SPS SPS SM Pure 2023 September — Question 7 9 marks

Exam BoardSPS
ModuleSPS SM Pure (SPS SM Pure)
Year2023
SessionSeptember
Marks9
TopicFactor & Remainder Theorem
TypeTwo polynomials, shared factor or separate conditions
DifficultyStandard +0.3 This is a straightforward application of the Factor Theorem requiring students to substitute x=3 into both polynomials, set equal to zero, and solve simultaneous equations. The factorization in part (b) is routine once p and q are found. All steps are standard textbook procedures with no novel insight required, making it slightly easier than average.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem

7. \(( x - 3 )\) is a common factor of \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\) where: $$\begin{aligned} & \mathrm { f } ( x ) = 2 x ^ { 3 } - 11 x ^ { 2 } + ( p - 15 ) x + q \\ & \mathrm {~g} ( x ) = 2 x ^ { 3 } - 17 x ^ { 2 } + p x + 2 q \end{aligned}$$
    1. Show that \(3 p + q = 90\) and \(3 p + 2 q = 99\) Fully justify your answer.
      1. (ii) Hence find the values of \(p\) and \(q\).
    2. \(\quad \mathrm { h } ( x ) = \mathrm { f } ( x ) + \mathrm { g } ( x )\) Using your values of \(p\) and \(q\), fully factorise \(\mathrm { h } ( x )\)

7.\\
$( x - 3 )$ is a common factor of $\mathrm { f } ( x )$ and $\mathrm { g } ( x )$ where:

$$\begin{aligned}
& \mathrm { f } ( x ) = 2 x ^ { 3 } - 11 x ^ { 2 } + ( p - 15 ) x + q \\
& \mathrm {~g} ( x ) = 2 x ^ { 3 } - 17 x ^ { 2 } + p x + 2 q
\end{aligned}$$
\begin{enumerate}[label=(\alph*)]
\item (i) Show that $3 p + q = 90$ and $3 p + 2 q = 99$

Fully justify your answer.\\
(a) (ii) Hence find the values of $p$ and $q$.
\item $\quad \mathrm { h } ( x ) = \mathrm { f } ( x ) + \mathrm { g } ( x )$

Using your values of $p$ and $q$, fully factorise $\mathrm { h } ( x )$
\end{enumerate}

\hfill \mbox{\textit{SPS SPS SM Pure 2023 Q7 [9]}}