Solve p(algebraic transform) = 0

3 The polynomial \(\mathrm { p } ( x )\) is defined by $$\mathrm { p } ( x ) = x ^ { 3 } - 3 a x + 4 a$$ where \(a\) is a constant.

1. Given that \(( x - 2 )\) is a factor of \(\mathrm { p } ( x )\), find the value of \(a\).
2. When \(a\) has this value,
   1. factorise \(\mathrm { p } ( x )\) completely,
   2. find all the roots of the equation \(\mathrm { p } \left( x ^ { 2 } \right) = 0\).

5. (a) Find, using algebra, all solutions of $$20 x ^ { 3 } - 50 x ^ { 2 } - 30 x = 0$$ (b) Hence find all real solutions of $$20 ( y + 3 ) ^ { \frac { 3 } { 2 } } - 50 ( y + 3 ) - 30 ( y + 3 ) ^ { \frac { 1 } { 2 } } = 0$$

1. (a) Find, using algebra, all real solutions of

$$2 x ^ { 3 } + 3 x ^ { 2 } - 35 x = 0$$ (b) Hence find all real solutions of $$2 ( y - 5 ) ^ { 6 } + 3 ( y - 5 ) ^ { 4 } - 35 ( y - 5 ) ^ { 2 } = 0$$

2. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. $$f ( x ) = a x ^ { 3 } + ( 6 a + 8 ) x ^ { 2 } - a ^ { 2 } x$$ where \(a\) is a positive constant. Given \(\mathrm { f } ( - 1 ) = 32\)

1. 1. show that the only possible value for \(a\) is 3
   2. Using \(a = 3\) solve the equation $$\mathrm { f } ( x ) = 0$$
2. Hence find all real solutions of
   1. \(3 y + 26 y ^ { \frac { 2 } { 3 } } - 9 y ^ { \frac { 1 } { 3 } } = 0\)
   2. \(3 \left( 9 ^ { 3 z } \right) + 26 \left( 9 ^ { 2 z } \right) - 9 \left( 9 ^ { z } \right) = 0\)

1. In this question you must show detailed reasoning. Solutions relying on calculator technology are not acceptable.

$$f ( x ) = x ^ { 2 } ( 2 x + 1 ) - 15 x$$

1. Solve $$\mathrm { f } ( x ) = 0$$
2. Hence solve $$y ^ { \frac { 4 } { 3 } } \left( 2 y ^ { \frac { 2 } { 3 } } + 1 \right) - 15 y ^ { \frac { 2 } { 3 } } = 0 \quad y > 0$$ giving your answer in simplified surd form.

1. In this question you should show all stages of your working.

Solutions relying on calculator technology are not acceptable.

1. Using algebra, find all solutions of the equation $$3 x ^ { 3 } - 17 x ^ { 2 } - 6 x = 0$$
2. Hence find all real solutions of $$3 ( y - 2 ) ^ { 6 } - 17 ( y - 2 ) ^ { 4 } - 6 ( y - 2 ) ^ { 2 } = 0$$

In this question you should show all stages of your working. Solutions relying on calculator technology are not acceptable.

1. Using algebra, find all solutions of the equation $$3x^3 - 17x^2 - 6x = 0$$ [3]
2. Hence find all real solutions of $$3(y - 2)^6 - 17(y - 2)^4 - 6(y - 2)^2 = 0$$ [3]