm final answer","content":"The constant term in the expansion is **$16$**.\n\n### Markscheme Guidance\n- **1 mark (M1)**: Correctly identifying the term independent of $x$ as $\\binom{4}{4} (3x)^0 (-2)^4$.\n- **1 mark (A1)**: Correctly evaluating this term to $16$."}]},{"id":"5223e7bd-a957-4fe3-b334-bc66a9604836:full_expansion","label":"Full Binomial Expansion","steps":[{"type":"text","order":0,"title":"Set up the expansion","content":"To find the constant term by fully expanding the expression, we use the Binomial Theorem:\n\n$$(a + b)^n = \\sum_{r=0}^{n} \\binom{n}{r} a^{n-r} b^r$$\n\nIn our case, we have:\n- $a = 3x$\n- $b = -2$\n- $n = 4$\n\nLet's write down the five terms of this expansion, corresponding to $r = 0, 1, 2, 3, 4$:\n\n$$\\begin{aligned} (3x - 2)^4 = &\\binom{4}{0}(3x)^4(-2)^0 + \\binom{4}{1}(3x)^3(-2)^1 + \\binom{4}{2}(3x)^2(-2)^2 \\\\ &+ \\binom{4}{3}(3x)^1(-2)^3 + \\binom{4}{4}(3x)^0(-2)^4 \\end{aligned}$$"},{"type":"text","order":1,"title":"Calculate each term","content":"Now we evaluate each part of the sum step by step:\n\n- **For $r=0$:** \n $$1 \\times 81x^4 \\times 1 = 81x^4$$\n- **For $r=1$:** \n $$4 \\times 27x^3 \\times (-2) = -216x^3$$\n- **For $r=2$:** \n $$6 \\times 9x^2 \\times 4 = 216x^2$$\n- **For $r=3$:** \n $$4 \\times 3x \\times (-8) = -96x$$\n- **For $r=4$:** \n $$1 \\times 1 \\times 16 = 16$$"},{"type":"text","order":2,"title":"Write out full polynomial","content":"Combining all the calculated terms gives us the complete expanded polynomial:\n\n$$(3x - 2)^4 = 81x^4 - 216x^3 + 216x^2 - 96x + 16$$"},{"type":"text","order":3,"title":"Identify the constant term","content":"The constant term is the term independent of $x$ (the term where the exponent of $x$ is zero).\n\nFrom our expanded polynomial, this is the final term:\n\n$$\\text{Constant term} = 16$$"}]},{"id":"5223e7bd-a957-4fe3-b334-bc66a9604836:manual_multiplication","label":"Manual Algebraic Expansion","steps":[{"type":"text","order":0,"title":"Rewrite the expression","content":"To expand $(3x - 2)^4$ manually, we can rewrite it as the product of two quadratic expressions:\n\n$$(3x - 2)^4 = (3x - 2)^2(3x - 2)^2$$"},{"type":"text","order":1,"title":"Expand the quadratic term","content":"Let's first expand the quadratic term $(3x - 2)^2$:\n\n$$\\begin{aligned} (3x - 2)^2 &= (3x - 2)(3x - 2) \\\\ &= (3x)(3x) + (3x)(-2) + (-2)(3x) + (-2)(-2) \\\\ &= 9x^2 - 6x - 6x + 4 \\\\ &= 9x^2 - 12x + 4 \\end{aligned}$$"},{"type":"text","order":2,"title":"Set up the full product","content":"Now substitute this result back into our original expression:\n\n$$(3x - 2)^4 = (9x^2 - 12x + 4)(9x^2 - 12x + 4)$$"},{"type":"text","order":3,"title":"Identify the constant term","content":"The question asks for the **constant term** (the term independent of $x$, containing no $x$ variables).\n\nWhen multiplying the two trinomials, the only product that results in a constant term is the multiplication of the constant term from the first bracket by the constant term from the second bracket:\n\n$$\\text{Constant Term} = 4 \\times 4 = 16$$"},{"type":"text","order":4,"title":"State the final answer","content":"The constant term in the expansion of $(3x - 2)^4$ is **$16$**."}]}],"generatedAt":"2026-06-11T03:21:26.615Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"99e59da2-94f2-44bf-8774-2d57bdeaf4e6","revisionId":"23483958-7802-4097-914d-fd5132d895e2","questionNumber":null,"questionType":"LA","specification":"","difficulty":null,"difficultyBand":"easy","tier":null,"totalMarks":3,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"fe600a5c-db7b-4145-8038-30358ca78cc3","label":null,"content":"Find the first three terms, in ascending powers of $x$, in the expansion of $(2 - x)^6$.","markscheme":"- $2^6 + \\binom{6}{1}2^5(-x) + \\binom{6}{2}2^4(-x)^2$ M1\n- Two of the three terms correct A1\n- $64 - 192x + 240x^2$ A1\n\nThe sign of the $x$ term must be negative; the $x^2$ term is positive because $(-x)^2 = 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