Answer:

Given:
Zeros of 2x^2-5x+7 are alpha and beta.

We need to find the polynomial whose zeros are 2alpha+3beta and 3alpha+2beta.

Explanation:

We know that if alpha and beta are the roots of the quadratic equation ax^2+bx+c=0, then the equation can be written in the form of a(x-alpha)(x-beta)=0.

Let's find the equation whose roots are alpha and beta.

The given quadratic equation is 2x^2-5x+7=0.
Using the quadratic formula, we can find the roots of this equation as:

alpha = (5+sqrt(11))/4
beta = (5-sqrt(11))/4

Now, we can write the equation in the form of a(x-alpha)(x-beta)=0 as:

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

Expanding this equation, we get:

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

Simplifying, we get:

2x^2-5x+7=0

Therefore, the equation whose roots are alpha and beta is 2x^2-5x+7=0.

Now, let's find the equation whose roots are 2alpha+3beta and 3alpha+2beta.

Let's assume that the equation is of the form ax^2+bx+c=0.

Using the sum and product of roots formula, we can find the values of b and c as:

b = -a(2alpha+3beta+3alpha+2beta) = -5a(alpha+beta)
c = a(2alpha+3beta)(3alpha+2beta) = 6a(alpha*beta)

We know that the equation whose roots are alpha and beta is 2x^2-5x+7=0. Therefore, alpha*beta = 7/2 and alpha+beta = 5/2.

Substituting these values in the expressions of b and c, we get:

b = -5a(5/2) = -25a/2
c = 6a(2alpha+3beta)(3alpha+2beta) = 6a(6alpha^2+13alpha*beta+4beta^2)

Substituting the values of alpha and beta, we get:

c = 6a(6[(5+sqrt(11))/4]^2+13[(5+sqrt(11))/4][(5-sqrt(11))/4]+4[(5-sqrt(11))/4]^2)

Simplifying, we get:

c = 153a/8

Therefore, the equation whose roots are 2alpha+3beta and 3alpha+2beta is:

ax^2-25ax/2+153a/8=0

Simplifying, we get:

2x^2-25x+153=0

Thus, the polynomial whose zeros are 2alpha+3beta and 3alpha+2beta is 2x^2-25x+153.