Exponential substitution equations

6. $$f ( x ) = - 6 x ^ { 3 } - 7 x ^ { 2 } + 40 x + 21$$

1. Use the factor theorem to show that \(( x + 3 )\) is a factor of \(\mathrm { f } ( x )\)
2. Factorise f(x) completely.
3. Hence solve the equation $$6 \left( 2 ^ { 3 y } \right) + 7 \left( 2 ^ { 2 y } \right) = 40 \left( 2 ^ { y } \right) + 21$$ giving your answer to 2 decimal places.

4 In this question you must show detailed reasoning.
The cubic polynomial \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = 2 x ^ { 3 } - 3 x ^ { 2 } - 11 x + 6\).

1. Use the factor theorem to show that \(( 2 x - 1 )\) is a factor of \(\mathrm { f } ( x )\).
2. Express \(\mathrm { f } ( x )\) in fully factorised form.
3. Hence solve the equation \(2 \times 8 ^ { y } - 3 \times 4 ^ { y } - 11 \times 2 ^ { y } + 6 = 0\).

4 In this question you must show detailed reasoning. The cubic polynomial \(6 x ^ { 3 } + k x ^ { 2 } + 57 x - 20\) is denoted by \(\mathrm { f } ( x )\). It is given that \(( 2 x - 1 )\) is a factor of \(\mathrm { f } ( x )\).

1. Use the factor theorem to show that \(k = - 37\).
2. Using this value of \(k\), factorise \(\mathrm { f } ( x )\) completely.
3. 1. Hence find the three values of \(t\) satisfying the equation \(6 \mathrm { e } ^ { - 3 t } - 37 \mathrm { e } ^ { - 2 t } + 57 \mathrm { e } ^ { - t } - 20 = 0\).
   2. Express the sum of the three values found in part (c)(i) as a single logarithm.