Find the zeros of the polynomial2 1 26x x  , and verify the relation between thecoefficients and zeros of the polynomial.

The polynomial given seems to be incorrectly written. However, assuming the polynomial is 2x^2 + x - 26, we can find the zeros of the polynomial by setting it equal to zero and solving for x.

Step 1: Set the polynomial equal to zero 2x^2 + x - 26 = 0

Step 2: Solve for x using the quadratic formula x = [-b ± sqrt(b^2 - 4ac)] / 2a Here, a = 2, b = 1, and c = -26

x = [-1 ± sqrt((1)^2 - 42(-26))] / 2*2 x = [-1 ± sqrt(1 + 208)] / 4 x = [-1 ± sqrt(209)] / 4

So, the zeros of the polynomial are x = [-1 + sqrt(209)] / 4 and x = [-1 - sqrt(209)] / 4

To verify the relation between the coefficients and zeros of the polynomial, we use the fact that the sum of the zeros is -b/a and the product of the zeros is c/a.

Sum of zeros = -b/a = -1/2 Product of zeros = c/a = -26/2 = -13

Let's check if these relations hold true:

Sum of zeros = [-1 + sqrt(209)] / 4 + [-1 - sqrt(209)] / 4 = -1/2 Product of zeros = ([-1 + sqrt(209)] / 4) * ([-1 - sqrt(209)] / 4) = -13

So, the relations between the coefficients and zeros of the polynomial are verified.