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Chapter 6: Problem 84

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. $$w+5, w^{3}+125$$

Short Answer

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

Step by step solution

01

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 \( w^3 + 125 \).
02

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].
03

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 鈭� 5w + 25`, confirming a is completely divided. Therefore, the first polynomial is a factor of the second polynomial.
04

Factor the Polynomial

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Perform the indicated operations. When possible write down only the answer. $$\frac{5 x-5 y}{x} \cdot \frac{1}{x-y}$$

Perform the indicated operations. When possible write down only the answer. $$\frac{\frac{3 x}{5}}{y}$$

Find the solution set to each equation. $$\frac{x}{5}=\frac{x+2}{3}$$

Find the solution set to each equation. $$\frac{3 w}{3 w-5}=\frac{w}{w+2}$$

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