Question

find the all the zeroes of the polynomial 3x⁴-13x³+17x²+5x-6 if one its zeroes are -1/√3 and 1/√3​

Answer 1

Given : Two zeroes of the polynomial 3x⁴ - 13x³ + 17x² + 5x - 6 are -1/√3 and 1/√3 .

Exigency To Find : Other Zeroes of Polynomial .

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❍ Let's Consider the given polynomial 3x⁴ - 13x³ + 17x² + 5x - 6 be p(x) .

⠀⠀⠀⠀⠀Given that ,

⠀⠀⠀⠀⠀⠀⠀⠀▪︎⠀⠀Two zeroes of polynomial 3x⁴ - 13x³ + 17x² + 5x-6 are -1/√3 and 1/√3 .

$\qquad \therefore \sf \bigg( x + \dfrac{1}{\sqrt{3}} \bigg) \bigg( x - \dfrac{1}{\sqrt{3}} \bigg)$

$\qquad :\implies \sf \bigg( x + \dfrac{1}{\sqrt{3}} \bigg) \bigg( x - \dfrac{1}{\sqrt{3}} \bigg)$

$\qquad\bigstar\:\: \sf Algebraic \:Indentity \:\: :\: ( a^2 - b^2 ) = ( a + b ) ( a - b )$

$\qquad :\implies \sf \bigg( x^2 - \dfrac{1^2}{(\sqrt{3})^2} \bigg)$

$\qquad :\implies \sf \bigg( x^2 - \dfrac{1}{(3} \bigg)$

$\qquad :\implies \sf \bigg( \dfrac{3x^2 - 1}{3} \bigg)$

$\qquad :\implies \sf \bigg( 3x^2 - 1 \bigg)$

$\qquad \therefore \underline {\:\bf\: 3x^2 - 1 \: \sf is \:the \:factor \:p(x) \:}.\:$

⠀⠀⠀⠀⠀Now ,

• By Dividing p(x) [ 3x⁴ - 13x³ + 17x² + 5x - 6 ] by 3x² - 1 we get ,

$\qquad \qquad \sf \qquad\quad x^2 - 5x - 6 \\ \begin{array}{cc}\sf3x^2 - 1 \big)&\sf \overline {3x^4 - 15x^3 + 17x^2 + 5x - 6 }\\\\ & \sf \underline {3x^4 \qquad - x^2 }\downarrow\\\\\\ & \sf \ \ \ \ -15 x^3 + 18x^2 + 5x - 6 \\\\ &\underline {\sf-15x^3 \qquad + 5x}\downarrow\\\\\\ & \sf \ \ \ \ 18x^2 - 6 \\\\ & \: \sf \underline{\:\:18x^2 - 6\:\:}\\\\ & \underline {\sf \ \ \qquad 0 \qquad  } \end{array}\\\\\\ \bullet \:\:\sf Divisor \rightarrow \bf\:3x^2 - 1\\ \\\bullet \:\:\sf Quotient \rightarrow\bf\:\:x^2 - 5x - 6 \\\\ \bullet \:\: \sf Remainder \rightarrow\:\bf \:0\:$

⠀⠀⠀⠀⠀Now , We have ,

$\qquad \dashrightarrow \sf x^2 - 5x - 6 \:\:$

$\qquad \dashrightarrow \sf x^2 - 3x - 2x - 6 \:\:$

$\qquad \dashrightarrow \sf x( x - 3) - 2 ( x - 3 ) \:\:$

$\qquad \dashrightarrow \sf ( x - 3) ( x - 2 ) \:\:$

$\qquad \dashrightarrow \sf x \:=\:3\:or\:x \:\;= 2\:\:\:\:$

$\qquad \therefore \pmb{\underline{\purple{\: }x \:=\:3\:or\: \:\; 2\:\:} }\bigstar$

$\qquad \therefore \underline { \sf Hence, \: Other \: two \: zeroes \:of \:Polynomial \:p(x)\:are \: \bf 3 \:and \: 2 }\:$