(a) If (x + 2) is a factor of g(x) = 2x\(^3\) +11x\(^2\) - x - 30, find the zeros of g(x).
(b) Solve 3(2\(^x\)) +3\(^{y - 2}\) = 25 and 2x - 3\(^{y + 1}\) = -19 simultaneously.
(a) Since \((x + 2)\) is a factor, divide \(g(x) = 2x^3 + 11x^2 - x - 30\) by \((x + 2)\).
Using synthetic division with root \(x = -2\):
\[2x^3 + 11x^2 - x - 30 = (x + 2)(2x^2 + 7x - 15)\]
Factorise the quadratic:
\[2x^2 + 7x - 15 = (2x - 3)(x + 5)\]
So \(g(x) = (x + 2)(2x - 3)(x + 5)\). The zeros are:
\[x = -2,\qquad x = \tfrac{3}{2},\qquad x = -5\]
(b) Let \(a = 2^x\) and \(b = 3^y\). The equations become:
\[3a + \tfrac{b}{9} = 25 \ \Rightarrow\ 27a + b = 225\]
\[a - 3b = -19\]
From the second equation \(a = 3b - 19\). Substitute:
\[27(3b - 19) + b = 225 \ \Rightarrow\ 82b - 513 = 225 \ \Rightarrow\ b = 9\]
So \(3^y = 9 \Rightarrow y = 2\), and \(a = 3(9) - 19 = 8\), so \(2^x = 8 \Rightarrow x = 3\).
Check: \(3(8) + 3^{0} = 25\) and \(8 - 3^{3} = -19\). Hence \(x = 3,\ y = 2\).