If (x + 2) and (x - 1) are factors of (f(x) = 6x^{4} +

Explanation

When f(x) is divided by (x - a), the remainder is f(a). When (x - a) is a factor, then f(a) = 0.

\(f(x) = 6x^{4} + mx^{3} - 13x^{2} + nx + 14\)

(a) When divided by (x + 2), \(f(-2) = 6(-2^{4}) + m(-2^{3}) - 13(-2^{2}) + n(-2) + 14 = 0\)

= \(96 - 8m - 52 - 2n + 14 = 0\)

\(58 = 8m + 2n .... (1)\)

When divided by (x - 1), \(f(1) = 6(1^{4}) + m(1^{3}) - 13(1^{2}) + n(1) + 14 = 0\)

\(6 + m - 13 + n + 14 = 0\)

\(m + n = -7 ... (2)\)

\(m = -7 - n \) (from equation 2)

\(\therefore 58 = 8(-7 - n) + 2n \)

\(-56 - 8n + 2n = 58\)

\(-6n = 58 + 56 = 114 \implies n = -19\)

\(m = -7 - (-19) = -7 + 19 = 12\)

\(\therefore \text{m and n = 12 and -19}\)

\(\therefore f(x) = 6x^{4} + 12x^{3} - 13x^{2} - 19x + 14\)

(b) When divided by (x + 1)

\(f(-1) = 6(-1^{4}) + 12(-1^{3}) - 13(-1^{2}) - 19(-1) + 14\)

= \(6 - 12 - 13 + 19 + 14 = 14\).