If (x + 1) and (x – 2) are factors of the polynomial \(g(x) = x^{4} + ax^{3} + bx^{2} – 16x – 12\), find the values of a and b.
Explanation
If (x + 1) and (x - 2) are factors of the polynomial \(g(x)\), then \(g(-1) & g(2) = 0\)
i.e. \(g(-1) = (-1)^{4} + a(-1)^{3} + b(-1)^{2} - 16(-1) - 12 = 0\)
\(1 - a + b + 16 - 12 = 0 \implies a - b = 5 ... (1)\)
\(g(2) = (2)^{4} + a(2)^{3} + b(2)^{2} - 16(2) - 12 = 0\)
\(16 + 8a + 4b - 32 - 12 = 0 \implies 8a + 4b = 28 ... (2)\)
From (1), \(a = 5 + b\).
(2) becomes : \(8(5 + b) + 4b = 28 \implies 40 + 8b + 4b = 28\)
\(40 + 12b = 28 \implies 12b = -12\)
\(b = -1\)
\(a = 5 + b = 5 + (-1) = 4\)
\((a, b) = (4, -1)\)