Problem 21 - Algebra 2

Question

Given a polynomial and one of its factors, find the remaining factors of the polynomial. Some factors may not be binomials. $$ x^{3}-6 x^{2}+11 x-6 ; x-2 $$

Step-by-Step Solution

Answer

The remaining factors are \((x-3)(x-1)\).

Step 1: Understand the Problem

We are given a cubic polynomial \(x^3 - 6x^2 + 11x - 6\) and one of its factors \(x-2\). Our task is to find the remaining factors of the polynomial.

Step 2: Use Polynomial Division

Since \(x-2\) is a factor, divide the polynomial \(x^3 - 6x^2 + 11x - 6\) by \(x-2\) using synthetic or long division to find the quotient, which will be another polynomial.

Step 3: Perform Synthetic Division Setup

For synthetic division, write down the coefficients of the polynomial: \(1, -6, 11, -6\). Use the root of the factor, \(x = 2\), for the division process.

Step 4: Execute Synthetic Division

1. Bring down the first coefficient (1). 2. Multiply it by 2 and add to the next coefficient (-6) resulting in -4. 3. Multiply -4 by 2 and add to 11 resulting in 3. 4. Multiply 3 by 2 and add to -6 resulting in 0. The result of the division is \(x^2 - 4x + 3\).

Step 5: Factor the Quotient

Now factor the quotient \(x^2 - 4x + 3\). Look for two numbers that multiply to 3 and add to -4. These numbers are -3 and -1, hence \(x^2 - 4x + 3 = (x-3)(x-1)\).

Step 6: Identify All Factors

The original polynomial \(x^3 - 6x^2 + 11x - 6\) can thus be written as \((x-2)(x-3)(x-1)\). These are all the factors of the polynomial, including the given factor \(x-2\).

Key Concepts

Cubic Polynomial

Cubic polynomials are expressions of degree three, which means the highest power of the variable is three. These types of polynomials can take the form \( ax^3 + bx^2 + cx + d \), where \( a \), \( b \), \( c \), and \( d \) are constants. In our given problem, the cubic polynomial is \( x^3 - 6x^2 + 11x - 6 \). Understanding cubic polynomials involves recognizing their overall structure and behavior.

• The leading term, \( x^3 \), dominates the behavior of the polynomial as \( x \) becomes very large or very small.
• The constant term, in this case, \(-6\), tells us the y-intercept of the graph of the polynomial.
• The solutions to the equation \( x^3 - 6x^2 + 11x - 6 = 0 \) are the roots of the polynomial, corresponding to where the graph crosses the x-axis.

Recognizing that \( x-2 \) is a factor means that 2 is one of these roots. Finding all factors of such a polynomial is essential to fully solving it.

Synthetic Division

Synthetic division is a simplified form of polynomial division, especially useful when dividing by linear factors of the form \( x-c \). It streamlines the process, making it faster and less error-prone compared to long division. To perform synthetic division with our polynomial \( x^3 - 6x^2 + 11x - 6 \) and the factor \( x-2 \), we:

• Write down the coefficients: 1, -6, 11, and -6.
• Use the root, \( x = 2 \).
• Bring down the first coefficient (1) straightforwardly.

Next, multiply the first coefficient (1) by the root (2), and add the result to the next coefficient (-6). Continue this process for all coefficients.The result, \( x^2 - 4x + 3 \), is a new polynomial we can factor further to find all the polynomial's roots.

Polynomial Division

Polynomial division is a technique to divide polynomials, similar to numerical long division but operating with variables and coefficients. It helps simplify complex polynomials or find remainders when non-factors are divided.For the given problem, we used synthetic division—a much quicker method—to achieve the same result as long division. The polynomial \( x^3 - 6x^2 + 11x - 6 \) divided by \( x - 2 \) yielded \( x^2 - 4x + 3 \). This resulting polynomial, often called the quotient, can then be further analyzed or factored. What's important in polynomial division is ensuring each step is performed correctly, as one small mistake can lead to incorrect solutions. Confident division is especially useful when solving problems, ensuring accurate simplification and solution-finding.

Factoring Quadratic Expressions

Factoring quadratic expressions is key to solving polynomials. After using synthetic division on our cubic polynomial, we arrived at a quadratic expression: \( x^2 - 4x + 3 \).To factor this expression, we look for two numbers that:

• Multiply to give the constant term, 3.
• Add up to give the linear coefficient, -4.

The two numbers meeting these criteria are -3 and -1, leading to the factorization: \( x^2 - 4x + 3 = (x-3)(x-1) \).Once factored, it's clear that the roots of the quadratic are 3 and 1, completing our factorization of the initial cubic polynomial. By understanding how to factor quadratics, you can effectively solve problems involving polynomials, providing a clear path to uncovering all roots.