Year: 2016
Paper: 2
Question Number: 2
Course: LFM Stats And Pure
Section: Polynomials
As in previous years the Pure questions were the most popular of the paper with questions 1, 3 and 7 the most popular of these. The least popular questions of the paper were questions 10, 11, 12 and 13 with fewer than 400 attempts for each of them. There were many examples of solutions in this paper that were insufficiently well explained, given that the answer to be reached had been provided in the question.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
Use the factor theorem to show that $a+b-c$ is a factor of
\[
(a+b+c)^3 -6(a+b+c)(a^2+b^2+c^2) +8(a^3+b^3+c^3)
\,.
\tag{$*$}
\]
Hence factorise ($*$) completely.
\begin{questionparts}
\item
Use the result above to solve the equation
\[
(x+1)^3 -3 (x+1)(2x^2 +5) +2(4x^3+13)=0\,.
\]
\item
By setting $d+e=c$, or otherwise, show that
$(a+b-d-e)$ is a factor of
\[
(a+b+d+e)^3 -6(a+b+d+e)(a^2+b^2+d^2+e^2) +8(a^3+b^3+d^3+e^3)
\,
\]
and factorise this expression completely.
Hence solve the equation
\[
(x+6)^3 - 6(x+6)(x^2+14) +8(x^3+36)=0\,.
\]
\end{questionparts}Suppose $c = a+b$ then
\begin{align*}
(a+b+c)^3 &-6(a+b+c)(a^2+b^2+c^2) +8(a^3+b^3+c^3) \\
&= (2(a+b))^3-6(2(a+b))(a^2+b^2+(a+b)^2) + 8(a^3+b^3+(a+b)^3) \\
&=16(a+b)^3 - 24(a+b)(a^2+b^2+ab)+8(a^3+b^3) \\
&= 8(a+b)(2(a+b)^2-3(a^2+b^2+ab)+(a^2-ab+b^2)) \\
&= 0
\end{align*}
Therefore $a+b-c$ is a factor. By symmetry $a-b+c$ and $-a+b+c$ are also factors. Since our polynomial is degree $3$ it must be
$K(a+b-c)(b+c-a)(c+a-b)$ for some $K$.
Since the coefficient of $a^3$ is $3$, $K = 3$. so we have: $3(a+b-c)(b+c-a)(c+a-b)$
\begin{questionparts}
\item
We want $x + a + b = x+1$, $x^2 + a^2 + b^2 = x^3+\frac52, x^3 + a^3 + b^3 = x^3+ \frac{13}{4}$. $a+b = 1, a^2 + b^2 = 5/2$ so $a = \frac32, b = -\frac12$
\begin{align*}
0 &= (x+1)^3 - 3(x+1)(2x^2+5)+2(4x^3+13) \\
&= 3(x +\frac{3}{2}+\frac{1}{2})(x - \frac{3}{2} - \frac{1}{2})(-x + \frac{3}{2} - \frac{1}{2}) \\
&= 3(x+2)(x-2)(1-x)
\end{align*}
and so the roots are $x = 1, 2, -2$
\item Letting $c = d+e$ we have
\begin{align*}
(a+b+d+e)^3 &-6(a+b+d+e)(a^2+b^2+d^2+e^2) +8(a^3+b^3+d^3+e^3) \\
&= (a+b+c)^3 -6(a+b+c)(a^2+b^2+c^2-2de) +8(a^3+b^3+c^3 - 3cde) \\
&= (a+b+c)^3 -6(a+b+c)(a^2+b^2+c^2)+8(a^3+b^3+c^3)+12(a+b+c)de - 24cde \\
&= \underbrace{(a+b+c)^3 -6(a+b+c)(a^2+b^2+c^2)+8(a^3+b^3+c^3)}_{\text{has a factor of }a+b-c} + 12(a+b-c)de
\end{align*}
Therefore there is a factor of $a+b-c$ or $a+b-d-e$. By symmetry we must have the factors:
$(a+b-d-e)(a-b-d+e)(a-b+d-e)$ and so the final expression must be:
$K(a+b-d-e)(a-b-d+e)(a-b+d-e)$
The coefficient of $a^3$ is $3$, therefore $K = 3$
We want $x+a+b+c = x + 6$, $x^2+a^2+b^2+c^2 = 14$ and $x^3 + a^3+b^3+c^3 = 36$, ie $a = 1,b=2,c=3$ would work, so
\begin{align*}
0 &= (x+6)^3 - 6(x+6)(x^2+14) +8(x^3+36) \\
&= 3(x+1-2-3)(x-1+2-3)(x-1-2+3) \\
&= 3x(x-4)(x-2)
\end{align*}
ie the roots are $x = 0, 2, 4$
\end{questionparts}
This question received poor responses in terms of the average mark per attempt. Application of the factor theorem to the first part often produced a neat solution. However, a number of candidates in the first part did not use the factor theorem as requested and so produced very complicated algebraic expressions that were more of a challenge to simplify. Where the result had been successfully shown and the expression factorised, many candidates were able to see the relevance to solving the equation in part (i). Those candidates who were able to follow through the method to successfully factorise the second expression were often able to complete the question.