The equation $5{x^3} + 48{x^2} + 100x + 2 = a$ has roots ${r_1}$, ${r_2}$ and ${r_3}$.

Given that ${r_1} + {r_2} + {r_3} + {r_1}{r_2}{r_3} = 0$, find the value of a.

${r_1} + {r_2} + {r_3} = \frac{{ - 48}}{5}$     (M1)(A1)

${r_1}{r_2}{r_3} = \frac{{a - 2}}{5}$     (M1)(A1)

$\frac{{ - 48}}{5} + \frac{{a - 2}}{5} = 0$     M1

$a = 50$     A1

Note:     Award M1A0M1A0M1A1 if answer of 50 is found using $\frac{{48}}{5}$ and $\frac{{2 - a}}{5}$.

[6 marks]