Question

find a cubic polynomial whose zeros are 1, - 2 and 3

koi hai toh maths ke thAnks dedo ​

Answer 1

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GIVEN :

Zeros of the polynomial :

α = 1, β = -2 and γ = 3

SOLUTION :

SUM OF ZEROS :

⟼ α + β + γ

= 1 - 2 + 3

= 2

SUM OF PRODUCT OF ZEROS TWO TAKEN AT A TIME :

⟼ αβ + βγ +γα

= -2 - 6 + 3

= -5

PRODUCT OF ZEROS :

⟼ αβγ

= (1)(-2)(3)

= -6

$\begin{gathered}\qquad \qquad \boxed {\begin{array}{cc} \bf{\underline {\bigstar\:\: For \: a \:cubic \: polynomial\::}}\\\\ \sf{ Whose \:\:zeroes \:\:are\:\:\alpha \:\;\: \beta\:\: γ} \\ \\ {x}^{3} - (α + β + γ) {x}^{2} + (αβ + βγ +γα)x - αβγ.\\\\ \end{array}} \end{gathered}$

By substituting the values,

The required Cubic polynomial is

$= {x}^{3} - (α + β + γ) {x}^{2} + (αβ + βγ +γα)x - αβγ. \\ \\ = {x}^{3} - (2) {x}^{2} + ( - 5)x - ( - 6). \\ \\ = {x}^{3} - 2 {x}^{2} - 5x + 6$

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Answer 2

Zeros of the polynomial :

α = 1, β = -2 and γ = 3

SOLUTION :

SUM OF ZEROS :

⟼ α + β + γ

= 1 - 2 + 3

= 2

SUM OF PRODUCT OF ZEROS TWO TAKEN AT A TIME :

⟼ αβ + βγ +γα

= -2 - 6 + 3

= -5

PRODUCT OF ZEROS :

⟼ αβγ

= (1)(-2)(3)

= -6