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Chapter 4: Problem 65

Find each product. $$ (5-3 x)(4+x) $$

Short Answer

Expert verified
\( (5-3x)(4+x) = -3x^2 - 7x + 20 \)

Step by step solution

01

Distribute each term in the first polynomial

To find the product \( (5-3x)(4+x) \), start by distributing each term in \( 5-3x \) to each term in \( 4+x \).
02

Multiply 5 by each term in the second polynomial

First multiply \( 5 \) by each term in \( 4 + x \): \[ 5 \cdot 4 = 20 \] \[ 5 \cdot x = 5x \]
03

Multiply -3x by each term in the second polynomial

Next, multiply \( -3x \) by each term in \( 4 + x \): \[ -3x \cdot 4 = -12x \] \[ -3x \cdot x = -3x^2 \]
04

Combine like terms

Add all the resulting products together: \[ 20 + 5x - 12x - 3x^2 \] Combine the like terms: \[ -3x^2 + 5x - 12x + 20 = -3x^2 - 7x + 20 \]

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Key Concepts

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

distributive property
In algebra, the distributive property is a core concept. This property tells us how to handle the multiplication of a single term by a polynomial. Essentially, we 'distribute' that single term to every term inside the polynomial. This is crucial when expanding expressions and simplifying equations.
In our exercise \( (5-3x)(4+x) \), we see this property in action. First, we distribute each term in \( 5-3x \) to each term in \( 4+x \).
  • Multiply \(5 \) by \(4\), then by \(x \).
  • Next, multiply \((-3x) \) by \(4 \), then by \(x \).

This gives us progressively simplified steps:
\( 5 \times 4 = 20 \)
\( 5 \times x = 5x \)
\((-3x) \times 4 = -12x \)
\((-3x) \times x = -3x^2 \). The results are then used in combining like terms.
combining like terms
Combining like terms is another fundamental algebraic process. This involves summarizing and condensing an expression by adding or subtracting terms that have the same variable raised to the same power.
In our example: \(20 + 5x -12x -3x^2\), we combine like terms to simplify the expression.
Look for terms with the same variable and exponent:
  • Here, \(5x \) and \(-12x \) are like terms.
  • Combine these to get \(-7x \).

This leaves us with: \(-3x^2 -7x + 20\). Always keep track of positive and negative signs to ensure accuracy.
algebraic expressions
Algebraic expressions consist of constants, variables, and operations (like addition and multiplication). These expressions can represent real-world quantities and are manipulated by using algebraic rules.
Take the initial expression: \( (5 - 3x)(4 + x) \). This starts as a simple product of two binomials. By distributing and combining like terms, we can simplify it to \(-3x^2 - 7x + 20\).
Understanding how to work with these pieces forms the foundation for solving more complex problems. Remember, the skills you build with simpler expressions will help you tackle more intricate algebraic equations in the future.

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Most popular questions from this chapter

Find each product. Recall that \(a^{2}=a \cdot a\) and \(a^{3}=a^{2} \cdot a\). $$ -4 r(3 r+2)(2 r-5) $$

Find each product. Recall that \(a^{2}=a \cdot a\) and \(a^{3}=a^{2} \cdot a\). $$ -2 x^{5}\left(3 x^{2}+2 x-5\right)(4 x+2) $$

Perform each indicated operation. \(\left(16 x^{3}-x^{2}+3 x\right)+\left(-12 x^{3}+3 x^{2}+2 x\right)\)

\(\frac{\left(3 p^{-2} q^{3}\right)^{2}\left(5 p^{-1} q^{-4}\right)^{-1}}{\left(p^{2} q^{-2}\right)^{-3}}\)

Find each product. Recall that \(a^{2}=a \cdot a\) and \(a^{3}=a^{2} \cdot a\). $$ (3 r-2 s)^{4} $$
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