To find the zeroes of a quadratic polynomial, we need to solve for the values of the variable that make the polynomial equal to zero. In this case, we have the polynomial t²-15, and we want to find the values of t that satisfy the equation t²-15=0.



One way to find the zeroes of a quadratic polynomial is by using the quadratic formula, which states that the solutions to the equation ax²+bx+c=0 are given by:

x = (-b ± sqrt(b²-4ac)) / 2a

In our case, a=1, b=0, and c=-15. Substituting these values into the quadratic formula, we get:

t = (-0 ± sqrt(0²-4(1)(-15))) / 2(1)

Simplifying the expression inside the square root, we get:

t = (-0 ± sqrt(60)) / 2

t = (± sqrt(60)) / 2

We can simplify this further by factoring out a 2 from under the square root:

t = (± sqrt(4*15)) / 2

t = (± 2sqrt(15)) / 2

t = ± sqrt(15)

Therefore, the zeroes of the polynomial t²-15 are t=±sqrt(15).



In a quadratic polynomial of the form ax²+bx+c, the sum of the zeroes is given by -b/a, and the product of the zeroes is given by c/a. We can verify these relationships using the zeroes we found above.

The sum of the zeroes is:

sqrt(15) + (-sqrt(15)) = 0

This matches the coefficient of the x term, which in our case is 0.

The product of the zeroes is:

sqrt(15) * (-sqrt(15)) = -15

This also matches the constant term of the polynomial, which is -15.

Therefore, we have verified the relationship between the zeroes and coefficients of the quadratic polynomial t²-15.