Year: 2023
Paper: 2
Question Number: 4
Course: LFM Stats And Pure
Section: Polynomials
Many candidates were able to express their reasoning clearly and presented good solutions to the questions that they attempted. There were excellent solutions seen for all of the questions. An area where candidates struggled in several questions was in the direction of the logic that was required in a solution. Some candidates failed to appreciate that separate arguments may be needed for the "if" and "only if" parts of a question and, in some cases, candidates produced correct arguments, but for the wrong direction. In several questions it was clear that candidates who used sketches or diagrams generally performed much better that those who did not. Sketches often also helped to make the solution clearer and easier to understand. Several questions on the STEP papers ask candidates to show a given result. Candidates should be aware that there is a need to present sufficient detail in their solutions so that it is clear that the reasoning is well understood.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item Show that, if $(x-\sqrt{2})^2 = 3$, then $x^4 - 10x^2 + 1 = 0$.
Deduce that, if $\mathrm{f}(x) = x^4 - 10x^2 + 1$, then $\mathrm{f}(\sqrt{2}+\sqrt{3}) = 0$.
\item Find a polynomial $\mathrm{g}$ of degree 8 with integer coefficients such that $\mathrm{g}(\sqrt{2}+\sqrt{3}+\sqrt{5}) = 0$. Write your answer in a form without brackets.
\item Let $a$, $b$ and $c$ be the three roots of $t^3 - 3t + 1 = 0$.
Find a polynomial $\mathrm{h}$ of degree 6 with integer coefficients such that $\mathrm{h}(a+\sqrt{2}) = 0$, $\mathrm{h}(b+\sqrt{2}) = 0$ and $\mathrm{h}(c+\sqrt{2}) = 0$. Write your answer in a form without brackets.
\item Find a polynomial $\mathrm{k}$ with integer coefficients such that $\mathrm{k}(\sqrt[3]{2}+\sqrt[3]{3}) = 0$. Write your answer in a form without brackets.
\end{questionparts}\begin{questionparts}
\item $\,$ \begin{align*}
&& 3 &= (x-\sqrt2)^2 \\
&&&= x^2 - 2\sqrt2 x + 2 \\
\Rightarrow && 2\sqrt2 x &= x^2-1 \\
\Rightarrow && 8x^2 &= x^4 - 2x^2 + 1 \\
\Rightarrow && 0 &= x^4 - 10x^2 + 1
\end{align*}
Noticing that $(\sqrt2+\sqrt3-\sqrt2)^2 = 3$ we note that $\sqrt2 + \sqrt3$ is a root of our quartic.
\item Suppose $x = \sqrt2 + \sqrt3 + \sqrt5$ then \begin{align*}
&& 0 &= (x - \sqrt5)^4 - 10(x-\sqrt5)^2 + 1 \\
&&&= x^4 - 4\sqrt5x^3 + 30x^2-20\sqrt5 x +25 - 10x^2+20\sqrt5x -50 + 1\\
&&&= (x^4+20x^2- 24) - 4\sqrt5 x^3 \\
\Rightarrow && 80x^6 &= (x^4+20x^2-24)^2 \\
&&&= x^8 + 40x^6 + 352x^4 - 960x^2+576 \\
\Rightarrow && 0 &= x^8-40x^6 + 352x^4-960x^2+576
\end{align*}
So take $g(x) = x^8-40x^6 + 352x^4-960x^2+576$.
\item Notice that if $p(t) = t^3-3t+1$ then $p(t -\sqrt2) = 0$ for $t = a,b,c$ so
\begin{align*}
&& 0 &= (t - \sqrt2)^3 -3(t - \sqrt2) + 1 \\
&&&= t^3-3\sqrt2 t^2 + 6t - 2\sqrt2 - 3t + 3\sqrt 2 + 1 \\
&&&= (t^3+3t+1) - \sqrt2 (3t^2+1) \\
\Rightarrow && 2(3t^2+1)^2 &= (t^3+3t+1)^2 \\
\Rightarrow && 2(9t^4+6t^2+1) &= t^6 + 6t^4+2t^3+9t^2+6t+1 \\
\Rightarrow && 0 &= t^6-12t^4+2t^3-3t^2+6t-1
\end{align*}
\item $\,$ \begin{align*}
&& t &= \sqrt[3]{2} + \sqrt[3]{3} \\
\Rightarrow && t^3 &= 2 + 3\sqrt[3]{12} + 3\sqrt[3]{18} + 3 \\
&&&= 5 + 3 \sqrt[3]{6}(\sqrt[3]{2} + \sqrt[3]{3}) \\
&&&= 5 + 3\sqrt[3]{6}t \\
\Rightarrow && 162t^3 &= (t^3-5)^3 \\
&&&= t^9-15t^6+75t^3 -125 \\
\Rightarrow && 0 &= t^9-15t^6-87t^3-125
\end{align*}
so $k(x) = x^9 - 15x^6-87x^3-125$
\end{questionparts}
There were a wide variety of different approaches to part (i), including some which identified what the four roots of a quartic with integer coefficients would have to be in order for the required condition to be met. In part (ii) many candidates were able to identify a valid approach to the question although some algebraic errors meant that some did not reach the correct final polynomial. As with part (i) there were a number of different approaches that were taken. In part (iii) most candidates recognised that a translation of the graph would provide a cubic with the correct roots. Many were then able to apply similar methods to the earlier parts of the question to obtain the required polynomial with integer coefficients. Many candidates did not attempt the final part of the question, but those who did were generally able to adapt the methods from the previous parts successfully to make good progress.