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Chapter 0: Problem 7

\((2 y+3)(y+5)\)

Short Answer

Expert verified
The short answer based on the given step-by-step solution is: When multiplying the binomials \((2y+3)(y+5)\), we use the distributive property (FOIL) and perform the following multiplications: 1. \((2y)(y) = 2y^2\) 2. \((2y)(5) = 10y\) 3. \((3)(y) = 3y\) 4. \((3)(5) = 15\) Then, combine and simplify the terms: \(2y^2 + 10y + 3y + 15 = 2y^2 + 13y + 15\). Therefore, the final expression is \(\boldsymbol{2y^2 + 13y + 15}\).

Step by step solution

01

Identify the terms in each binomial

In the expression \((2y+3)(y+5)\), we have two binomials. Identify the terms in each binomial: - First binomial: \(2y\) and \(3\) - Second binomial: \(y\) and \(5\)
02

Use the distributive property (FOIL)

Now, we will use the distributive property (FOIL) to multiply both terms in the first binomial by both terms in the second binomial. FOIL stands for First, Outer, Inner, Last. 1. First terms: \((2y)(y)\) 2. Outer terms: \((2y)(5)\) 3. Inner terms: \((3)(y)\) 4. Last terms: \((3)(5)\)
03

Perform the multiplications

Compute the product for each pair of terms from Step 2: 1. \((2y)(y) = 2y^2\) 2. \((2y)(5) = 10y\) 3. \((3)(y) = 3y\) 4. \((3)(5) = 15\)
04

Combine the results

Now, combine the product terms obtained in Step 3 to form the final expression: \(2y^2 + 10y + 3y + 15\)
05

Simplify the expression

Combine like terms to simplify the expression: \(2y^2 + (10y + 3y) + 15\) \(2y^2 + 13y + 15\) The simplified expression is \(\boldsymbol{2y^2 + 13y + 15}\).

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Key Concepts

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

Distributive Property
The distributive property plays a crucial role in multiplying binomials like \((2y+3)(y+5)\). It essentially allows you to simplify expressions by distributing terms across parentheses.

Here's how it works:
  • When you have terms in parentheses, you can distribute each term in the first parenthesis across each term in the second parenthesis.
  • This step is essential because it breaks down complex expressions into more manageable parts.
For example, multiplying the binomials: \((2y+3)(y+5)\) requires distributing the terms \(2y\) and \(3\) across \(y\) and \(5\).

Once you understand this property, you'll find it easier to handle any polynomial multiplication task by breaking it down into simpler steps.
FOIL Method
The FOIL method is a specialized application of the distributive property, particularly for binomials. FOIL stands for First, Outer, Inner, and Last.

In the expression \((2y+3)(y+5)\):
  • First: Multiply the first terms from both binomials, \((2y)(y) = 2y^2\).
  • Outer: Multiply the outer terms, \((2y)(5) = 10y\).
  • Inner: Multiply the inner terms, \((3)(y) = 3y\).
  • Last: Multiply the last terms, \((3)(5) = 15\).
These steps enable you to cover all possible combinations of terms in the binomials, ensuring no term is left out.

Practicing the FOIL method can greatly enhance your ability to solve similar problems quickly.
Simplifying Expressions
Simplifying expressions involves combining like terms to create a more concise expression. This is often the final step when multiplying binomials.

In our previous example, after using the FOIL method, we obtained:\(2y^2 + 10y + 3y + 15\).
  • The goal now is to combine the terms containing the same variables to simplify the expression.
  • Combine the \(y\) terms: \(10y + 3y = 13y\).
The simplified expression, therefore, becomes \(2y^2 + 13y + 15\).

Simplifying an expression not only makes it easier to understand but also prepares it for further operations or evaluations.
Polynomial Arithmetic
Polynomial arithmetic refers to operations involving polynomials, including addition, subtraction, multiplication, and division.

When dealing with multiplication, such as our binomial example, you are performing polynomial arithmetic.
  • This involves distributing terms and simplifying expressions, as seen with the distributive property and FOIL method.
  • Understanding polynomial arithmetic helps you work with expressions containing variables raised to different powers.
By practicing these operations, you build foundational skills for more advanced algebraic concepts.

Mastering polynomial arithmetic is essential as it frequently appears in many areas of algebra and calculus.

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Most popular questions from this chapter

Convert each expression in Exercises into its technology formula equivalent as in the table in the text.\(2.1 x^{-3}-x^{-1}+\frac{x^{2}-3}{2}\)

Find all possible real solutions of each equation \(3 x^{6}-x^{4}-12 x^{2}+4=0\)

Convert each expression in Exercises into its technology formula equivalent as in the table in the text.\(2^{2 x^{2}-x+1}\)

\(\sqrt{(x+1)^{3}}+\sqrt{(x+1)^{5}}\)

\(2 x^{2}-3 x-2\)
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