Problem 15
Question
In Exercises 15–58, find each product. $$(x+1)\left(x^{2}-x+1\right)$$
Step-by-Step Solution
Verified Answer
The product of \((x+1)\) and \(\left(x^{2}-x+1\right)\) is \(x^{3}+1\)
1Step 1: Distribution Step
Multiply each term in polynomial \(x+1\) with each term in polynomial \((x^{2}-x+1)\). So, the equation transforms into: \(x(x^{2}-x+1) + 1(x^{2}-x+1)\)
2Step 2: Opening the Brackets
Open the brackets to get: \(x^{3}-x^{2}+x + x^{2}-x+1\)
3Step 3: Combining Like Terms
Combine the like terms, that is those terms containing \(x^{2}\) and \(x\), to simplify the expression: \(x^{3} - x^{2}+ x^{2} + x - x + 1 \) which simplifies to \(x^{3} + 1\)
Key Concepts
Distributive PropertyCombining Like TermsAlgebraic Expressions
Distributive Property
The distributive property is at the heart of multiplying polynomials. It allows us to break down a seemingly complex multiplication problem into smaller, more manageable pieces. This property states that multiplying a sum by a number is the same as multiplying each addend individually by the number and then adding the results. In the context of the given exercise, when you have \[ (x+1)(x^2-x+1) \] the distributive property helps by allowing the multiplication of each term in \(x+1\) by each term in \(x^2-x+1\). Here's a more detailed breakdown:
- Multiply \(x\) by each of the terms in \((x^2-x+1)\).
- Then, multiply \(1\) by each of the terms in \((x^2-x+1)\).
Combining Like Terms
Once you have expanded the polynomial using the distributive property, you'll often find several 'like terms'. 'Like terms' are terms whose variables and their exponents are the same. Combining these terms is a crucial step in simplifying your expression.After using the distributive property on \[ x(x^2-x+1) + 1(x^2-x+1) \] , you end up with an expression:
- \(x^3 - x^2 + x + x^2 - x + 1\)
- Combine \(-x^2\) and \(+x^2\) to cancel out.
- Combine \(+x\) and \(-x\) to cancel out.
Algebraic Expressions
Algebraic expressions, like the one in the given exercise \[ (x+1)(x^2-x+1) \] , are mathematical phrases that consist of variables, coefficients, and operators (like addition, subtraction, and multiplication). Understanding these elements is key to mastering polynomial multiplication.
Key Components of Algebraic Expressions:
- Variables: In our example, \(x\) is the variable. Variables stand for unknown or changeable values.
- Coefficients: These are numbers that multiply the variables. In the term \(-x^2\), the coefficient is \(-1\).
- Operators: Signs like \(+\) and \(-\) are operators that show the relationships between terms.
Other exercises in this chapter
Problem 15
Multiply or divide as indicated. $$\frac{x-2}{3 x+9} \cdot \frac{2 x+6}{2 x-4}$$
View solution Problem 15
$$\text { Factor by grouping.}$$ $$3 x^{3}-2 x^{2}-6 x+4$$
View solution Problem 15
Use the product rule to simplify the expressions in Exercises \(13-22\) In Exercises \(17-22,\) assume that variables represent nonnegative real Numbers. $$\sqr
View solution Problem 16
Evaluate each algebraic expression for the given value or values of the variable(s). $$\frac{2 x+y}{x y-2 x}, \text { for } x=-2 \text { and } y=4$$
View solution