Pre-U Pre-U 9795/1 Specimen — Question 2 5 marks

Exam BoardPre-U
ModulePre-U 9795/1 (Pre-U Further Mathematics Paper 1)
SessionSpecimen
Marks5
TopicRoots of polynomials
TypeSum of powers of roots
DifficultyStandard +0.3 This is a straightforward application of Newton's identities and Vieta's formulas for a cubic polynomial. The first sum is immediate from coefficients, the second requires one algebraic manipulation (squaring the sum), and the third uses a standard identity. While it's a Further Maths topic, the execution is mechanical with no problem-solving insight required, making it slightly easier than average.
Spec4.05a Roots and coefficients: symmetric functions

2 The equation \(x ^ { 3 } - 14 x ^ { 2 } + 16 x + 21 = 0\) has roots \(\alpha , \beta , \gamma\). Determine the values of \(\alpha + \beta + \gamma\), \(\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 }\) and \(\alpha ^ { 3 } + \beta ^ { 3 } + \gamma ^ { 3 }\).

\(\alpha + \beta + \gamma = 14\): B1
\(\alpha^2 + \beta^2 + \gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\): M1
\(= 14^2 - 2 \times 16 = 164\) ft: A1
\(\alpha^3 + \beta^3 + \gamma^3 = 14\sum\alpha^2 - 16\sum\alpha - 21\sum 1\)
For substituting roots into equation & adding: M1
\(= 2009\) cao: A1
Total: 5 marks
$\alpha + \beta + \gamma = 14$: B1

$\alpha^2 + \beta^2 + \gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)$: M1

$= 14^2 - 2 \times 16 = 164$ **ft**: A1

$\alpha^3 + \beta^3 + \gamma^3 = 14\sum\alpha^2 - 16\sum\alpha - 21\sum 1$

For substituting roots into equation & adding: M1

$= 2009$ **cao**: A1

**Total: 5 marks**
2 The equation $x ^ { 3 } - 14 x ^ { 2 } + 16 x + 21 = 0$ has roots $\alpha , \beta , \gamma$. Determine the values of $\alpha + \beta + \gamma$, $\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 }$ and $\alpha ^ { 3 } + \beta ^ { 3 } + \gamma ^ { 3 }$.

\hfill \mbox{\textit{Pre-U Pre-U 9795/1  Q2 [5]}}