3. If (x + 2) and (x -1) are factors of f(x) =... | WAEC SSCE - Further Mathematics - 2025 (Essay)

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3. If (x + 2) and (x -1) are factors of f(x) = 6x\(^4\) + mx\(^3\) - 13x\(^2\) + n x + 14, find the:

a. value of m and n. Leave your answer in this format 'm,n.'

b. remainder when f(x) is divided by (x + 1)

Answer

a. (x + 2) and (x -1) are factors of f(x) = 6x\(^4\) + mx\(^3\) - 13x\(^2\) + n x + 14

Then, f(-2) and f(1) are = 0

f(-2) = 6(-2)\(^4\) + m(-2)\(^3\) - 13(-2)\(^2\) + n (-2) + 14 = 0

= 6 x 16 - 8m - 13 x 2 - 2n + 14 = 0 = 96 - 8m - 52 - 2n + 14 = 0

= 8m + 2n = 58: divide through by 2

= 4m + n = 29 - - -- - - - - - -(i)

f(1) = 6x\(^4\) + mx\(^3\) - 13x\(^2\) + n x + 14 = 0

f(1) = 6(1)\(^4\) + m(1)\(^3\) - 13(1)\(^2\) + n (1) + 14 = 0

= 6 + m - 13 + n + 14 = 0

= m + n = -7 - - - - - - - - -(ii)

Solving eqn i and ii simultaneously

from eqn i - - - - - n = 58 - 4m

put n = 58 - 4m into eqn ii

m + n = -7 = m + 29 - 4m = -7

- 3m = -36

m = \(\frac{36}{3}\) = 12

put m = 12 into eqn ii

m + n = -7 = 12 + n = -7

n = -7 - 12 = -19.

b. 6x\(^4\) + mx\(^3\) - 13x\(^2\) + n x + 14 = 0 becomes 6x\(^4\) + 12x\(^3\) - 13x\(^2\) - 19x + 14 = 0

f(x) = 6x\(^4\) + 12x\(^3\) - 13x\(^2\) - 19x + 14 = 0

remainder when f(x) is divided by (x + 1)

Let (x + 1) = 0 then, x = -1

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

= 6 - 12 - 13 + 19 + 14 = 14

Therefore, the remainder = 14.

Answer Details