Problem 84 - Algebra for College Students

Question

For each pair of polynomials, use division to determine whether the first polynomial is a factor of the second. Use synthetic division when possible. If the first polynomial is a factor, then factor the second polynomial. See Example 7. $$w+5, w^{3}+125$$

Step-by-Step Solution

Answer

The first polynomial is a factor of the second. The factorization is \((w + 5)(w^2 - 5w + 25)\).

Step 1: Identify the Polynomials

The first polynomial is given as \( w + 5 \) and the second polynomial is \( w^3 + 125 \). We need to use division to determine if \( w + 5 \) is a factor of \( w^3 + 125 \).

Step 2: Set Up Synthetic Division

To use synthetic division, we first identify the root of the divisor. For \( w + 5 \), the root is \( -5 \). Set up the coefficients of the dividend \( w^3 + 125 \) which are [1, 0, 0, 125].

Step 3: Perform Synthetic Division

Synthetic division setup with root \( -5 \) and coefficients [1, 0, 0, 125]:\begin{array}{r|rrrr}-5 & 1 & 0 & 0 & 125 \text{} & \text{} & -5 & 25 & -125 \text{}\text{} & 1 & -5 & 25 & 0 \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}& \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text?\text?\text?\text?\text?\text} & 1 & -5 & 25 & 0 \text?\text?\text?\text?\text?\text?\text?\text?\text?\text?\text} \text?\text?\text?\text?\text}\text?\text} \text}\text?\text? Remove the leading zeros from the partial quotient: Row 3: -5 \ 1 [-5 25]} The example's final quotient is `w^2 − 5w + 25`, confirming a is completely divided. Therefore, the first polynomial is a factor of the second polynomial.

Step 4: Factor the Polynomial

Since the quotient obtained is \( w^2 - 5w + 25 \) with a remainder of 0, the factorization of the original polynomial is \((w + 5)(w^2 - 5w + 25)\).

Key Concepts

Synthetic Division

Synthetic division is a simplified form of polynomial division. It's primarily used when dividing by linear polynomials like \(x - c\). Instead of the traditional long division, synthetic division streamlines the process and reduces the number of computations.First, identify the root of the divisor polynomial. For example, if dividing by \(w + 5\), the root is \(-5\). Write down the coefficients of the polynomial you're dividing. For \(w^3 + 125\), list them as [1, 0, 0, 125].Next, set up the synthetic division format: the root on the left and the coefficients on the right. Here's the layout:\(-5\) | 1 0 0 125Perform synthetic division step-by-step:

• Bring down the first coefficient to the bottom row.
• Multiply the root by this value and place the result under the next coefficient.
• Add the values in the column, write the sum below.
• Repeat these steps for all coefficients.

The final row gives the coefficients of the quotient polynomial. If the last number is 0, there's no remainder, confirming that the divisor is a factor. In our example, the final quotient is \(w^2 - 5w + 25\), showing \(w + 5\) is a factor of \(w^3 + 125\).

Factorization

Factorization is the process of breaking down a polynomial into simpler polynomials that, when multiplied together, give the original polynomial. Identifying factors can simplify polynomial problems.In our exercise, after using synthetic division, we derived the quotient polynomial \(w^2 - 5w + 25\). This quotient combined with the divisor \(w + 5\) gives the factorization:\((w + 5)(w^2 - 5w + 25)\).When verifying polynomial factorization:

• Always represent the polynomial correctly.
• Ensure the synthetic division result has a remainder of zero.
• Combine each factor properly.

Factorization simplifies solving equations, graphing functions, and finding roots (solutions). For the polynomial \(w^3 + 125\), we recognize that it can also be expressed as \((w)^3 + (5)^3\), a sum of cubes that comfortably fits into factorization patterns.

Polynomials

Polynomials are algebraic expressions made up of variables and coefficients, involving terms like constants or powers of a variable. These terms are combined using addition, subtraction, and multiplication.In our problem, the polynomials dealt with were \(w + 5\) and \(w^3 + 125\). Understanding polynomials is crucial because they are fundamental in algebra and used extensively in calculus, geometry, and applied mathematics.Key polynomial concepts:

• Degree: The highest power of the variable in the polynomial.
• Standard form: Writing the polynomial with terms in descending order of degree.
• Leading coefficient: The coefficient of the term with the highest degree.

For our example, \(w^3 + 125\):

• The degree is 3.
• It's written in standard form since the terms descend by power.
• The leading coefficient is 1 (the coefficient of \(w^3\)).

Recognizing these attributes aids in operations such as division and factorization. Thoroughly understanding the nature of polynomials prepares students for more advanced mathematical concepts, ensuring a solid mathematical foundation.