Solve p(exponential) = 0

7 The polynomial \(\mathrm { p } ( x )\) is defined by $$\mathrm { p } ( x ) = a x ^ { 3 } - 11 x ^ { 2 } - 19 x - a$$ where \(a\) is a constant. It is given that \(( x - 3 )\) is a factor of \(\mathrm { p } ( x )\).

1. Find the value of \(a\).
2. When \(a\) has this value, factorise \(\mathrm { p } ( x )\) completely.
3. Hence find the exact values of \(y\) that satisfy the equation \(\mathrm { p } \left( \mathrm { e } ^ { y } + \mathrm { e } ^ { - y } \right) = 0\).
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

5 The polynomial \(\mathrm { p } ( x )\) is defined by $$\mathrm { p } ( x ) = 2 x ^ { 3 } + a x ^ { 2 } - 3 x - 4$$ where \(a\) is a constant. It is given that ( \(x - 4\) ) is a factor of \(\mathrm { p } ( x )\).

1. Find the value of \(a\) and hence factorise \(\mathrm { p } ( x )\).
2. Show that the equation \(\mathrm { p } \left( \mathrm { e } ^ { 3 y } \right) = 0\) has only one real root and find its exact value.

4 The polynomial \(\mathrm { p } ( x )\) is defined by $$\mathrm { p } ( x ) = a x ^ { 3 } + 23 x ^ { 2 } - a x - 8$$ where \(a\) is a constant. It is given that \(( 2 x + 1 )\) is a factor of \(\mathrm { p } ( x )\).

1. Find the value of \(a\) and hence factorise \(\mathrm { p } ( x )\) completely.
2. Hence solve the equation \(\mathrm { p } \left( \mathrm { e } ^ { 4 y } \right) = 0\), giving your answer correct to 3 significant figures.

4 The polynomial \(\mathrm { f } ( x )\) is defined by $$f ( x ) = 12 x ^ { 3 } + 25 x ^ { 2 } - 4 x - 12$$

1. Show that \(\mathrm { f } ( - 2 ) = 0\) and factorise \(\mathrm { f } ( x )\) completely.
2. Given that $$12 \times 27 ^ { y } + 25 \times 9 ^ { y } - 4 \times 3 ^ { y } - 12 = 0$$ state the value of \(3 ^ { y }\) and hence find \(y\) correct to 3 significant figures.

4 The polynomial \(\mathrm { p } ( x )\) is defined by $$\mathrm { p } ( x ) = a x ^ { 3 } + a x ^ { 2 } - 15 x - 18$$ where \(a\) is a constant. It is given that ( \(x - 2\) ) is a factor of \(\mathrm { p } ( x )\).

1. Find the value of \(a\).
2. Using this value of \(a\), factorise \(\mathrm { p } ( x )\) completely.
3. Hence solve the equation \(\mathrm { p } \left( \mathrm { e } ^ { \sqrt { } y } \right) = 0\), giving the answer correct to 2 significant figures.

8. $$f ( x ) = 2 x ^ { 3 } - 5 x ^ { 2 } - 23 x - 10$$

1. Find the remainder when \(\mathrm { f } ( x )\) is divided by ( \(x - 3\) ).
2. Show that ( \(x + 2\) ) is a factor of \(\mathrm { f } ( x )\).
3. Hence fully factorise \(\mathrm { f } ( x )\).
4. Hence solve $$2 \left( 3 ^ { 3 t } \right) - 5 \left( 3 ^ { 2 t } \right) - 23 \left( 3 ^ { t } \right) = 10$$ giving your answer to 3 decimal places.

4. Let \(f ( x )\) be given by: $$f ( x ) = x ^ { 3 } + x ^ { 2 } - 12 x - 18$$ a) Use the factor theorem to show that ( \(x + 3\) ) is a factor of \(f ( x )\)
b) Factorise \(f ( x )\) into a linear and a quadratic factor and hence find exact values for all of the solutions of the equation \(f ( x ) = 0\), showing detailed reasoning with your working
c) Hence write down the one solution to the equation $$e ^ { 3 x } + e ^ { 2 x } - 12 e ^ { x } - 18 = 0$$ in the form \(\ln ( a + \sqrt { b } )\)
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6

1. It is given that $$f ( x ) = x ^ { 3 } - x ^ { 2 } + x - 6$$ Use the factor theorem to show that \(( x - 2 )\) is a factor of \(\mathrm { f } ( x )\).
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2. Find the quadratic factor of \(\mathrm { f } ( x )\).
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3. Hence, show that there is only one real solution to \(\mathrm { f } ( x ) = 0\)
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4. Find the exact value of \(x\) that solves $$\mathrm { e } ^ { 3 x } - \mathrm { e } ^ { 2 x } + \mathrm { e } ^ { x } - 6 = 0$$ \(7 \quad\) Curve \(C\) has equation \(y = x ^ { 2 }\)
   \(C\) is translated by vector \(\left[ \begin{array} { l } 3
   0 \end{array} \right]\) to give curve \(C _ { 1 }\)
   Line \(L\) has equation \(y = x\)
   \(L\) is stretched by scale factor 2 parallel to the \(x\)-axis to give line \(L _ { 1 }\)
   Find the exact distance between the two intersection points of \(C _ { 1 }\) and \(L _ { 1 }\)