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Determine the constant term in the binomial expansion of \$P(t)\$.","markscheme":"- Identifies that the constant term in the binomial expansion is the term independent of \$t\$, which corresponds to \$\\binom{5}{5} \\left(\\frac{1}{2}t\\right)^0 (-3)^5\$ or simply \$(-3)^5\$ M1\n- Evaluates the constant term to be \$-243\$ A1","marks":2,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"binomial_formula","label":"Binomial Formula Approach","steps":[{"type":"text","order":0,"title":"Identify the general term formula","content":"To find the constant term in the binomial expansion of $P(t) = \\left(\\frac{1}{2}t - 3\\right)^5$, we use the general term formula of a binomial expansion:\n\n$$T_{r+1} = \\binom{n}{r} a^{n-r} b^r$$\n\nHere, we identify:\n- $n = 5$\n- $a = \\frac{1}{2}t$\n- $b = -3$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$T_{r+1} = \\binom{\\htmlData{anchor=n-val}{\\color{@sky}{5}}}{r} \\left(\\htmlData{anchor=a-val}{\\color{@amber}{\\frac{1}{2}t}}\\right)^{5-r} (\\htmlData{anchor=b-val}{\\color{@purple}{-3}})^r$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"n-val"},{"color":"amber","style":"text-color","anchor":"a-val"},{"color":"purple","style":"text-color","anchor":"b-val"}],"connections":[]},"order":1,"title":"Set up the general term","description":"Substitute $n = 5$, $a = \\frac{1}{2}t$, and $b = -3$ into the general term formula."},{"type":"text","order":2,"title":"Find the constant term condition","content":"The constant term is the term independent of $t$, which means the power of $t$ must be $0$.\n\nFrom our general term:\n$$T_{r+1} = \\binom{5}{r} \\left(\\frac{1}{2}\\right)^{5-r} t^{5-r} (-3)^r$$\n\nFor the exponent of $t$ to be zero, we require:\n$$5 - r = 0 \\implies r = 5$$"},{"type":"text","order":3,"title":"Substitute and evaluate","content":"Now substitute $r = 5$ into the general term expression to find the constant value:\n\n$$\\begin{aligned} T_{6} &= \\binom{5}{5} \\left(\\frac{1}{2}t\\right)^0 (-3)^5 \\\\ &= 1 \\cdot 1 \\cdot (-3)^5 \\\\ &= -243 \\end{aligned}$$\n\nThis calculation awards:\n- M1 for identifying that the constant term is $(-3)^5$ or $\\binom{5}{5} \\left(\\frac{1}{2}t\\right)^0 (-3)^5$.\n- A1 for evaluating the term to $-243$."}]},{"id":"evaluating_zero","label":"Evaluating at t = 0","steps":[{"type":"text","order":0,"title":"Understand the constant term","content":"The constant term in a polynomial expression (such as the binomial expansion of $P(t)$) is the term that is completely independent of the variable $t$.\n\nA very elegant and efficient way to isolate this constant term is to evaluate the polynomial at $t = 0$, because any term containing $t$ will become zero, leaving only the constant term."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$P(\\htmlData{anchor=t-val}{\\color{@sky}{t}}) = \\left(\\frac{1}{2}\\htmlData{anchor=t-var}{\\color{@sky}{t}} - 3\\right)^5$"},{"latex":"$$P(\\htmlData{anchor=t-zero-val}{\\color{@sky}{0}}) = \\left(\\frac{1}{2}(\\htmlData{anchor=t-zero-sub}{\\color{@sky}{0}}) - 3\\right)^5$"},{"latex":"$$\\phantom{P(0)} = (\\htmlData{anchor=term-zero}{\\color{@amber}{0}} - 3)^5$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"t-zero-sub"},{"color":"amber","style":"text-color","anchor":"term-zero"}],"connections":[{"to":"t-zero-val","from":"t-val","color":"sky","label":"Evaluate at t = 0","style":"solid"},{"to":"t-zero-sub","from":"t-var","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Substitute $t = 0$","description":"We substitute $t = 0$ into our pressure function $P(t)$ to eliminate all the terms that depend on $t$."},{"type":"text","order":2,"title":"Evaluate the constant term","content":"Now we simplify and evaluate the remaining power:\n\n$$P(0) = (-3)^5$$\n\nCalculating $(-3)^5$:\n\n$$(-3)^5 = -243$$\n\nThis gives us our constant term of $-243$.\n\n### Markscheme Breakdown\n- **M1**: Realising that the constant term is obtained when $t=0$, which corresponds to evaluating $(-3)^5$.\n- **A1**: Correctly calculating the final value of $-243$.","calculator":[{"gdc":{"expression":"(-3)^5","instructions":"In the Run-Matrix menu, enter `(-3)^5` and press `EXE`."},"ti84":{"expression":"(-3)^5","instructions":"Type `(-3)^5` on the main screen and press `ENTER`."},"desmos":{"expression":"(-3)^5","instructions":"Type `(-3)^5` into the input bar to see the result directly."},"description":"You can verify this calculation easily on any scientific or graphing calculator."}]}]},{"id":"full_expansion","label":"Full Binomial Expansion","steps":[{"type":"text","order":0,"title":"Recall the binomial theorem","content":"To find the constant term in the binomial expansion of $P(t) = \\left(\\frac{1}{2}t - 3\\right)^5$, we can write out the full binomial expansion using the binomial theorem:\n\n$$(a + b)^n = \\sum_{k=0}^{n} \\binom{n}{k} a^{n-k} b^k$$\n\nIn our expression:\n- $a = \\frac{1}{2}t$\n- $b = -3$\n- $n = 5$\n\nLet's write down the coefficients for $n = 5$ using Pascal's triangle or $\\binom{5}{k}$:\n$$\\binom{5}{0}=1, \\quad \\binom{5}{1}=5, \\quad \\binom{5}{2}=10, \\quad \\binom{5}{3}=10, \\quad \\binom{5}{4}=5, \\quad \\binom{5}{5}=1$$"},{"type":"text","order":1,"title":"Write the full expansion","content":"Let's expand $\\left(\\frac{1}{2}t - 3\\right)^5$ fully step-by-step using these coefficients:\n\n$$\\begin{aligned} \\left(\\frac{1}{2}t - 3\\right)^5 = &\\ \\binom{5}{0}\\left(\\frac{1}{2}t\\right)^5(-3)^0 \\\\ &+ \\binom{5}{1}\\left(\\frac{1}{2}t\\right)^4(-3)^1 \\\\ &+ \\binom{5}{2}\\left(\\frac{1}{2}t\\right)^3(-3)^2 \\\\ &+ \\binom{5}{3}\\left(\\frac{1}{2}t\\right)^2(-3)^3 \\\\ &+ \\binom{5}{4}\\left(\\frac{1}{2}t\\right)^1(-3)^4 \\\\ &+ \\binom{5}{5}\\left(\\frac{1}{2}t\\right)^0(-3)^5 \\end{aligned}$$"},{"type":"text","order":2,"title":"Simplify each term","content":"Now we simplify each of the six terms:\n\n$$\\begin{aligned} T_1 &= 1 \\cdot \\frac{1}{32}t^5 \\cdot 1 = \\frac{1}{32}t^5 \\\\ T_2 &= 5 \\cdot \\frac{1}{16}t^4 \\cdot (-3) = -\\frac{15}{16}t^4 \\\\ T_3 &= 10 \\cdot \\frac{1}{8}t^3 \\cdot 9 = \\frac{45}{4}t^3 \\\\ T_4 &= 10 \\cdot \\frac{1}{4}t^2 \\cdot (-27) = -\\frac{135}{2}t^2 \\\\ T_5 &= 5 \\cdot \\frac{1}{2}t \\cdot 81 = \\frac{405}{2}t \\\\ T_6 &= 1 \\cdot 1 \\cdot (-243) = -243 \\end{aligned}$$\n\nPutting them all together, the fully expanded function is:\n\n$$P(t) = \\frac{1}{32}t^5 - \\frac{15}{16}t^4 + \\frac{45}{4}t^3 - \\frac{135}{2}t^2 + \\frac{405}{2}t - 243$$"},{"type":"text","order":3,"title":"Identify the constant term","content":"The constant term is the term that is independent of $t$ (meaning it has a power of $t^0$):\n\n$$\\text{Constant term} = T_6 = -243$$\n\n**Marks awarded:**\n- **1 Mark (M1)**: For identifying the term independent of $t$, which is $\\binom{5}{5} \\left(\\frac{1}{2}t\\right)^0 (-3)^5$ or simply $(-3)^5$.\n- **1 Mark (A1)**: For evaluating this term to get $-243$."}]}],"generatedAt":"2026-06-09T05:15:32.871Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"0bd8a7ef-2ccf-4abc-831d-d46ee7a5dd57:binomial_formula","label":"Binomial Formula Approach","steps":[{"type":"text","order":0,"title":"Identify the general term formula","content":"To find the constant term in the binomial expansion of $P(t) = \\left(\\frac{1}{2}t - 3\\right)^5$, we use the general term formula of a binomial expansion:\n\n$$T_{r+1} = \\binom{n}{r} a^{n-r} b^r$$\n\nHere, we identify:\n- $n = 5$\n- $a = \\frac{1}{2}t$\n- $b = -3$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$T_{r+1} = \\binom{\\htmlData{anchor=n-val}{\\color{@sky}{5}}}{r} \\left(\\htmlData{anchor=a-val}{\\color{@amber}{\\frac{1}{2}t}}\\right)^{5-r} (\\htmlData{anchor=b-val}{\\color{@purple}{-3}})^r$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"n-val"},{"color":"amber","style":"text-color","anchor":"a-val"},{"color":"purple","style":"text-color","anchor":"b-val"}],"connections":[]},"order":1,"title":"Set up the general term","description":"Substitute $n = 5$, $a = \\frac{1}{2}t$, and $b = -3$ into the general term formula."},{"type":"text","order":2,"title":"Find the constant term condition","content":"The constant term is the term independent of $t$, which means the power of $t$ must be $0$.\n\nFrom our general term:\n$$T_{r+1} = \\binom{5}{r} \\left(\\frac{1}{2}\\right)^{5-r} t^{5-r} (-3)^r$$\n\nFor the exponent of $t$ to be zero, we require:\n$$5 - r = 0 \\implies r = 5$$"},{"type":"text","order":3,"title":"Substitute and evaluate","content":"Now substitute $r = 5$ into the general term expression to find the constant value:\n\n$$\\begin{aligned} T_{6} &= \\binom{5}{5} \\left(\\frac{1}{2}t\\right)^0 (-3)^5 \\\\ &= 1 \\cdot 1 \\cdot (-3)^5 \\\\ &= -243 \\end{aligned}$$\n\nThis calculation awards:\n- M1 for identifying that the constant term is $(-3)^5$ or $\\binom{5}{5} \\left(\\frac{1}{2}t\\right)^0 (-3)^5$.\n- A1 for evaluating the term to $-243$."}]},{"id":"0bd8a7ef-2ccf-4abc-831d-d46ee7a5dd57:evaluating_zero","label":"Evaluating at t = 0","steps":[{"type":"text","order":0,"title":"Understand the constant term","content":"The constant term in a polynomial expression (such as the binomial expansion of $P(t)$) is the term that is completely independent of the variable $t$.\n\nA very elegant and efficient way to isolate this constant term is to evaluate the polynomial at $t = 0$, because any term containing $t$ will become zero, leaving only the constant term."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$P(\\htmlData{anchor=t-val}{\\color{@sky}{t}}) = \\left(\\frac{1}{2}\\htmlData{anchor=t-var}{\\color{@sky}{t}} - 3\\right)^5$"},{"latex":"$$P(\\htmlData{anchor=t-zero-val}{\\color{@sky}{0}}) = \\left(\\frac{1}{2}(\\htmlData{anchor=t-zero-sub}{\\color{@sky}{0}}) - 3\\right)^5$"},{"latex":"$$\\phantom{P(0)} = (\\htmlData{anchor=term-zero}{\\color{@amber}{0}} - 3)^5$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"t-zero-sub"},{"color":"amber","style":"text-color","anchor":"term-zero"}],"connections":[{"to":"t-zero-val","from":"t-val","color":"sky","label":"Evaluate at t = 0","style":"solid"},{"to":"t-zero-sub","from":"t-var","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Substitute $t = 0$","description":"We substitute $t = 0$ into our pressure function $P(t)$ to eliminate all the terms that depend on $t$."},{"type":"text","order":2,"title":"Evaluate the constant term","content":"Now we simplify and evaluate the remaining power:\n\n$$P(0) = (-3)^5$$\n\nCalculating $(-3)^5$:\n\n$$(-3)^5 = -243$$\n\nThis gives us our constant term of $-243$.\n\n### Markscheme Breakdown\n- **M1**: Realising that the constant term is obtained when $t=0$, which corresponds to evaluating $(-3)^5$.\n- **A1**: Correctly calculating the final value of $-243$.","calculator":[{"gdc":{"expression":"(-3)^5","instructions":"In the Run-Matrix menu, enter `(-3)^5` and press `EXE`."},"ti84":{"expression":"(-3)^5","instructions":"Type `(-3)^5` on the main screen and press `ENTER`."},"desmos":{"expression":"(-3)^5","instructions":"Type `(-3)^5` into the input bar to see the result directly."},"description":"You can verify this calculation easily on any scientific or graphing calculator."}]}]},{"id":"0bd8a7ef-2ccf-4abc-831d-d46ee7a5dd57:full_expansion","label":"Full Binomial Expansion","steps":[{"type":"text","order":0,"title":"Recall the binomial theorem","content":"To find the constant term in the binomial expansion of $P(t) = \\left(\\frac{1}{2}t - 3\\right)^5$, we can write out the full binomial expansion using the binomial theorem:\n\n$$(a + b)^n = \\sum_{k=0}^{n} \\binom{n}{k} a^{n-k} b^k$$\n\nIn our expression:\n- $a = \\frac{1}{2}t$\n- $b = -3$\n- $n = 5$\n\nLet's write down the coefficients for $n = 5$ using Pascal's triangle or $\\binom{5}{k}$:\n$$\\binom{5}{0}=1, \\quad \\binom{5}{1}=5, \\quad \\binom{5}{2}=10, \\quad \\binom{5}{3}=10, \\quad \\binom{5}{4}=5, \\quad \\binom{5}{5}=1$$"},{"type":"text","order":1,"title":"Write the full expansion","content":"Let's expand $\\left(\\frac{1}{2}t - 3\\right)^5$ fully step-by-step using these coefficients:\n\n$$\\begin{aligned} \\left(\\frac{1}{2}t - 3\\right)^5 = &\\ \\binom{5}{0}\\left(\\frac{1}{2}t\\right)^5(-3)^0 \\\\ &+ \\binom{5}{1}\\left(\\frac{1}{2}t\\right)^4(-3)^1 \\\\ &+ \\binom{5}{2}\\left(\\frac{1}{2}t\\right)^3(-3)^2 \\\\ &+ \\binom{5}{3}\\left(\\frac{1}{2}t\\right)^2(-3)^3 \\\\ &+ \\binom{5}{4}\\left(\\frac{1}{2}t\\right)^1(-3)^4 \\\\ &+ \\binom{5}{5}\\left(\\frac{1}{2}t\\right)^0(-3)^5 \\end{aligned}$$"},{"type":"text","order":2,"title":"Simplify each term","content":"Now we simplify each of the six terms:\n\n$$\\begin{aligned} T_1 &= 1 \\cdot \\frac{1}{32}t^5 \\cdot 1 = \\frac{1}{32}t^5 \\\\ T_2 &= 5 \\cdot \\frac{1}{16}t^4 \\cdot (-3) = -\\frac{15}{16}t^4 \\\\ T_3 &= 10 \\cdot \\frac{1}{8}t^3 \\cdot 9 = \\frac{45}{4}t^3 \\\\ T_4 &= 10 \\cdot \\frac{1}{4}t^2 \\cdot (-27) = -\\frac{135}{2}t^2 \\\\ T_5 &= 5 \\cdot \\frac{1}{2}t \\cdot 81 = \\frac{405}{2}t \\\\ T_6 &= 1 \\cdot 1 \\cdot (-243) = -243 \\end{aligned}$$\n\nPutting them all together, the fully expanded function is:\n\n$$P(t) = \\frac{1}{32}t^5 - \\frac{15}{16}t^4 + \\frac{45}{4}t^3 - \\frac{135}{2}t^2 + \\frac{405}{2}t - 243$$"},{"type":"text","order":3,"title":"Identify the constant term","content":"The constant term is the term that is independent of $t$ (meaning it has a power of $t^0$):\n\n$$\\text{Constant term} = T_6 = -243$$\n\n**Marks awarded:**\n- **1 Mark (M1)**: For identifying the term independent of $t$, which is $\\binom{5}{5} \\left(\\frac{1}{2}t\\right)^0 (-3)^5$ or simply $(-3)^5$.\n- **1 Mark (A1)**: For evaluating this term to get $-243$."}]}],"generatedAt":"2026-06-09T05:15:42.856Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"6aad7e90-e6a1-4e9b-bcdd-dac4fb902d3f","revisionId":"3d01b55c-88a6-406b-bcd0-0bcf95bcc69c","questionNumber":null,"questionType":"LA","specification":"","difficulty":2,"difficultyBand":"easy","tier":null,"totalMarks":2,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"5223e7bd-a957-4fe3-b334-bc66a9604836","label":null,"content":"Find the constant term in the expansion of \$(3x - 2)^4\$.","markscheme":"- Identifies that the constant term is the term independent of $x$, given by the term $\\binom{4}{4} (3x)^0 (-2)^4$ M1\n- Evaluates the constant term as $1 \\times 1 \\times 16 = 16$ 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":"71fe1e35-de1f-4e32-b9b5-0508b7c5ee72","revisionId":"0f74f61d-2394-45c3-918c-b770810887cc","questionNumber":null,"questionType":"LA","specification":"","difficulty":null,"difficultyBand":"easy","tier":null,"totalMarks":2,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"7b95e2ee-e1c3-4119-b0c4-cd6639c73854","label":null,"content":"Expand $(1+x)^4$.","markscheme":"- Use coefficients $\\binom40,\\binom41,\\binom42,\\binom43,\\binom44$ M1\n- $1+4x+6x^2+4x^3+x^4$ 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1.3 Sequences and series

Question 5

1.3 Sequences and series Questions

1.3 Sequences and series

Question 5

1.3 Sequences and series Questions