Math Problem Statement

For the polynomial P(x) = 2x^4 + 3x^3 - 8x^2 - 3x + 6, give all the possible rational zeros and supply the table using Descartes' Rule of Signs.

Solution

Let's solve the given problem step by step.

The polynomial is:

P(x)=2x4+3x38x23x+6P(x) = 2x^4 + 3x^3 - 8x^2 - 3x + 6

Part (a) - Possible Rational Zeros (p/q):

The Rational Root Theorem states that any rational solution pq\frac{p}{q} of the polynomial is a fraction where:

  • pp is a factor of the constant term (6),

  • qq is a factor of the leading coefficient (2).

  • Factors of the constant term (6): ±1,±2,±3,±6\pm 1, \pm 2, \pm 3, \pm 6

  • Factors of the leading coefficient (2): ±1,±2\pm 1, \pm 2

So, the possible rational zeros are: pq=±1,±2,±3,±6,±12,±32\frac{p}{q} = \pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}

Part (b) - Descartes' Rule of Signs:

  • Positive Roots: According to Descartes' Rule of Signs, the number of positive real roots is determined by counting the number of sign changes in P(x)P(x): P(x)=2x4+3x38x23x+6P(x) = 2x^4 + 3x^3 - 8x^2 - 3x + 6 Sign changes: 2x43x32x^4 \to 3x^3 (no change), 3x38x23x^3 \to -8x^2 (change), 8x23x-8x^2 \to -3x (no change), 3x6-3x \to 6 (change).

    • Number of sign changes = 2.
    • So, there are either 2 or 0 positive real roots.
  • Negative Roots: To find the number of negative roots, evaluate P(x)P(-x): P(x)=2(x)4+3(x)38(x)23(x)+6=2x43x38x2+3x+6P(-x) = 2(-x)^4 + 3(-x)^3 - 8(-x)^2 - 3(-x) + 6 = 2x^4 - 3x^3 - 8x^2 + 3x + 6 Sign changes: 2x43x32x^4 \to -3x^3 (change), 3x38x2-3x^3 \to -8x^2 (no change), 8x23x-8x^2 \to 3x (change), 3x63x \to 6 (no change).

    • Number of sign changes = 2.
    • So, there are either 2 or 0 negative real roots.
  • Imaginary Roots: Imaginary roots occur when there are fewer real roots than the degree of the polynomial. Since the polynomial has degree 4, the total number of roots (real and complex) must equal 4. Depending on how many real roots exist, the rest will be imaginary.

Thus, the possibilities are:

  • If there are 2 positive and 2 negative real roots, there are 0 imaginary roots.
  • If there are 0 positive and 2 negative real roots, there are 2 imaginary roots.
  • If there are 2 positive and 0 negative real roots, there are 2 imaginary roots.
  • If there are 0 positive and 0 negative real roots, there are 4 imaginary roots.

Filling in the table:

PositiveNegativeImaginary
2 or 02 or 02 or 4

Would you like further clarification on any part of the solution? Here are some questions to expand on this:

  1. How can we test which of the possible rational roots is an actual root of the polynomial?
  2. What is the significance of Descartes' Rule of Signs in finding real roots?
  3. How do you factor a polynomial of degree 4 after finding its roots?
  4. What are imaginary roots, and how do they relate to the Fundamental Theorem of Algebra?
  5. Can a polynomial have more imaginary roots than real roots?

Tip: The Rational Root Theorem is most useful when combined with synthetic division or factoring to confirm the actual roots.

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Math Problem Analysis

Mathematical Concepts

Polynomial functions
Rational zeros theorem
Descartes' Rule of Signs

Formulas

Rational Root Theorem: p/q = factors of the constant term / factors of the leading coefficient

Theorems

Rational Root Theorem
Descartes' Rule of Signs

Suitable Grade Level

High School - Advanced Algebra