Question

given that x minus root 5 is a factor of the cubic polynomial x cube minus 3 root 5 x square + 13 x minus 3 root 5 find all the zeros of the polynomial

Answer 1

Answer:

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Sol :

given that x-√5 is a factor of the cubic polynomial x3-3√5x2+13x-3√5

x-√5 ) x3-3√5x2+13x-3√5 ( x2 -2√5x + 3

x3- √5x2 ( substract )

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- 2√5x2+13x

- 2√5x2+10x ( substract )

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3x - 3√5

3x - 3√5 ( substract )

------------------------

0

∴ The quotient is x2 -2√5x + 3 = 0

Using roots of quadratic formula

a = 1, b = 2√5, c = 3

x = (-b ± √(b2 - 4ac) ) / 2a

x = (2√5 ± √((2√5)2 - 12) ) / 2

∴ the other zeros are x = √5 ± √2.

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Step-by-step explanation:

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Here is your answer,

______________________________________________________________

p(x) = x³ - 3√5x² + 13x - 3√5

x-√5 is a root of p(x)

We will divide p(x) by x-√5,

                                                                   

                    x-√5 )   x³ - 3√5x² + 13x - 3√5(  x² - 2√5x + 3

                                x³ -  √5x²

                             (-)   (+)       

                                   -2√5x²  +  13x

                                   -2√5x²  +   10x

                                   (+)       (-)         

                                                      3x - 3√5

                                                      3x - 3√5

                                                   (-)    (+)     

                                                            0      

 

Now we get a quadratic equation as quotient , we will find the roots of the quotient :

                   x² - 2√5x + 3

     Roots = -b  ± √(b²- 4ac)

                         2a

               = -(-2√5) ± √[(-2√5)² - 4*1*3]

                              2*1

               = 2√5 ± √(20 - 12)

                            2

               = 2√5 ± √8

                         2

               = 2√5 ± 2√2

                         2

               

               √5 + √2 and  √5 - √2

All the roots are  √5 , √5 + √2 and √5 - √2.

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Answer 2

AnswEr :

$\bigstar\:\boxed{\sf\underbrace{\sf Dividend}_{\large f(x)} = \underbrace{\sf Divisor}_{\large p(x)}\times\underbrace{\sf Quotient}_{\large q(x)} + \underbrace{\sf Remainder}_{\large r(x)}}$

★ Here, we Have :

• f(x) = x³ – 3√5x² + 13x – 3√5
• p(x) = x – √5

We will Divide f(x) by p(x) to find the q(x) that will be other Roots of the Given Polynomial.

$\boxed{\begin{array}{l | c | r}\sf x-\sqrt{5}&\sf x^3-3\sqrt{5}x^2+13x-3\sqrt{5}&\sf x^2-2\sqrt{5}x+3\\&\sf x^3-\sqrt{5}x^2\qquad\qquad\qquad\\&( - )\:\:\:\:( + )\qquad\:\:\:\qquad\qquad\\&\rule{90}{0.8}\quad\:\:\:\\&\sf\:\:-2\sqrt{5}x^2+13x\qquad\\ &\sf-\:2\sqrt{5}x^2+10x\qquad\!\!\\ & ( + )\qquad( - )\qquad\\&\quad\rule{85}{0.8}\\&\qquad\qquad\sf3x-3\sqrt{5}\\ &\sf\qquad\qquad\sf3x-3\sqrt{5}\\&\qquad\qquad(-)\quad(+)\:\\&\qquad\quad\rule{65}{0.8}\\ &\sf\qquad\qquad0\end{array}}$

$\rule{150}{1}$

• we get q(x) = x² – 2√5x + 3, we will find roots from this now.

$:\implies\tt x^2 - 2\sqrt{5}x + 3 = 0\\\\\\:\implies\tt x = \dfrac{-\:b\pm\sqrt{b^2-4ac}}{2a}\\\\\\:\implies\tt x = \dfrac{-\:( - 2\sqrt{5}) \pm\sqrt{( - 2 \sqrt{5})^2-4 \times 1 \times 3}}{2\times 1}\\\\\\:\implies\tt x = \dfrac{2 \sqrt{5} \pm \sqrt{20 - 12}}{2}\\\\\\:\implies\tt x = \dfrac{2 \sqrt{5} \pm \sqrt{8}}{2}\\\\\\:\implies\tt x = \dfrac{2 \sqrt{5} \pm2 \sqrt{2}}{2}\\\\\\:\implies\tt x =\sqrt{5} \pm \sqrt{2}\\\\\\:\implies\underline{\boxed{\tt x = (\sqrt{5} +\sqrt{2}) \quad or \quad (\sqrt{5} -\sqrt{2})}}$

⠀

$\therefore\:\underline{\sf{Roots \:are \: \sqrt{5}, (\sqrt{5}+\sqrt{2}) \:and \:(\sqrt{5}-\sqrt{2})}.}$