Answer:

All zeroes of the POLYNOMIAL 2x^4 - 9x^3 + 5x^2 + 3x - 1 are 2 + √3, 2 - √3, 1 and -1/2.

Step-by-step EXPLANATION:

In these type of questions, where two zeroes are given and you have to find all the zeroes. First, convert the zeroes into x related equations

Like,

x = 2 + √3                                   x = 2- √3

x - 2 - √3 = 0                              x -2 + √3 = 0

Second step, multiply both the zeroes

(x - 2 - √3)(x - 2 + √3)

= (x - 2)^2 - (√3)^2                                                  (a^2 - b^2 = a-b x a+b)

= x^2 + 4 - 4x - 3                                                     [(a-b)^2 = a^2 + b^2 - 2ab]

= x^2 - 4x + 1

After this, DIVIDE the product from the given polynomial

2x^4 - 9x^3 + 5x^2 + 3x - 1/x^2 - 4x + 1

= 2x^2 - x - 1

Next, use split the middle TERM technique in the quotient formed

2x^2 - x - 1

= 2x^2 + x - 2x - 1

= x(2x + 1) - 1(2x + 1)

= (x - 1)(2x + 1)

The remainders here are the remaining two zeroes of the polynomial

x - 1 = 0                     2x + 1 = 0

x = 1                          2x = -1

                                x = -1/2

And the last step, notice the question whether it is ASKING all zeroes or the remaining zeroes and write accordingly. Here, it is asked about all the zeroes so we write all of them

Therefore, all zeroes of the polynomial 2x^4 - 9x^3 + 5x^2 + 3x - 1 are 2 + √3, 2 - √3, 1 and -1/2.

Hope I helped