Show equation reduces to polynomial

6 It is given that \(2 \ln ( 4 x - 5 ) + \ln ( x + 1 ) = 3 \ln 3\).

1. Show that \(16 x ^ { 3 } - 24 x ^ { 2 } - 15 x - 2 = 0\).
2. By first using the factor theorem, factorise \(16 x ^ { 3 } - 24 x ^ { 2 } - 15 x - 2\) completely.
3. Hence solve the equation \(2 \ln ( 4 x - 5 ) + \ln ( x + 1 ) = 3 \ln 3\).

7

1. Use the factor theorem to show that ( \(2 x + 3\) ) is a factor of $$8 x ^ { 3 } + 4 x ^ { 2 } - 10 x + 3$$
2. Show that the equation \(2 \cos 2 \theta = \frac { 6 \cos \theta - 5 } { 2 \cos \theta + 1 }\) can be expressed as $$8 \cos ^ { 3 } \theta + 4 \cos ^ { 2 } \theta - 10 \cos \theta + 3 = 0 .$$
3. Solve the equation \(2 \cos 2 \theta = \frac { 6 \cos \theta - 5 } { 2 \cos \theta + 1 }\) for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\).
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

7. (a) Use the factor theorem to show that \(( x + 1 )\) is a factor of \(x ^ { 3 } - x ^ { 2 } - 10 x - 8\).
(b) Find all the solutions of the equation \(x ^ { 3 } - x ^ { 2 } - 10 x - 8 = 0\).
(c) Prove that the value of \(x\) that satisfies $$2 \log _ { 2 } x + \log _ { 2 } ( x - 1 ) = 1 + \log _ { 2 } ( 5 x + 4 )$$ is a solution of the equation $$x ^ { 3 } - x ^ { 2 } - 10 x - 8 = 0$$ (d) State, with a reason, the value of \(x\) that satisfies equation (I).

2 Answer part (i) of this question on the insert provided.
The insert shows the graph of \(y = \frac { 1 } { x }\).

1. On the insert, on the same axes, plot the graph of \(y = x ^ { 2 } - 5 x + 5\) for \(0 \leqslant x \leqslant 5\).
2. Show algebraically that the \(x\)-coordinates of the points of intersection of the curves \(y = \frac { 1 } { x }\) and \(y = x ^ { 2 } - 5 x + 5\) satisfy the equation \(x ^ { 3 } - 5 x ^ { 2 } + 5 x - 1 = 0\).
3. Given that \(x = 1\) at one of the points of intersection of the curves, factorise \(x ^ { 3 } - 5 x ^ { 2 } + 5 x - 1\) into a linear and a quadratic factor. Show that only one of the three roots of \(x ^ { 3 } - 5 x ^ { 2 } + 5 x - 1 = 0\) is rational.

9 The polynomial \(f ( x )\) is given by $$f ( x ) = x ^ { 3 } + 6 x ^ { 2 } + x - 4 .$$

1. (a) Show that ( \(\mathrm { x } + 1\) ) is a factor of \(\mathrm { f } ( \mathrm { x } )\).
   (b) Hence find the exact roots of the equation \(f ( x ) = 0\).
2. (a) Show that the equation $$2 \log _ { 2 } ( x + 3 ) + \log _ { 2 } x - \log _ { 2 } ( 4 x + 2 ) = 1$$ can be written in the form \(f ( x ) = 0\).
   (b) Explain why the equation $$2 \log _ { 2 } ( x + 3 ) + \log _ { 2 } x - \log _ { 2 } ( 4 x + 2 ) = 1$$ has only one real root and state the exact value of this root.

1. \(\mathrm { f } ( x ) = x ^ { 3 } + a x ^ { 2 } - a x + 48\), where \(a\) is a constant

Given that \(\mathrm { f } ( - 6 ) = 0\)

1. 1. show that \(a = 4\)
   2. express \(\mathrm { f } ( x )\) as a product of two algebraic factors. Given that \(2 \log _ { 2 } ( x + 2 ) + \log _ { 2 } x - \log _ { 2 } ( x - 6 ) = 3\)
2. show that \(x ^ { 3 } + 4 x ^ { 2 } - 4 x + 48 = 0\)
3. hence explain why $$2 \log _ { 2 } ( x + 2 ) + \log _ { 2 } x - \log _ { 2 } ( x - 6 ) = 3$$ has no real roots.

3. \begin{figure}[h] \end{figure} Figure 2 shows the curves with equations \(y = 7 - 2 x - 3 x ^ { 2 }\) and \(y = \frac { 2 } { x }\).
The two curves intersect at the points \(P , Q\) and \(R\).

1. Show that the \(x\)-coordinates of \(P , Q\) and \(R\) satisfy the equation $$3 x ^ { 3 } + 2 x ^ { 2 } - 7 x + 2 = 0$$ Given that \(P\) has coordinates \(( - 2 , - 1 )\),
2. find the coordinates of \(Q\) and \(R\).

6

1. Use the Factor Theorem to show that \(4 x - 3\) is a factor of $$16 x ^ { 3 } + 11 x - 15$$
2. Given that \(x = \cos \theta\), show that the equation $$27 \cos \theta \cos 2 \theta + 19 \sin \theta \sin 2 \theta - 15 = 0$$ can be written in the form $$16 x ^ { 3 } + 11 x - 15 = 0$$
3. Hence show that the only solutions of the equation $$27 \cos \theta \cos 2 \theta + 19 \sin \theta \sin 2 \theta - 15 = 0$$ are given by \(\cos \theta = \frac { 3 } { 4 }\).