Question

If three zeroes of a polynomial x4-x3-3x2+3x are 0,root over 3 and root over -3. Find the fourth zero

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

$\sf{If \: \alpha , \beta , \gamma , \delta \: are \: the \: roots \: of \: the \: equation \: a {x}^{4} + b {x}^{3} + c {x}^{2} + dx + e = 0 \: then }$

$\displaystyle \sf{ \sum \alpha = - \frac{b}{a} \: }$

$\displaystyle \sf{ \sum \alpha \beta = \frac{c}{a} \: }$

$\displaystyle \sf{ \sum \alpha \beta \gamma = - \frac{d}{a} \: }$

$\displaystyle \sf{ \alpha \beta \gamma \delta = \frac{e}{a} \: }$

GIVEN

$\sf{Three \: zeroes \: of \: the \: equation }$

$\sf{ \: {x}^{4} - {x}^{3} - 3 {x}^{2} + 3x = 0 \: are \:0 \: , \sqrt{3} \: , - \sqrt{3} }$

TO DETERMINE

The fourth zero of the equation

CALCULATION

Comparing the given equation with

$\sf{ \: a {x}^{4} + b {x}^{3} + c {x}^{2} + dx + e = 0 }$

$\sf{a = 1 , b = - 1 , c = - 3 ,d = 3, e = 0 }$

$\sf{Let \: \: \delta \: \: be \: the \: Fourth \: root \: of \: the \: equation \: }$

By the above mentioned formula we get

$\displaystyle \: \sf{0 + \sqrt{3} - \sqrt{3} + \delta \: = - \: \frac{ - 1}{1} \: }$

$\implies \: \displaystyle \: \sf{ \delta \: = 1 \: }$

So the required fourth zero is 1

Answer 2

Given : 0 , √3 , - √3 are zeroes of  x⁴ - x³ - 3x² + 3x = 0  

To Find : the fourth zero

Solution:  

Another method than 1st solution :

0 , √3 , - √3 are zeroes of  x⁴ - x³ - 3x² + 3x = 0  

=> (x - 0)(x - √3)(x - (-√3)) is factor of  x⁴ - x³ - 3x² + 3x

=> x ( x² - 3) is  factor of  x⁴ - x³ - 3x² + 3x

(x⁴ - x³ - 3x² + 3x)/(x ( x² - 3))

= ( x³ - x² - 3x + 3)/ ( x² - 3)

= (x²(x - 1) - 3(x - 1))/ ( x² - 3)

= (x² - 3)(x - 1)/ ( x² - 3)

= x - 1

x - 1 = 0

=> x  = 1

fourth zero is 1

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