**Q) Find the zeroes of the polynomial f(t) = t ^{2} + 4 √3 t – 15 and verify the relationship between the zeroes and the coefficients of the polynomial.**

**Ans: **In the given polynomial equation, to find zeroes, we will start with f(x) = 0

Therefore, t^{2} + 4√3 t – 15 = 0

**Step 1:** Let’s start calculating the zeroes of the polynomial.

The factors of 15 are 5 and 3, but since we need to arrive at 4√3, we can take factors of 15 as 5√3 and √3, so that product is 15 and difference is 4√3. With this input, let’s start solving f(t) = 0

t^{2} + 4 √3 t – 15 = 0

∴ t^{2} + 5 √3 t – √3 t – 15 = 0

∴ t(t + 5 √3) – √3 (t + 5√3) = 0

∴ (t + 5 √3) (t – √3) = 0

**Hence t = √3 and t = – 5√3 the zeroes of the given polynomial.**

**Step 2:** Next we have to verify the relationship between the zeroes and the coefficients of the polynomial.

To do this, we need to find the sum of zeroes and the product of zeroes

We know that, if α and β be the zeros of the polynomial, then

Sum of zeroes, α + β =

and Product of Zeroes, α × β =

We will find the values of both sides and if these are matched, the relationship between the zeroes and the coefficients will get verified.

Since, we have already calculated values of the zeroes of polynomial, let’s calculate values of the coefficients now.

When we compare polynomial t^{2} + 4 √3 t – 15 = 0 with standard quadratic equation ax^{2} + bx + c = 0, we get

a = 1, b = 4 √3 and c = – 15

**Step 3:** Let’s start verifying the relationships:

The sum of the zeros:

α + β = √3 – 5√3 = – 4√3

and = – 4 √3

Hence, the relation of sum of the zeros (α + β = ) is verified.

Next, we take Product of zeroes

α × β = √3 (- 5√3) = – 15

and = -15

Hence, the relationship of product of zeros (α × β = ) is also verified.

**Thus, x = √3 and – 5√3 are the zeroes of the polynomial.**

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