Math Problem Statement

do that exact same process I showed you with the correct answer from that question with this question

Find a formula for the polynomial P(x) with degree 3, real coefficients, zeros at x = 3 + 2i and x = 5, and y-intercept at (0, 260).

Solution

To find the polynomial P(x)P(x) based on the given conditions, let's go through a structured solution process similar to what you requested:

  1. Identify the Roots:

    • Given a root at x=3+2ix = 3 + 2i. Since we need real coefficients, the complex conjugate x=32ix = 3 - 2i must also be a root.
    • There’s another root at x=5x = 5.
  2. Form the Polynomial:

    • The polynomial P(x)P(x) can be expressed in terms of its roots as: P(x)=k(x(3+2i))(x(32i))(x5)P(x) = k (x - (3 + 2i))(x - (3 - 2i))(x - 5)
    • Simplify the product of the first two terms using the difference of squares: (x(3+2i))(x(32i))=(x3)2(2i)2=(x3)2+4(x - (3 + 2i))(x - (3 - 2i)) = (x - 3)^2 - (2i)^2 = (x - 3)^2 + 4
    • Expanding (x3)2+4(x - 3)^2 + 4: (x3)2+4=x26x+9+4=x26x+13(x - 3)^2 + 4 = x^2 - 6x + 9 + 4 = x^2 - 6x + 13
    • Thus, the polynomial becomes: P(x)=k(x26x+13)(x5)P(x) = k (x^2 - 6x + 13)(x - 5)
  3. Expand the Polynomial:

    • Now, expand (x26x+13)(x5)(x^2 - 6x + 13)(x - 5): P(x)=k(x35x26x2+30x+13x65)=k(x311x2+43x65)P(x) = k (x^3 - 5x^2 - 6x^2 + 30x + 13x - 65) = k (x^3 - 11x^2 + 43x - 65)
    • So we have: P(x)=k(x311x2+43x65)P(x) = k(x^3 - 11x^2 + 43x - 65)
  4. Determine the Constant kk:

    • Use the yy-intercept at (0,260)(0, 260), which means P(0)=260P(0) = 260.
    • Substitute x=0x = 0 into the polynomial: P(0)=k(65)=260P(0) = k(-65) = 260
    • Solve for kk: k=26065=4k = \frac{260}{-65} = -4
  5. Write the Final Polynomial:

    • Substitute k=4k = -4 back into P(x)P(x): P(x)=4(x311x2+43x65)P(x) = -4(x^3 - 11x^2 + 43x - 65)
    • Expanding the terms, we get: P(x)=4x3+44x2172x+260P(x) = -4x^3 + 44x^2 - 172x + 260

So, the polynomial P(x)P(x) is: P(x)=4x3+44x2172x+260P(x) = -4x^3 + 44x^2 - 172x + 260

Would you like further details on any of these steps?

Related Questions

  1. What if the polynomial had a different yy-intercept? How would that change the result?
  2. How does the requirement for real coefficients affect the choice of complex roots?
  3. How would this process differ if the degree of the polynomial were 4 instead of 3?
  4. Can you verify the polynomial by plugging in the zeros?
  5. What role does the complex conjugate root play in constructing polynomials with real coefficients?

Tip

When constructing a polynomial with complex roots, always remember to include the complex conjugate to ensure real coefficients.

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Math Problem Analysis

Mathematical Concepts

Polynomials
Complex Numbers
Roots and Factors

Formulas

Polynomial with roots: P(x) = k(x - root1)(x - root2)...(x - rootn)
Difference of squares: (a + b)(a - b) = a^2 - b^2

Theorems

Conjugate Root Theorem

Suitable Grade Level

Grade 11-12