To find the zeros of the polynomial t²-15, we need to solve the equation t²-15=0.

We can factor the quadratic as (t+√15)(t-√15)=0.

Therefore, the zeros of the polynomial are t=√15 and t=-√15.

Now, let's verify the relationship between the zeros and the coefficients.

- The sum of the zeros is √15 + (-√15) = 0.
- The product of the zeros is (√15)(-√15) = -15.

We can also see that the coefficients of the polynomial are a=1, b=0, and c=-15.

- The sum of the zeros is -b/a = 0/1 = 0.
- The product of the zeros is c/a = -15/1 = -15.

Therefore, we can see that the sum of the zeros is equal to the negative of the coefficient of the x-term divided by the coefficient of the x²-term, and the product of the zeros is equal to the constant term divided by the coefficient of the x²-term.



To find the zeros of the polynomial 3x²-x-4, we need to solve the equation 3x²-x-4=0.

We can use the quadratic formula x = (-b ± √(b²-4ac))/(2a) to find the zeros.

Plugging in the coefficients a=3, b=-1, and c=-4, we get:

x = (-(-1) ± √((-1)²-4(3)(-4)))/(2(3))
x = (1 ± √49)/6
x = (1 ± 7)/6

Therefore, the zeros of the polynomial are x=4/3 and x=-1.

Now, let's verify the relationship between the zeros and the coefficients.

- The sum of the zeros is 4/3 + (-1) = 1/3.
- The product of the zeros is (4/3)(-1) = -4/3.

We can also see that the coefficients of the polynomial are a=3, b=-1, and c=-4.

- The sum of the zeros is -b/a = 1/3.
- The product of the zeros is c/a = -4/3.

Therefore, we can see that the sum of the zeros is equal to the negative of the coefficient of the x-term divided by the coefficient of the x²-term, and the product of the zeros is equal to the constant term divided by the coefficient of the x²-term.

Solution of (i)
let p(t)=t²-15
zero of the polynomial is the value of t where p(t)=0
putting p(t)=0
t²-15=0
(t)²-(√15)²=0
using a²-b²=(a-b)(a+b)
(t-√15)(t+√15)=0
so we have the vale of t=√15 and t=-√15
sum of zeros=-b/a=0/1=0
√15-√15=0
product of zeros=c/a=-15/1=-15
√15×(-√15)=-15

solution of (ii)
by factorise this equation we get
3x²+3x-4x-4=0
3x(x+1)-4(x+1)=0
(x+1)(3x-4)=0
x+1=0. 3x-4=0
x=-1. x=4/3
sum of zeros=-b/a=-(-1)/3=1/3
-1+4/3=-3+4/3=1/3
product of zeros=c/a=-4/3
-1×4/3=-4/3