JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Complex Numbers

f (x)

and

g (x)

are two polynomials such that the polynomial

P (x) = f (x^{3}) + xg (x^{3})

is divisible by

x^{2} + x + 1

, then

P (1)

is equal to ______

Answer: 0

Solution

The roots of

x^{2} + x + 1 = 0

are the non-real cube roots of unity

ω

and

ω^{2}

, both satisfying

ω^{3} = 1

. Divisibility of

P (x)

x^{2} + x + 1

forces

P (ω) = P (ω^{2}) = 0

, and since

ω^{3} = (ω^{2})^{3} = 1

f (1) + ω g (1) = 0, f (1) + ω^{2} g (1) = 0

Subtracting gives

(ω - ω^{2}) g (1) = 0

, and

ω \neq = ω^{2}

, so

g (1) = 0

and then

f (1) = 0

. Therefore

P (1) = f (1) + 1 \cdot g (1) = 0

P (1) = 0

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Integer answer