Determining the Value of k for Polynomial Factors and Zeroes

Determining the Value of k for Polynomial Factors and Zeroes

Consider the polynomial 2x4x3 – 14x25x6. We aim to determine the value of k such that the polynomial can be factorized appropriately. This problem involves understanding polynomial factors, zeroes, and the process of polynomial division.

Context and Problem Setup

Given the polynomial 2x4x3 – 14x25x6, we recognize it as a polynomial of degree 4. A key factor of this polynomial is (x22xk), which is a quadratic polynomial of degree 2.

Factorization Process

The factorization process involves finding the other factor which, when multiplied by (x22xk), yields the original polynomial. Let's denote the unknown linear factor as (2x^2 bx c).

The product of these factors should be equal to the original polynomial:

[ 2x^4x^3 – 14x^25x6 (x^22xk)(2x^2axb) ]

By comparing coefficients, we can determine the values of (k), (a), (b), and (c).

Step-by-Step Solution

1. **Coefficient of (x^3):**

[ 1 a4 a -3 ]

Substituting (a -3), we get:

[ 2k – 6 -14 ]

Solving for (k), we find:

[ 2k – 8 -14 Rightarrow 2k -6 Rightarrow k -3 ]

2. **Coefficient of (x^2):**

[ -14 b2k2a -14 Rightarrow b2(-3)2(-3) -14 Rightarrow b - 8 -14 Rightarrow b -6 ]

3. **Coefficient of (x):**

[ 5 2bak 5 Rightarrow 2(-6)(-3)k 5 Rightarrow k -3 ]

Thus, the correct value of (k) is -3.

4. **Verifying the Polynomial:**

The polynomial can be written as:

[ 2x^4 x^3 – 14x^2 5x 6 x^22x-32x^2 – 3x - 2 ]

Factoring it further, we get:

[ (x^22x3)(x-3)(x-1) ]

The zeroes of the polynomial are -3, 1, 2, and -1/2.

General Method for Biquadratic Polynomials

A given polynomial of biquadratic degree 4, (px 2x^4 x^3 – 14x^2 5x 6), has one of its factors as a quadratic polynomial (qx x^2 2x k). The other factors will be two linear factors (ax b).

To find (a) and (b), we consider the product of the three factors:

[ x^2 2x k xa xb x^4 2abx^3 {k2abab} x^2 {kab 2ab} x kab ]

By comparing coefficients, the relationships are derived as:

[ 2ab 1/2 ]

[ k2abab -7 ]

[ kab2ab 5/2 ]

[ kab 3 ]

Solving these equations, we ultimately find the value of (k) to be -3.

This step-by-step factorization and comparison process is crucial for understanding and applying polynomial theory effectively.