Problem 15
Question
\(\left(14 h^{3}-6 h^{2}+12 h\right) \div(-2 h)\)
Step-by-Step Solution
Verified Answer
-7h^{2} + 3h - 6
1Step 1: Rewrite the Expression
Rewrite the given expression as a single fraction: \[ \frac{14h^{3} - 6h^{2} + 12h}{-2h} \]
2Step 2: Distribute the Division
Split the fraction into three separate fractions: \[ \frac{14h^{3}}{-2h} - \frac{6h^{2}}{-2h} + \frac{12h}{-2h} \]
3Step 3: Simplify Each Fraction
Simplify each fraction individually: \[ \frac{14h^{3}}{-2h} = -7h^{2} \] \[ \frac{6h^{2}}{-2h} = -3h \] \[ \frac{12h}{-2h} = -6 \]
4Step 4: Combine Simplified Terms
Combine the simplified fractions to get the final expression: \[ -7h^{2} + 3h - 6 \]
Key Concepts
Understanding Algebraic ExpressionsSimplifying FractionsIntroduction to Polynomials
Understanding Algebraic Expressions
Algebraic expressions are mathematical phrases that can contain numbers, variables, and operators. In our exercise, we see an expression like \(14h^{3}-6h^{2}+12h\). Each part of this expression is called a term.
A term can be a constant (like \(-6\)) or can involve variables (like \(14h^{3}\)).
The power of the variable (the exponent) tells us how many times the variable is multiplied by itself. For instance, \(h^{3}\) means \((h \times h \times h)\). Simplifying algebraic expressions involves combining like terms and reducing where possible.
Understanding how to manipulate these terms is crucial as it forms the basis of algebra.
A term can be a constant (like \(-6\)) or can involve variables (like \(14h^{3}\)).
The power of the variable (the exponent) tells us how many times the variable is multiplied by itself. For instance, \(h^{3}\) means \((h \times h \times h)\). Simplifying algebraic expressions involves combining like terms and reducing where possible.
Understanding how to manipulate these terms is crucial as it forms the basis of algebra.
Simplifying Fractions
Simplifying fractions means making the fraction as simple as possible. In our exercise, this appears in step 2. To simplify, we break down the expression into smaller parts:
\[ \frac{14h^{3} - 6h^{2} + 12h}{-2h} \].
\br> This becomes \[ \frac{14h^{3}}{-2h} - \frac{6h^{2}}{-2h} + \frac{12h}{-2h} \]. Simplifying each fraction separately:
- \(\frac{14h^{3}}{-2h} = -7h^{2}\textrm{ (14 divided by -2, and } h^{3}/h = h^{2})\)
- \(\frac{6h^{2}}{-2h} = -3h \textrm{ (6 divided by -2, and } h^{2}/h = h free final}\)
- \(\frac{12h}{-2h} = -6 \).
Simplifying helps in reducing complicated fractions to manageable numbers.
Besides division, this often involves canceling common factors across the numerator and the denominator. This process makes it easier to work with large or complex fractions.
\[ \frac{14h^{3} - 6h^{2} + 12h}{-2h} \].
\br> This becomes \[ \frac{14h^{3}}{-2h} - \frac{6h^{2}}{-2h} + \frac{12h}{-2h} \]. Simplifying each fraction separately:
- \(\frac{14h^{3}}{-2h} = -7h^{2}\textrm{ (14 divided by -2, and } h^{3}/h = h^{2})\)
- \(\frac{6h^{2}}{-2h} = -3h \textrm{ (6 divided by -2, and } h^{2}/h = h free final}\)
- \(\frac{12h}{-2h} = -6 \).
Simplifying helps in reducing complicated fractions to manageable numbers.
Besides division, this often involves canceling common factors across the numerator and the denominator. This process makes it easier to work with large or complex fractions.
Introduction to Polynomials
A polynomial is a mathematical expression consisting of variables (also known as indeterminates), coefficients, and the operations of addition, subtraction, multiplication, and non-negative integer exponents.
Our exercise involves a polynomial expression \(14h^{3} - 6h^{2} + 12h\). To identify as a polynomial, an expression must exhibit a combination of terms, like those we see here.
Key aspects of polynomials:
Our exercise involves a polynomial expression \(14h^{3} - 6h^{2} + 12h\). To identify as a polynomial, an expression must exhibit a combination of terms, like those we see here.
Key aspects of polynomials:
- They are classified by their degree: the highest power of the variable. In our case, the degree is 3.
- They are composed of terms, each term made up of a coefficient and a variable raised to a power.
- Simplifying polynomials involves combining like terms and distributing operations like addition and subtraction.
This understanding of polynomials is foundational in algebra because they generalize arithmetic operations and are used in more complex calculations.
Other exercises in this chapter
Problem 14
A measurement is \(0.000003 \times 10^{4} \mathrm{~g}\). When this measurement is rewritten in scientific notation, will the exponent be greater or less than 4
View solution Problem 14
\(\left(5 y^{7}\right)\left(6 y^{3}\right)\)
View solution Problem 15
A student simplified \(3 x+8 x^{2}\) as \(11 x^{3}\). Explain why this is not correct.
View solution Problem 15
\(\frac{x^{9}}{x^{2}}\)
View solution