Answer: (a) \(y^4-2y^3+4y^2+2y+11=0\)
(b) \(S_2=-4\)
(c) \(S_4=-76\)
(a)
Let \(y=2x+1\). Then \(x=\frac12(y-1)\). Substitute into the original equation:
\(\left(\frac{y-1}{2}\right)^4+\left(\frac{y-1}{2}\right)^3+\left(\frac{y-1}{2}\right)^2+\frac{y-1}{2}+1=0.\)
That is
\(\frac1{16}(y-1)^4+\frac18(y-1)^3+\frac14(y-1)^2+\frac12(y-1)+1=0.\)
Expanding and simplifying gives
\(\frac1{16}y^4-\frac18y^3+\frac14y^2+\frac18y+\frac{11}{16}=0.\)
Multiplying by \(16\),
\(y^4-2y^3+4y^2+2y+11=0.\)
(b)
For the transformed quartic \(y^4-2y^3+4y^2+2y+11=0\), the sum of roots is \(S_1=2\), and the sum of products in pairs is \(4\). Hence
\(S_2=(\text{sum of roots})^2-2(\text{sum of products in pairs})=2^2-2(4)=-4.\)
(c)
Each transformed root \(y\) satisfies
\(y^4-2y^3+4y^2+2y+11=0.\)
Sum this equation over the four roots:
\(S_4-2S_3+4S_2+2S_1+44=0.\)
Using \(S_1=2\), \(S_2=-4\), and \(S_3=-22\),
\(S_4-2(-22)+4(-4)+2(2)+44=0,\)
so \(S_4+76=0\), and therefore
\(S_4=-76.\)
Mark scheme
| Part | Marks | Expected result | Notes |
|---|
| (a) | B1 | \(y=2x+1\Rightarrow x=\frac12(y-1)\) | Correct substitution used in equation. |
| (a) | M1 A1 | \(\frac1{16}(y-1)^4+\frac18(y-1)^3+\frac14(y-1)^2+\frac12(y-1)+1=0\) | Expands. |
| (a) | A1 | \(\frac1{16}y^4-\frac18y^3+\frac14y^2+\frac18y+\frac{11}{16}=0\Rightarrow y^4-2y^3+4y^2+2y+11=0\) | Simplifies, AG. |
| (b) | M1 | \(S_2=2^2-2(4)\) | Uses \((\alpha+\beta+\gamma+\delta)^2=\alpha^2+\beta^2+\gamma^2+\delta^2+2(\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta)\). |
| (b) | A1 | \(S_2=-4\) | |
| (c) | M1 | \(S_4-2S_3+4S_2+2S_1+44=0\) | Sums over roots. |
| (c) | A1 | \(S_4=-76\) | |