Problem 11
Question
Make a term-coefficient-degree table like that shown in Example 1 for each polynomial. $$ 8 x^{2}+x-7 $$ (TABLE NOT COPY)
Step-by-Step Solution
Verified Answer
Create a table with terms \(8x^2\), \(x\), \(-7\); coefficients 8, 1, -7; and degrees 2, 1, 0 respectively.
1Step 1: Identify Each Term in the Polynomial
First, we need to identify each term in the polynomial \(8x^2 + x - 7\). The polynomial has three terms: \(8x^2\), \(x\), and \(-7\).
2Step 2: Record Coefficients of Each Term
Next, we determine the coefficient of each term. In \(8x^2\), the coefficient is 8. In \(x\), the coefficient is implicitly 1. In \(-7\), the constant term has a coefficient of -7.
3Step 3: Determine the Degree of Each Term
Now, identify the degree of each term. The degree of \(8x^2\) is 2 (highest power of \(x\)), the degree of \(x\) is 1, and \(-7\) is a constant term with a degree of 0.
4Step 4: Create the Term-Coefficient-Degree Table
Using the information from the previous steps, create a table:| Term | Coefficient | Degree ||---------|-------------|--------|| \(8x^2\) | 8 | 2 || \(x\) | 1 | 1 || \(-7\) | -7 | 0 |This table organizes each term with its respective coefficient and degree.
Key Concepts
Term Identification in PolynomialsUnderstanding CoefficientsDetermining the Degree of a TermCreating a Polynomial Table
Term Identification in Polynomials
Polynomials are algebraic expressions that consist of several terms. To effectively work with polynomials, the first step is to identify each term. A term is a part of the polynomial that can be as simple as a constant number or as complex as a product of numbers and variables raised to a power.
When we look at the polynomial expression \(8x^2 + x - 7\), we can breakdown each component:
When we look at the polynomial expression \(8x^2 + x - 7\), we can breakdown each component:
- \(8x^2\)
- \(x\)
- \(-7\)
Understanding Coefficients
Coefficients are the numerical factors in terms of a polynomial. They multiply the variable or variables in the term.
In the polynomial \(8x^2 + x - 7\), each term has a specific coefficient:
In the polynomial \(8x^2 + x - 7\), each term has a specific coefficient:
- In \(8x^2\), the coefficient is 8.
- In \(x\), even though the number isn't written, the coefficient is 1. This is because \(x\) is the same as \(1x\).
- In the term \(-7\), the whole term is a constant, so it’s technically its own coefficient, which is -7.
Determining the Degree of a Term
The degree of a term is the exponent of the variable within that term. For polynomials, the degree is an important indicator of the polynomial’s behavior and complexity.
Looking again at the polynomial \(8x^2 + x - 7\), the degree of each term is as follows:
Looking again at the polynomial \(8x^2 + x - 7\), the degree of each term is as follows:
- For \(8x^2\), the degree is 2, since the variable is raised to the power of 2.
- For \(x\), the degree is 1 because it can be rewritten as \(x^1\).
- For the constant term \(-7\), the degree is 0, because there is no variable present (or the variable could be considered as \(x^0\), which equals 1).
Creating a Polynomial Table
A polynomial table effectively organizes information about the terms in a polynomial, including their coefficients and degrees. This structure simplifies comparisons and calculations by displaying data in one glance.
When creating a polynomial table for \(8x^2 + x - 7\), insert the terms, their coefficients, and their degrees into a structured list or spreadsheet, like so:
When creating a polynomial table for \(8x^2 + x - 7\), insert the terms, their coefficients, and their degrees into a structured list or spreadsheet, like so:
- Term: \(8x^2\), Coefficient: 8, Degree: 2
- Term: \(x\), Coefficient: 1, Degree: 1
- Term: \(-7\), Coefficient: -7, Degree: 0
Other exercises in this chapter
Problem 11
Complete each solution. $$ \begin{aligned} (2 x+5)(3 x-2) &=2 x(3 x)-\quad(2)+\quad(3 x)-(2) \\ &=6 x^{2}-\quad+10 \\ &=6 x^{2}+\quad-10 \end{aligned} $$
View solution Problem 11
Fill in the blanks to add (subtract) the polynomials. $$ \left(6 x^{2}+2 x+3\right)+\left(4 x^{2}-7 x+1\right) $$ $$ \begin{aligned} &=\left(6 x^{2}+\quad\right
View solution Problem 11
Complete each solution to simplify each expression. \(\left(y^{5} y^{3}\right)^{-5}=(\quad)^{-5}=y \quad=\frac{1}{y}\)
View solution Problem 12
Complete each solution. $$ \text { True or false: } 6 x+4+\frac{-3}{x+2}=6 x+4-\frac{3}{x+2} $$
View solution