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0 votes
15.5k views
in Mathematics by (34.9k points)

If ƒ(x) and g(x) are two polynomials such that the polynomial P(x) = ƒ(x3) + xg(x3) is divisible by x2 + x + 1, then P(l) is equal to........

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2 Answers

+1 vote
by (34.4k points)

Answer is 0

P(x) = ƒ(x3) + xg(x3)

P(1) = ƒ(1) + g(1) ...(1)

Now P(x) is divisible by x2 + x + 1

⇒ P(x) = Q(x)(x2 + x + 1)

P(w) = 0 = P(w2) where w, w2 are non-real cube roots of units

P(x) = ƒ(x3) + xg(x3)

P(w) = ƒ(w3) + wg(w3) = 0

ƒ(1) + wg(1) = 2 ...(2)

P(w2) = ƒ(w6) + w2g(w6) = 0

ƒ(1) + w2g(1) = 0 ...(3)

(2) + (3)

⇒ 2ƒ(1) + (w + w2)g(1) = 0

2ƒ(1) = g(1) ...(4)

(2) – (3)
⇒ (w – w2)g(1) = 0

g(1) = 0 = ƒ(1) from (4)

from (1) P(1) = ƒ(1) + g(1) = 0

0 votes
by (15 points)
P(x) = ƒ(x3) + xg(x3) P(1) = ƒ(1) + g(1) ...(1) Now P(x) is divisible by x2 + x + 1 ⇒ P(x) = Q(x)(x2 + x + 1) P(w) = 0 = P(w2) where w, w2 are non-real cube roots of units P(x) = ƒ(x3) + xg(x3) P(w) = ƒ(w3) + wg(w3) = 0 ƒ(1) + wg(1) = 2 ...(2) P(w2) = ƒ(w6) + w2g(w6) = 0 ƒ(1) + w2g(1) = 0 ...(3) (2) + (3) ⇒ 2ƒ(1) + (w + w2)g(1) = 0 2ƒ(1) = g(1) ...(4) (2) – (3) ⇒ (w – w2)g(1) = 0 g(1) = 0 = ƒ(1) from (4) from (1) P(1) = ƒ(1) + g(1) = 0

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