Question

a monic cubic polynomial px satisfied the condition p(1). p(3) =17. p(5). find p(2) ..

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Answer 1

Answer :

we know that general form of a cubic polynomial is p(x) = ax³ + bx² + cx + d

given that p(1) = 1

⇒ p(1) = a + b + c + d = 1

p(2) = 2

⇒ p(2) = 8a + 4b + 2c + d = 2 and since a + b + c + d = 1

⇒ 7a + 3b + c = 1

p(3) = 3

⇒ 27a + 9b + 3c + d = 3

since a + b + c + d = 1

⇒ 26a + 8b + 2c = 2

since 7a + 3b + c = 1 and 14a + 6b + 2c = 2

⇒ 12a + 2b = 0

⇒ 6a + b = 0 ⇒ b = -6a

substituting b = -6a in 7a + 3b + c = 1

we get -11a + c = 1

p(4) = 5

⇒ 64a + 16b + 4c + d = 5

since a + b + c + d = 1

⇒ 63a + 15b + 3c = 4

substituting b = -6a in 63a + 15b + 3c = 4 we get

63a - 90a + 3c = 4 ⇒ -27a + 3c = 4

solving equations -11a + c = 1 and -27a + 3c = 4

we get a = 1/6 and c = 17/6 and b = -6a = -1

substituting all the values in a + b + c + d = 1 we get d = -1

so the given polynomial is x³/6 - x² + 17x/6 - 1

⇒ p(6) = 6² - 6² + 17 - 1 = 16

$\rule{200}{2}$

Answer 2

Answer:

we know that general form of a cubic polynomial is p(x) = ax³ + bx² + cx + d

given that p(1) = 1

⇒ p(1) = a + b + c + d = 1

p(2) = 2

⇒ p(2) = 8a + 4b + 2c + d = 2 and since a + b + c + d = 1

⇒ 7a + 3b + c = 1

p(3) = 3

⇒ 27a + 9b + 3c + d = 3

since a + b + c + d = 1

⇒ 26a + 8b + 2c = 2

since 7a + 3b + c = 1 and 14a + 6b + 2c = 2

⇒ 12a + 2b = 0

⇒ 6a + b = 0 ⇒ b = -6a

substituting b = -6a in 7a + 3b + c = 1

we get -11a + c = 1

p(4) = 5

⇒ 64a + 16b + 4c + d = 5

since a + b + c + d = 1

⇒ 63a + 15b + 3c = 4

substituting b = -6a in 63a + 15b + 3c = 4 we get

63a - 90a + 3c = 4 ⇒ -27a + 3c = 4

solving equations -11a + c = 1 and -27a + 3c = 4

we get a = 1/6 and c = 17/6 and b = -6a = -1

substituting all the values in a + b + c + d = 1 we get d = -1

so the given polynomial is x³/6 - x² + 17x/6 - 1

⇒ p(6) = 6² - 6² + 17 - 1 = 16