Question

obtain all zeros of the polynomial 6x^4 -23x^2+13x^3-39x+15, if two of its zeros are √3 and -√3​

Answer 1

Answer:

form it in sequence

two zeroes are given so alpha and beta are +-root 3

so qudratic equation formed is x2-3

So divide it with the given polynomial u will get it

Answer 2

Given :

• The given polynomial : 6x⁴ - 23x² + 13x⅗ - 39x + 15

Rearranging the equation, we get : 6x⁴ + 13x³ - 23x² - 39x + 15

• Two of it's zeroes are ±√3

To Find :

• All the zeroes of the Given Polynomial.

We are given that two of it's zeroes are √3 and -√3.

So the equation would be x² - 3 = 0

Let us divide this equation with the given polynomial to find the remaining zeroes.

Have a look at the attachment!

We got the Quotient as 6x² + 13x - 5

Let us split the middle term of this equation.

⟹ 6x² + 13x - 5

⟹ 6x² - 15x + 2x + 5

⟹ 3x ( 2x - 5 ) + 1 ( 2x - 5 )

⟹ ( 2x - 5 ) ( 3x + 1 )

__________________

$2x - 5 = 0$

$2x = 5$

$x = \frac{5}{2}$

One of the zero is found.

$3x + 1 = 0$

$3x = - 1$

$x = \frac{ - 1}{3}$

The remaining zeroes are also found.

Hence, the zeroes of the Polynomial are $\sf{ \frac{5}{2}, \frac{ - 1}{ 3} + \sqrt{3} \: and - \sqrt{3} }$