P(x) =x2-6x+a if 3alpha +2beta=20 find value of 'a', alpha and beta are zeroes of polynomial

To find the value of 'a', we need to use the information given in the question.

The polynomial P(x) is given as P(x) = x^2 - 6x + a.

We are also given that 3alpha + 2beta = 20, where alpha and beta are the zeroes of the polynomial.

Since alpha and beta are the zeroes of the polynomial, we can write the equation as follows:

(x - alpha)(x - beta) = 0.

Expanding this equation, we get:

x^2 - (alpha + beta)x + alpha*beta = 0.

Comparing this equation with the given polynomial P(x), we can equate the coefficients:

-6 = -(alpha + beta) and a = alpha*beta.

From the equation 3alpha + 2beta = 20, we can solve for alpha and beta.

Rearranging the equation, we get:

3alpha = 20 - 2beta.

Dividing both sides by 3, we have:

alpha = (20 - 2beta)/3.

Substituting this value of alpha in the equation -6 = -(alpha + beta), we get:

-6 = -((20 - 2beta)/3 + beta).

Simplifying this equation, we can solve for beta.

Multiplying both sides by 3, we have:

-18 = -(20 - 2beta + 3beta).

Expanding and rearranging the equation, we get:

-18 = -20 + beta.

Adding 20 to both sides, we have:

2 = beta.

Now that we have the value of beta, we can substitute it back into the equation alpha = (20 - 2beta)/3 to find the value of alpha.

Substituting beta = 2, we have:

alpha = (20 - 2*2)/3 = 16/3.

Finally, we can substitute the values of alpha and beta into the equation a = alpha*beta to find the value of 'a'.

Substituting alpha = 16/3 and beta = 2, we have:

a = (16/3)*2 = 32/3.

Therefore, the value of 'a' is 32/3.