Find a polynomial p1(x) of degree 2 which is 0 at x = 2 and x = 3, but p1(1) = 1.Similarly, find quadratic polynomials p2(x) such that p2(1) = 0 = p2(3), p2(2) = 1;and, p3(x), such that p3(1) = 0 = p3(2), p3(3) = 1;

To find the polynomials, we can use the formula for a polynomial of degree 2, which is p(x) = ax^2 + bx + c. We know that the polynomial is 0 at certain points, so we can use those points to create a system of equations and solve for a, b, and c.

1. For p1(x), we know that p1(2) = 0 and p1(3) = 0. This gives us the equations:

   4a + 2b + c = 0 9a + 3b + c = 0

   We also know that p1(1) = 1, which gives us the equation:

   a + b + c = 1

   Solving this system of equations, we find that a = 1, b = -3, and c = 3. So, p1(x) = x^2 - 3x + 3.

2. For p2(x), we know that p2(1) = 0 and p2(3) = 0. This gives us the equations:

   a + b + c = 0 9a + 3b + c = 0

   We also know that p2(2) = 1, which gives us the equation:

   4a + 2b + c = 1

   Solving this system of equations, we find that a = 1, b = -3, and c = 2. So, p2(x) = x^2 - 3x + 2.

3. For p3(x), we know that p3(1) = 0 and p3(2) = 0. This gives us the equations:

   a + b + c = 0 4a + 2b + c = 0

   We also know that p3(3) = 1, which gives us the equation:

   9a + 3b + c = 1

   Solving this system of equations, we find that a = 1, b = -3, and c = 2. So, p3(x) = x^2 - 3x + 2.