Math Problem Statement

answer these quesetions. Provide steps for working the problem correctly( Step 1, Step 2, Step3, etc./use words). Also provide the concepts assessed for each problem

Write the complete linear factorization of the polynomial f(x) = 3x^4 + 8x^3 + 6x^2 + 3x - 2. Define 'extraneous solution.'

Solution

Let's solve the problems one by one as requested. I'll explain the steps thoroughly and identify the concepts assessed in each problem.


Question 3: Write the complete linear factorization of the polynomial

f(x)=3x4+8x3+6x2+3x2f(x) = 3x^4 + 8x^3 + 6x^2 + 3x - 2

Concept Assessed: Polynomial factorization, synthetic division, identifying roots, and linear factors.

Step-by-Step Solution:

  1. Step 1: Identify potential rational roots
    Use the Rational Root Theorem. Possible rational roots are factors of the constant term (-2) divided by factors of the leading coefficient (3):
    ±1,±2,±13,±23\pm 1, \pm 2, \pm \frac{1}{3}, \pm \frac{2}{3}

  2. Step 2: Test roots using synthetic division
    Test potential roots (e.g., x=1,1,2,2,x = 1, -1, 2, -2, \dots) to find one that results in a remainder of zero.
    After testing, you find that x=1x = -1 is a root.

  3. Step 3: Perform synthetic division
    Divide f(x)f(x) by x+1x + 1 using synthetic division:
    f(x)=(x+1)(3x3+5x2+x2)f(x) = (x + 1)(3x^3 + 5x^2 + x - 2)

  4. Step 4: Repeat the process for the cubic polynomial
    Apply the Rational Root Theorem to 3x3+5x2+x23x^3 + 5x^2 + x - 2.
    After testing, you find x=1x = 1 is a root. Perform synthetic division again:
    3x3+5x2+x2=(x1)(3x2+8x+2)3x^3 + 5x^2 + x - 2 = (x - 1)(3x^2 + 8x + 2)

  5. Step 5: Factor the quadratic
    Factor 3x2+8x+23x^2 + 8x + 2 using factoring methods:
    3x2+8x+2=(3x+2)(x+1)3x^2 + 8x + 2 = (3x + 2)(x + 1)

  6. Step 6: Write the complete linear factorization
    Combine all factors:
    f(x)=(x+1)(x1)(3x+2)(x+1)f(x) = (x + 1)(x - 1)(3x + 2)(x + 1)

    Alternatively, simplify repeated factors:
    f(x)=(x+1)2(x1)(3x+2)f(x) = (x + 1)^2(x - 1)(3x + 2)


Question 4: Define "extraneous solution"

Concept Assessed: Understanding solutions in equations, especially those introduced in the process of solving.

Definition:

An extraneous solution is a solution that arises during the process of solving an equation but does not satisfy the original equation.

Example Explanation:

Extraneous solutions often appear when:

  1. Both sides of an equation are squared.
  2. Variables are introduced or manipulated, such as in rational equations where denominators are cleared.

Always substitute solutions back into the original equation to verify if they are valid.


Concepts Assessed for Each Problem:

  1. Question 3: Polynomial factorization concepts:

    • Rational Root Theorem
    • Synthetic division
    • Quadratic factoring
    • Writing complete linear factorizations
  2. Question 4: Understanding extraneous solutions:

    • The nature of invalid solutions introduced during algebraic manipulation

Relative Questions:

  1. How does the Rational Root Theorem help in finding polynomial roots?
  2. Why is synthetic division preferred over long division for testing roots?
  3. How do extraneous solutions appear in radical equations?
  4. What are the differences between factoring quadratics and factoring higher-degree polynomials?
  5. Why must roots always be verified in the original equation?

Tip:

When solving equations, always check your solutions in the original equation to ensure none are extraneous.

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Math Problem Analysis

Mathematical Concepts

Polynomial Factorization
Rational Root Theorem
Synthetic Division
Understanding Extraneous Solutions

Formulas

Factorization of f(x): f(x) = (x + 1)^2(x - 1)(3x + 2)
Rational Root Theorem: Possible roots = ± factors of constant term / factors of leading coefficient

Theorems

Rational Root Theorem
Fundamental Theorem of Algebra

Suitable Grade Level

Grades 10-12