Question

find all zeroes of polynomials f(x) = 2x^4-2x^3-7x^2+3x6 , if two of its zeroes are (-√3/2) and (√3/2)​

Answer 1

We're given with a polynomial f( x ) : 2x⁴ – 2x³ – 7x² + 3x + 6 & if two of it's zeroes are (– √³⁄₂) and ( √³⁄₂).

To find: All zeroes of polynomial, we've to find out two more zeroes of the given polynomial.

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Now,

➟ (x – √³⁄₂) (x + √³⁄₂)

➟ (x² – ³⁄₂)

➟ (2x² – 3)

Hence, (2x² – 3) is a factor of polynomial f(x).

⠀⠀⠀⠀⠀

✇ ${\pink{\underline{\mathcal{\pmb{USING\;DIVISION\;ALGORITHM :\::}}}}}$

⌑ f( x ) = g( x ) × q( x ) + r( x ) ⌑

• f( x ) = 2x⁴ – 2x³ – 7x² + 3x + 6
• g( x ) = (2x² – 3)
• q( x ) = (x² – x – 2)
• r( x ) = 0

⠀

$\dashrightarrow\sf 2x^4 - 2x^3 - 7x^2 + 3x + 6 = (2x^2 - 3) (x^2- x - 2) + 0 \\\\\\\dashrightarrow\sf 2x^4 - 2x^3 - 7x^2 + 3x + 6 = (\sqrt{2x} +\sqrt{3}) (\sqrt{2x} - \sqrt{3}) (x^2 + 1x - 2x - 2) \\\\\\\dashrightarrow\sf 2x^4 - 2x^3 - 7x^2 + 3x + 6 = (\sqrt{2x} + \sqrt{3}) (\sqrt{2x} - \sqrt{3})\Big\{x (x + 1) - 2(x + 1)\Big\} \\\\\\\dashrightarrow\sf 2x^4 - 2x^3- 7x^2 + 3x + 6 = (\sqrt{2x} + \sqrt{3}) (\sqrt{2x} - \sqrt{3}) (x + 1) (x - 2) \\\\\\\dashrightarrow\underline{\boxed{\pmb{\frak{\pink{-\sqrt{\dfrac{3}{2}}, \sqrt{\dfrac{3}{2}}, -1 \;\&\;2}}}}}\;\bigstar$

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∴ Hence, –√³⁄₂, √³⁄₂, –1 and 2 are the zeroes of given polynomial.

Answer 2

Given : Two zeroes of the polynomial 2x⁴ - 2x³ + 7x² + 3x + 6 are -√3/2 and √3/2 .

Exigency To Find : All   Zeroes of Polynomial in which two zeroes are given and we have to find out other two zeroes of Polynomial ?

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❍ Let's Consider the given polynomial 2x⁴ - 2x³ - 7x² + 3x + 6 be p(x) .

⠀⠀⠀⠀⠀Given that ,

⠀⠀⠀⠀⠀⠀⠀⠀▪︎⠀⠀Two zeroes of polynomial 2x⁴ - 2x³ - 7x² + 3x + 6 are -√3/2 and √3/2.

$\qquad \therefore \sf \bigg( x + \sqrt{\dfrac{3}{2}} \bigg) \bigg( x - \sqrt{\dfrac{3}{2}} \bigg)$

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

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

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

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

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

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

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

⠀⠀⠀⠀⠀Now ,

⠀⠀⠀⠀⠀By Dividing p(x) [ 2x⁴ - 2x³ - 7x² + 3x + 6 ] by 2x² - 3 we get ,

$\qquad \qquad \sf \qquad\quad x^2 - x - 2 \\ \begin{array}{cc}\sf 2x^2 - 3 \big)&\sf \overline {2x^4 - 2x^3 - 7x^2 + 3x + 6 }\\\\ & \sf \underline {2x^4 \qquad - 3x^2 }\downarrow\\\\\\ & \sf \ \ \ \ -2 x^3 -4x^2 + 3x + 6 \\\\ &\underline {\sf-2x^3 \qquad + 3x}\downarrow\\\\\\ & \sf \ \ \ \ -4x^2 + 6 \\\\ & \: \sf \underline{\:\:-4x^2 + 6 \:\:}\\\\ & \underline {\sf \ \ \qquad 0 \qquad  } \end{array}\\\\\\ \bullet \:\:\sf Divisor \rightarrow \bf\:2x^2 + 3 \\ \\\bullet \:\:\sf Quotient \rightarrow\bf\:\:x^2 - x - 2 \\\\ \bullet \:\: \sf Remainder \rightarrow\:\bf \:0\:$

⠀⠀⠀⠀⠀Now , We have ,

$\qquad \dashrightarrow \sf x^2 - x - 2 = 0 \:\:\\\\\qquad \dashrightarrow \sf x^2 - x - 2 = 0 \:\:\\\\\qquad \dashrightarrow \sf x^2 - 2x + x - 2 = 0 \:\:\\\\\qquad \dashrightarrow \sf x( x - 2 )+ 1 ( x - 2 ) = 0\:\:\\\\\qquad \dashrightarrow \sf ( x + 1 )( x - 2 ) = 0\:\:\\\\\qquad \dashrightarrow \sf x\:=\:-1 \:\:or \:\:2\:\:\\\\\qquad \dashrightarrow \underline {\boxed{\pmb{\frak{ \pink{ \:\:x\:=\:-1 \:\:or \:\:2\:\:}}}}}\:\:\bigstar \:$

$\qquad \therefore \underline{\sf Hence, \: All \:zeroes \:of \:Polynomial \:p(x) \: are \:\pmb{\bf \sqrt{3/2} , \:-\sqrt{3/2} \:,\: -1 \:\& \: 2 \:}\:, respectively \:.}$