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Basics & Logarithms: 27x Divisible Equal

JEE Maths question with a full step-by-step solution.

Question
If 8x3+lx2−27x+m8x^3 + lx^2 - 27x + m is divisible by 2x2−x−62x^2 - x - 6, then l+ml + m is equal to:
A22
B−18-18
C−16-16correct
Dnone
Solution
Step 1: Factorize the divisor
2x2−x−6=(2x+3)(x−2)2x^2 - x - 6 = (2x+3)(x-2)
 The roots x=2x = 2 and x=−32x = -\dfrac{3}{2} must also be roots of the cubic. Step 2: Substitute x=2x = 2
8(8)+4l−54+m=0  ⟹  4l+m=−10⋯(1)8(8) + 4l - 54 + m = 0 \implies 4l + m = -10 \quad \cdots (1)
 Step 3: Substitute x=−32x = -\dfrac{3}{2}
8 ⁣(−278)+l ⁣(94)−27 ⁣(−32)+m=0  ⟹  9l+4m=−54⋯(2)8\!\left(-\frac{27}{8}\right) + l\!\left(\frac{9}{4}\right) - 27\!\left(-\frac{3}{2}\right) + m = 0 \implies 9l + 4m = -54 \quad \cdots (2)
 Step 4: Solve the system From (1)(1), m=−10−4lm = -10 - 4l. Substituting into (2)(2):
9l+4(−10−4l)=−54  ⟹  −7l−40=−54  ⟹  l=2,m=−189l + 4(-10-4l) = -54 \implies -7l - 40 = -54 \implies l = 2, \quad m = -18
l+m=2+(−18)=−16l + m = 2 + (-18) = -16
 Answer: (3)
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