Question

Verify that 3, -1 and -⅓ are the zeroes of the cubic polynomial p(x) = 3x3 - 5x2 -11x - 3 and also verify the relation between the zeroes and coefficients​

Answer 1

$\sf \Large Solution!!$

$\sf \longrightarrow x=3$

$\sf \to p(x)=3x^{3}-5x^{2}-11x-3$

$\sf \to p(3)=3(3)^{3}-5(3)^{2}-11(3)-3$

$\sf \to \;\;=81-45-33-3$

$\sf \to p(3)=0\;\big(3\: is\:a\:zero\:of\:p(x)\big)$

____________________________

$\sf \longrightarrow x=-1$

$\sf \to p(x)=3x^{3}-5x^{2}-11x-3$

$\sf \to p(-1)=3(-1)^{3}-5(-1)^{2}-11(-1)-3$

$\sf \to \;\;=-3-5+11-3$

$\sf \to p(-1)=0\;\big(-1\: is\:a\:zero\:of\:p(x)\big)$

____________________________

$\sf \longrightarrow x=\dfrac{-1}{3}$

$\sf \to p(x)=3x^{3}-5x^{2}-11x-3$

$\sf \to p\bigg(\dfrac{-1}{3}\bigg)=3\bigg(\dfrac{-1}{3}\bigg)^{3}-5\bigg(\dfrac{-1}{3}\bigg)^{2}-11\bigg(\dfrac{-1}{3}\bigg)-3$

$\sf \to \;\;=\dfrac{-1}{9}-\dfrac{5}{9}+\dfrac{11}{3}-3$

$\sf \to \;\;=\dfrac{-1-5+33-27}{9}$

$\sf \to \;\;=\dfrac{0}{9}$

$\sf \to p\bigg(\dfrac{-1}{9}\bigg)=0\;\bigg(\dfrac{-1}{3}\:is\:a\:zero\:of\:p(x)\bigg)$

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$\sf \large Verification\:of\:relation\:between\:zeros\:and\:coefficients$

$\sf \to p(x)=ax^{3}+bx^{2}+cx+d$

$\sf \to For\:p(x)=3x^{3}-5x^{2}-11x-3,$

$\sf \to a=3$

$\sf \to b=-5$

$\sf \to c=-11$

$\sf \to d=-3$

$\sf Zeroes:$

$\sf \to \alpha =3$

$\sf \to \beta =-1$

$\sf \to \gamma =\dfrac{-1}{3}$

$\sf We\:have,$

$\sf \longrightarrow \alpha +\beta +\gamma=\dfrac{-b}{a}$

$\sf \to (3)+(-1)+\bigg(\dfrac{-1}{3}\bigg)$

$\sf \to \dfrac{9-3-1}{3}$

$\sf \to \dfrac{5}{3}\:or\:\bold{\dfrac{-(-5)}{3}}=\dfrac{-b}{a}$

$\sf$

$\sf \longrightarrow \alpha \beta +\beta \gamma+\gamma \alpha =\dfrac{c}{a}$

$\sf \to (3)(-1)+(-1)\bigg(\dfrac{-1}{3}\bigg)+\bigg(\dfrac{-1}{3}\bigg)(3)$

$\sf \to -3+\dfrac{1}{3}-\dfrac{3}{3}$

$\sf \to \dfrac{-9+1-3}{3}$

$\sf \to \bold{\dfrac{-11}{3}}=\dfrac{c}{a}$

$\sf$

$\sf \longrightarrow \alpha \beta \gamma =\dfrac{-d}{a}$

$\sf \to (3)(-1)\bigg(\dfrac{-1}{3}\bigg)$

$\sf \to \dfrac{3}{3}\:or\:\bold{\dfrac{-(-3)}{3}}=\dfrac{-d}{a}$