Math Problem Statement

Find a formula for the polynomial 𝑃(π‘₯) with degree 3 real coefficients zeros at π‘₯=1+4𝑖 and π‘₯=3 𝑦-intercept at (0,204) 𝑃(π‘₯)=

Solution

To find the polynomial P(x)P(x) with the given properties, we will use the following information:

  1. Degree 3: The polynomial has a degree of 3, so it can be expressed in the form P(x)=a(xβˆ’r1)(xβˆ’r2)(xβˆ’r3)P(x) = a(x - r_1)(x - r_2)(x - r_3), where r1,r2,r_1, r_2, and r3r_3 are its roots.
  2. Real coefficients: The coefficients of the polynomial are real. If there is a complex root, its complex conjugate must also be a root.
  3. Zeros at x=1+4ix = 1 + 4i and x=3x = 3: The conjugate of 1+4i1 + 4i, which is 1βˆ’4i1 - 4i, must also be a root due to the real coefficients.
  4. yy-intercept at (0,204)(0, 204): This means P(0)=204P(0) = 204.

Given this, the roots of the polynomial are x=1+4ix = 1 + 4i, x=1βˆ’4ix = 1 - 4i, and x=3x = 3. The polynomial can be written as:

P(x)=a(xβˆ’(1+4i))(xβˆ’(1βˆ’4i))(xβˆ’3)P(x) = a(x - (1 + 4i))(x - (1 - 4i))(x - 3)

Step 1: Expand the Factors

First, we multiply the factors involving the complex roots:

(xβˆ’(1+4i))(xβˆ’(1βˆ’4i))(x - (1 + 4i))(x - (1 - 4i))

=(xβˆ’1βˆ’4i)(xβˆ’1+4i)= (x - 1 - 4i)(x - 1 + 4i)

=((xβˆ’1)βˆ’4i)((xβˆ’1)+4i)= ((x - 1) - 4i)((x - 1) + 4i)

=(xβˆ’1)2βˆ’(4i)2= (x - 1)^2 - (4i)^2

=(xβˆ’1)2+16= (x - 1)^2 + 16

=x2βˆ’2x+1+16= x^2 - 2x + 1 + 16

=x2βˆ’2x+17= x^2 - 2x + 17

Now, the polynomial becomes:

P(x)=a(x2βˆ’2x+17)(xβˆ’3)P(x) = a(x^2 - 2x + 17)(x - 3)

Step 2: Expand the Remaining Terms

P(x)=a(x3βˆ’3x2βˆ’2x2+6x+17xβˆ’51)P(x) = a(x^3 - 3x^2 - 2x^2 + 6x + 17x - 51)

=a(x3βˆ’5x2+23xβˆ’51)= a(x^3 - 5x^2 + 23x - 51)

Step 3: Determine aa Using the y-Intercept

Since the polynomial has a y-intercept of 204204, we substitute x=0x = 0 and set P(0)=204P(0) = 204:

P(0)=a(03βˆ’5(0)2+23(0)βˆ’51)=204P(0) = a(0^3 - 5(0)^2 + 23(0) - 51) = 204

βˆ’51a=204-51a = 204

a=βˆ’20451=βˆ’4a = -\frac{204}{51} = -4

Final Polynomial

P(x)=βˆ’4(x3βˆ’5x2+23xβˆ’51)P(x) = -4(x^3 - 5x^2 + 23x - 51)

P(x)=βˆ’4x3+20x2βˆ’92x+204P(x) = -4x^3 + 20x^2 - 92x + 204

Thus, the polynomial is:

P(x)=βˆ’4x3+20x2βˆ’92x+204\boxed{P(x) = -4x^3 + 20x^2 - 92x + 204}

Would you like a more detailed explanation or have any questions?


Questions

  1. What are complex conjugate roots, and why must they appear in polynomials with real coefficients?
  2. How do we determine the constant aa in a polynomial given a specific point?
  3. Why is it necessary to expand the polynomial when finding the final expression?
  4. How can we verify that the polynomial has the correct y-intercept?
  5. What is the significance of the degree of a polynomial?
  6. How does the Fundamental Theorem of Algebra relate to this problem?
  7. Can a polynomial with real coefficients have an odd number of non-real roots?
  8. How can you determine the end behavior of a polynomial function?

Tip

When dealing with polynomials, always check for complex conjugate pairs to ensure the coefficients remain real.

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Math Problem Analysis

Mathematical Concepts

Polynomials
Complex Numbers
Roots of Polynomials
Coefficient Realism

Formulas

Quadratic Formula
Cubic Polynomial Expansion

Theorems

Fundamental Theorem of Algebra

Suitable Grade Level

Grades 10-12