How can I find a second-degree polynomial p such that p(2) = 5, p(3) = 3, and p(2) = 2?

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Finding the Second-Degree Polynomial

We are looking for a second-degree polynomial of the form: 

        p(x) = ax2 + bx + c

We have the following constraints based on the question:

  • p(2) = 5
  • p(3) = 3
  • p(2) = 2

There’s an inconsistency in the conditions provided. If we are to maintain that p(2) = 5 and p(2) = 2 at the same time, this cannot hold true for any polynomial. Therefore, we should clarify the conditions before proceeding further.

Assuming the conditions meant to define two separate evaluations, we’ll proceed with:

  • p(2) = 5
  • p(3) = 3

To find a specific polynomial, we can set up a system of equations by substituting the x-values into our equation:

  1. For p(2) = 5:

    a(22) + b(2) + c = 5

    4a + 2b + c = 5
  2. For p(3) = 3:

    a(32) + b(3) + c = 3

    9a + 3b + c = 3

Now we have two equations:

  • 1. 4a + 2b + c = 5
  • 2. 9a + 3b + c = 3

We can eliminate c by subtracting equation (1) from equation (2): 

        (9a + 3b + c) - (4a + 2b + c) = 3 - 5
        
 5a + b = -2

Now we can express b in terms of a: 

        b = -2 - 5a

We now have a single variable equation for b. Next, we can substitute this back into one of our original equations to express everything in terms of a: 

        4a + 2(-2 - 5a) + c = 5
        
 4a - 4 - 10a + c = 5
        
 -6a + c = 9

Thus, we can express c in terms of a: 

        c = 6a + 9

With expressions for b and c, we can represent the polynomial as: 

        p(x) = ax2 + (-2 - 5a)x + (6a + 9)

Further evaluation will give you the polynomial based on your choice for a. An example would be setting a = 1: 

        p(x) = x2 - 7x + 15

This polynomial will meet the conditions for p(2) and p(3) derived from your equations.

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