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Verify that the numbers given alongside that cubic polynomials

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Verify that the numbers given alongside that cubic polynomials below are its zeroes . Also verify the relationship between the zeroes and coefficient in each case.
$2 x^{3} + x^{2} - 5 x + 2, (\frac{1}{2}, 1. - 2)$

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Verify that the numbers given alongside that cubic polynomials below are its zeroes . Also verify the relationship between the zeroes and coefficient in each case.

2 x^{3} + x^{2} - 5 x + 2, (\frac{1}{2}, 1. - 2)

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Text solutionVerified

2 x^{3} + x^{2} - 5 x + 2, (\frac{1}{2}, 1. - 2)

2 x^{3} + x^{2} - 5 x + 2

Compare with this

a x^{3} + b x^{2} + c x + d

x = \frac{1}{2}

p (\frac{1}{2}) = 2 (\frac{1}{2})^{3} + (\frac{1}{2})^{2} - 5 (\frac{1}{2}) + 2

= \frac{1}{4} + \frac{1}{4} - \frac{5}{2} + 2

= \frac{1 + 1 - 10 + 8}{4} = \frac{0}{4} = 0

x = 1

p (1) = 2 (1)^{3} + (1)^{2} - 5 (1) + 2

= 2 + 1 - 5 + 2 = 0

x = 2

P (- 2) = 2 (- 2)^{3} + (- 2)^{2} - 5 (- 2) + 2 = - 16 + 4 + 10 + 2 = 0

∴ α = \frac{1}{2}, β = 1

and

γ = - 2

are zeroes of

2 x^{3} + x^{2} - 5 x + 2

∴

a = 2,b = 1 , c = -5 , d= 2

∴ α = \frac{1}{2}, β = 1, an d γ = - 2

α + β + γ = \frac{1}{2} + 1 + (- 2)

= \frac{1 + 2 - 4}{2} = - \frac{1}{2} = - \frac{b}{a}

α β + β γ + γ α = (1/2) (1) + (1) (- 2) + (- 2) (1/2) = \frac{- 5}{2} = \frac{c}{a}

α β γ = (1/2) (1) (- 2) = - 1 = \frac{- 2}{2} = \frac{- d}{a}

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Question Text	Verify that the numbers given alongside that cubic polynomials below are its zeroes . Also verify the relationship between the zeroes and coefficient in each case. $2 x^{3} + x^{2} - 5 x + 2, (\frac{1}{2}, 1. - 2)$
Answer Type	Text solution:1
Upvotes	150

Verify that the numbers given alongside that cubic polynomials below are its zeroes . Also verify the relationship between the zeroes and coefficient in each case.2x3+x2−5x+2,(21​,1.−2)

Text solutionVerified

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Verify that the numbers given alongside that cubic polynomials below are its zeroes . Also verify the relationship between the zeroes and coefficient in each case.
$2 x^{3} + x^{2} - 5 x + 2, (\frac{1}{2}, 1. - 2)$