Question

find all zeroes of the polynomial 3y^4 + 6y^3 - 2y^2 - 10y - 5 , if two of its zeroes are root of 5/3 and - root of 5/3

Answer 1

Given: The polynomial 3y^4 + 6y^3 - 2y^2 - 10y - 5,  its zeroes are root of 5/3 and - root of 5/3.

To find: The remaining two zeroes?

Solution:

• Now the polynomial given is  3y^4 + 6y^3 - 2y^2 - 10y - 5.

• Its roots are: √5/3 and -√5/3

        (y - √5/3)(y + √5/3) = 0

        y^2 - √5/3^2 = 0

        y^2 - 5/3 = 0

        3y^2 - 5 = 0

• Now dividing the polynomial by 3y^2 - 5 , we get:

        3y^2 - 5 ) 3y^4 + 6y^3 - 2y^2 - 10y - 5 ( y^2 + 2y + 1

                         3y^4             - 5y^2

                        (-                          +)

                                    6y^3 + 3y^2 - 10y - 5

                                    6y^3             + 10y

                                    (-                       -    )

                                                3y^2          -  5

                                                3y^2          -  5

                                                (-                    + )

                                                           0

• Now the quotient is y^2 + 2y + 1

• Factorizing it, we get:

        y^2 + y + y + 1

        y(y+1) + 1(y+1)

        (y+1)(y+1)

        y = -1,-1

• So the remaining two zeroes are same, that is -1.

Answer:

      So the zeroes of the given polynomial are -1, -1, √5/3 and -√5/3.

 

Answer 2

Given polynomial is 3y⁴+6y³-2y²-10y-5 ---(1)

The degree of the given polynomial is 4, so it has almost 4 zeroes.

The two zeroes of given polynomials are

$\sqrt{ \frac{5}{3}} \: and \: -\sqrt{ \frac{5}{3}}$

$The \: equation \: of \: the \: polynomial \\whose\: zeroes \: are \: \sqrt{ \frac{5}{3}} \: and \\ -\sqrt{ \frac{5}{3}}\:is$

$\Big( y + \sqrt{ \frac{5}{3}} \Big)\Big( y - \sqrt{ \frac{5}{3}} \Big)$

$= y^{2} - \frac{5}{3} \: --(2)$

$Divide \: (1) \:with \: (2) , \:we \: get$

$3y^{4} + 6y^{3} - 2y^{2} - 10y - 5$

$= \Big( y^{2} - \frac{5}{3} \Big) (3y^{2}+6y+3)\: \blue { ( See \:the \: attachment )}$

$=3\Big( y^{2} - \frac{5}{3} \Big) (y^{2}+2y+1) \\= 3\Big( y^{2} - \frac{5}{3} \Big)( y+1)^{2} \\= 3</p><p>\Big( y + \sqrt{ \frac{5}{3}} \Big)\Big( y - \sqrt{ \frac{5}{3}} \Big)(y+1)(y+1)$

$\therefore The \:zeroes \: of \: the \\polynomial \: are \: \sqrt{ \frac{5}{3}} \: , -\sqrt{ \frac{5}{3}}, -1 \: and \: -1$

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