Standard +0.3 This is a standard Further Maths technique requiring the substitution x = t - 2 to shift roots by 2, then expanding and simplifying the resulting polynomial. While it involves algebraic manipulation across multiple steps, it's a routine textbook exercise with a well-established method that students are explicitly trained to execute.
1 The roots of the equation \(4 x ^ { 4 } - 2 x ^ { 3 } - 3 x + 2 = 0\) are \(\alpha , \beta , \gamma\) and \(\delta\). By using a suitable substitution, find a quartic equation whose roots are \(\alpha + 2 , \beta + 2 , \gamma + 2\) and \(\delta + 2\) giving your answer in the form \(a t ^ { 4 } + b t ^ { 3 } + c t ^ { 2 } + d t + e = 0\), where \(a , b , c , d\), and \(e\) are integers.
Attempt to expand \((u-2)^3\) following a linear substitution. 4 terms using \(\binom{n}{r}2^r u^{n-r}\). Note: \(u^3 - 6u^2 + 12u - 8\). Binomial coefficients might not be correct or evaluated. Might be by expanding brackets.
Attempt to expand \((u-2)^4\) following a linear substitution. 5 terms using \(\binom{n}{r}2^r u^{n-r}\). Note: \(u^4 - 8u^3 + 24u^2 - 32u + 16\). Binomial coefficients might not be correct or evaluated. Might be by expanding brackets.
\(4(u-2)^4 - 2(u-2)^3 - 3(u-2) + 2 (= 0)\)
M1
Forming (LHS of) equation in \(u\) (could be using their expansions)
Final answer can be in terms of \(x\). Final answer needs to be an equation with coefficients simplified. ISW attempts to cancel down following correct equation.
[5]
## Question 1:
| Answer | Mark | Guidance |
|--------|------|----------|
| $u = x + 2$ | **B1** | Correct substitution stated or used |
| $(u-2)^3 = u^3 - 3\times2u^2 + 3\times2^2u - 2^3$ | **M1** | Attempt to expand $(u-2)^3$ following a linear substitution. 4 terms using $\binom{n}{r}2^r u^{n-r}$. Note: $u^3 - 6u^2 + 12u - 8$. Binomial coefficients might not be correct or evaluated. Might be by expanding brackets. |
| $(u-2)^4 = u^4 - 4\times2u^3 + 6\times2^2u^2 - 4\times2^3u + 2^4$ | **M1** | Attempt to expand $(u-2)^4$ following a linear substitution. 5 terms using $\binom{n}{r}2^r u^{n-r}$. Note: $u^4 - 8u^3 + 24u^2 - 32u + 16$. Binomial coefficients might not be correct or evaluated. Might be by expanding brackets. |
| $4(u-2)^4 - 2(u-2)^3 - 3(u-2) + 2 (= 0)$ | **M1** | Forming (LHS of) equation in $u$ (could be using their expansions) |
| $\therefore 4(u^4 - 8u^3 + 24u^2 - 32u + 16) - 2(u^3 - 6u^2 + 12u - 8) - 3(u-2) + 2 = 0$ $\therefore 4u^4 - 34u^3 + 108u^2 - 155u + 88 = 0$ | **A1** | Final answer can be in terms of $x$. Final answer needs to be an equation with coefficients simplified. ISW attempts to cancel down following correct equation. |
**[5]**
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1 The roots of the equation $4 x ^ { 4 } - 2 x ^ { 3 } - 3 x + 2 = 0$ are $\alpha , \beta , \gamma$ and $\delta$. By using a suitable substitution, find a quartic equation whose roots are $\alpha + 2 , \beta + 2 , \gamma + 2$ and $\delta + 2$ giving your answer in the form $a t ^ { 4 } + b t ^ { 3 } + c t ^ { 2 } + d t + e = 0$, where $a , b , c , d$, and $e$ are integers.
\hfill \mbox{\textit{OCR Further Pure Core AS 2023 Q1 [5]}}