If (x + 2) and (x – 1) are factors of (f(x) = 6x^{4} + mx^{3} – 13x^{2} + nx + 14), find the
(a) values of m and n.
(b) remainder when f(x) is divided be (x + 1).
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).