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Evaluate f((3−5i)/2) for Cubic Polynomial | Complex JEE

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

Question

f(x) = 2x^3 + 2x^2 - 7x + 72

, then

f\!\left(\dfrac{3 - 5i}{2}\right)

1

2

3

4

correct

Solution

Step 1: Build the minimal (quadratic) equation satisfied by

x = \dfrac{3 - 5i}{2}

. Multiply by

2

2x = 3 - 5i

, so

2x - 3 = -5i

. Step 2: Square both sides to eliminate

i

(2x - 3)^2 = (-5i)^2 = 25i^2 = -25

Expand the left side:

4x^2 - 12x + 9 = -25 \implies 4x^2 - 12x + 34 = 0 \implies 2x^2 - 6x + 17 = 0

x

satisfies

2x^2 - 6x + 17 = 0

Step 3: Reduce

f(x)

using this relation by polynomial division. Divide

2x^3 + 2x^2 - 7x + 72

2x^2 - 6x + 17

2x^3 + 2x^2 - 7x + 72 = (2x^2 - 6x + 17)(x + 4) + r

Expand

(2x^2 - 6x + 17)(x + 4) = 2x^3 + 8x^2 - 6x^2 - 24x + 17x + 68 = 2x^3 + 2x^2 - 7x + 68

. So the remainder is

r = 72 - 68 = 4

f(x) = (2x^2 - 6x + 17)(x + 4) + 4

Step 4: Since

2x^2 - 6x + 17 = 0

at the given

x

, the first term vanishes:

f\!\left(\frac{3 - 5i}{2}\right) = 0\cdot(x + 4) + 4 = 4

Correct answer: (4)

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