Problem 67
Question
Write each polynomial in descending powers of the variable and with no missing powers. See Example 15. $$ 6 x^{5}+x^{3}-3 x+15 $$
Step-by-Step Solution
Verified Answer
The polynomial rearranged in descending order is \( 6x^5 + x^3 - 3x + 15 \).
1Step 1: Identify the Terms of the Polynomial
The given polynomial is \( 6x^5 + x^3 - 3x + 15 \). It consists of four terms: \( 6x^5 \), \( x^3 \), \( -3x \), and \( 15 \).
2Step 2: Order the Terms by Power
The terms need to be reordered so that they appear in descending order of the power of \( x \). We will list them starting from the highest power to the lowest.
3Step 3: Rearrange the Terms
Place the terms in descending order: \( 6x^5 \), \( x^3 \), \( -3x \), and finally \( 15 \). This ensures the polynomial is in the correct order.
4Step 4: Write Down the Ordered Polynomial
The polynomial, arranged in descending powers of \( x \) with no missing terms, is \( 6x^5 + x^3 - 3x + 15 \). No terms are missing, as the polynomial already includes powers \( x^5 \), \( x^3 \), \( x^1 \), and the constant term.
Key Concepts
Descending OrderPolynomial TermsPower of Variables
Descending Order
When working with polynomials, organizing them in descending order is a crucial step. This means arranging the polynomial terms from the highest power of the variable to the lowest.
For example, in the polynomial \(6x^5 + x^3 - 3x + 15\), each term is identified by the power of the variable \(x\). Here, \(6x^5\) has the highest exponent (5), making it the first term. The sequence continues with \(x^3\), then \(-3x\), and finally the constant term \(15\), which can be thought of as \(15x^0\).
Ordering in this fashion not only standardizes the polynomial, making it easier to understand and analyze, but it also facilitates other operations such as addition, subtraction, and differentiation.
For example, in the polynomial \(6x^5 + x^3 - 3x + 15\), each term is identified by the power of the variable \(x\). Here, \(6x^5\) has the highest exponent (5), making it the first term. The sequence continues with \(x^3\), then \(-3x\), and finally the constant term \(15\), which can be thought of as \(15x^0\).
Ordering in this fashion not only standardizes the polynomial, making it easier to understand and analyze, but it also facilitates other operations such as addition, subtraction, and differentiation.
Polynomial Terms
Polynomials are expressions consisting of terms. Each term is composed of a constant (numerical coefficient) and a variable raised to an exponent. A polynomial can have one or more terms.
In the polynomial \(6x^5 + x^3 - 3x + 15\), there are four terms:
In the polynomial \(6x^5 + x^3 - 3x + 15\), there are four terms:
- \(6x^5\) – the coefficient \(6\) is multiplied by the variable \(x\) raised to the power of 5.
- \(x^3\) – a lone \(x\) is raised to the power of 3, with an implicit coefficient of 1.
- \(-3x\) – here, \(-3\) is the coefficient, and \(x\) is the variable raised to the power 1.
- \(15\) – a constant term with a coefficient of 15 and \(x^0\) (since any number to the zero power is 1).
Power of Variables
The power of a variable in a polynomial expression is essentially the exponent associated with that variable. It indicates how many times you multiply the variable by itself.
In the Polynomial \(6x^5 + x^3 - 3x + 15\), each term's power is crucial for understanding its role and position in the polynomial.
In the Polynomial \(6x^5 + x^3 - 3x + 15\), each term's power is crucial for understanding its role and position in the polynomial.
- The term \(6x^5\) specifies that \(x\) is raised to the power of 5.
- \(x^3\) indicates \(x\) to the power of 3.
- \(-3x\) has \(x\) raised to the first power, commonly written just as \(x\).
- The term \(15\) represents a constant term with no \(x\), and it can be seen as \(x^0\) because \(x^0=1\).
Other exercises in this chapter
Problem 66
Multiply. \(\left(m+\frac{2}{9}\right)\left(m-\frac{1}{9}\right)\)
View solution Problem 66
Simplify each expression. $$ (4 y)^{0} $$
View solution Problem 67
Simplify each expression. Write each result using positive exponents only. $$ \frac{\left(-2 x y^{-3}\right)^{-3}}{\left(x y^{-1}\right)^{-1}} $$
View solution Problem 67
Mixed Practice Multiply. $$ (x+y)(4 x-y) $$
View solution