Trying to find the right answer to this..

p(t) = 2 + t + 2t 2 - t 3 ∴ p(0) = 2 + 0 + 2(0) 2 - (0) 3 = 2, p(1) = 2 + 1 + 2(1) 2 - (1) 3 = 2 + 1 + 2 - 1 = 4, and, p(2) = 2 + 2 + 2(2) 2 - (2) 3 = 2 + 2 + 8 - 8 = 4.

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Polynomials

Short Answer Type

41. Find the value of the polynomial 5x - 4x² + 3 at
x = 2.

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42. Find p(0), p(1) and p(2) for each of the following polynomials:
p(y) = y² - y + 1

125 Views

43. Find p(0), p(1) and p(2) for each of the following polynomials:
p(t) = 2 + t + 2t² - t³

p(t) = 2 + t + 2t² - t³∴ p(0) = 2 + 0 + 2(0)² - (0)³ = 2,
p(1) = 2 + 1 + 2(1)² - (1)³= 2 + 1 + 2 - 1 = 4,
and, p(2) = 2 + 2 + 2(2)² - (2)³= 2 + 2 + 8 - 8 = 4.

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44. Find p(0), p(1) and p(2) for each of the following polynomials:
p(x) = x³

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45. Find p(0), p(1) and p(2) for each of the following polynomials:
p(x) = (x - 1)(x + 1).

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46. Verify whether the following are zeroes of the polynomial, indicated against them.

p(x) = 3x + 1, x =

negative 1 third

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47. Verify whether the following are zeroes of the polynomial, indicated against them.
p(x) = 5x -

straight pi

, x =

4 over 5

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48. Verify whether the following are zeroes of the polynomial, indicated against them.
p(x) = x² - 1, x = 1, - 1

252 Views

49. Verify whether the following are zeroes of the polynomial, indicated against them.
p(x) = (x + 1)(x - 2), x = - 1, 2

183 Views

50. Verify whether the following are zeroes of the polynomial, indicated against them.
p(x) = x², x = 0

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Previous Year Papers

Test Series

Test Yourself

Short Answer Type

43. Find p(0), p(1) and p(2) for each of the following polynomials:p(t) = 2 + t + 2t2 - t3

43. Find p(0), p(1) and p(2) for each of the following polynomials:
p(t) = 2 + t + 2t² - t³