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Chapter 2: Problem 4

Expand and combine like terms. $$ (2 a+3)(3 a-2) $$

Short Answer

Expert verified
Answer: The expanded form of the expression is \(6a^2 + 5a - 6\).

Step by step solution

01

Use the FOIL method or distributive property to expand the expression

To expand the expression \((2a+3)(3a-2)\), we will use the FOIL method. The FOIL method stands for First, Outer, Inner, Last, meaning that you multiply the first terms in each bracket, then the outer terms, then the inner terms, and finally the last terms in each bracket. It can be visualized as follows: First: \((2a)(3a)\) Outer: \((2a)(-2)\) Inner: \((3)(3a)\) Last: \((3)(-2)\) Now, multiply each pair: First: \((2a)(3a) = 6a^2\) Outer: \((2a)(-2) = -4a\) Inner: \((3)(3a) = 9a\) Last: \((3)(-2) = -6\)
02

Combine like terms

We now combine any like terms: $$ 6a^2 - 4a + 9a - 6 $$ Combine the \(-4a\) and \(9a\) terms: $$ 6a^2 + 5a - 6 $$ So the final expanded expression is: $$ (2a+3)(3a-2) = 6a^2 + 5a - 6 $$

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Key Concepts

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

The FOIL Method in Algebra
The FOIL method is a powerful tool for expanding two binomials. It's a step-by-step approach that helps you systematically tackle the multiplication problem. Here's how it works. FOIL stands for:
  • First: Multiply the first terms of each binomial. For our example, multiply the first term of each binomial: \((2a)(3a) = 6a^2\).
  • Outer: Multiply the outer terms. In this case, \((2a)\) and \((-2)\) are the outer terms, giving us \(-4a\).
  • Inner: Multiply the inner terms. These are \((3)\) and \((3a)\), so we have \(9a\).
  • Last: Finally, multiply the last terms \((3)\) and \((-2)\), which equals \(-6\).
Once each multiplication is complete, you'll have four parts to combine. This method keeps things organized and ensures you don't miss any terms! It's especially useful for beginners learning to expand polynomials efficiently.
Get comfortable with FOIL and you will find binomials less daunting.
Understanding the Distributive Property
The distributive property is fundamental in algebra and acts like a guiding principle for expansion. It states that for any numbers or terms, \(a(b + c) = ab + ac\). This principle helps us break down complex expressions and distribute terms efficiently. In the example \((2a+3)(3a-2)\), you can distribute each part of the first binomial across the second binomial:
  • Distribute \((2a)\): Multiply \((2a)\) by each term in the second binomial, resulting in \(6a^2\) and \(-4a\).
  • Distribute \((3)\): Likewise, multiply \((3)\) by each term in the second binomial, which gives \(9a\) and \(-6\).
The distributive property ensures each term is correctly combined, allowing you to expand a product into a sum of terms. It's less about memorization and more about understanding that each term multiplies across all other terms.
Combining Like Terms
Combining like terms is a simple yet crucial step in simplifying expressions. Like terms are terms that include the same variable raised to the same power. They can be added or subtracted easily once identified. In the expression from our exercise \(6a^2 - 4a + 9a - 6\), we have:
  • \(-4a + 9a\): These are like terms because both have the variable \(a\). By adding them together, we get \(5a\).
After simplifying, our final expression is \(6a^2 + 5a - 6\).
This process can seem trivial but is important for streamlining algebraic expressions. It allows you to see the expression's true form and prepare it for further operations or evaluations. Always look for like terms when simplifying!

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Most popular questions from this chapter

Factor the expressions in exercises. $$ x^{2}-19 x+90 $$

Simplify each expression. Assume any factors you cancel are not zero. $$ \frac{\frac{4 a b^{3}}{3}}{\frac{2 b}{a^{2}}} $$

A contractor is managing three different job sites. It costs her $$\$ c$$ to employ a carpenter, $$\$ p$$ to employ a plumber, and $$\$ e$$ to employ an electrician. The total cost to employ carpenters, plumbers, and electricians at each site is Cost at site \(1=12 c+2 p+4 e\) Cost at site \(2=14 c+5 p+3 e\) Cost at site \(3=17 c+p+5 e\). Write expressions in terms of \(c, p,\) and \(e\) for: (a) The total employment cost for all three sites. (b) The difference between the employment cost at site 1 and site 3 . (c) The amount remaining in the contractor's budget after accounting for the employment cost at all three sites, given that originally the budget is \(50 c+10 p+20 e\)

Simplify each expression. Assume any factors you cancel are not zero. $$ \frac{\frac{1}{c^{2}}-\frac{1}{d^{2}}}{\frac{d-c}{c^{2} d}} $$

Which of the following expressions is equivalent to \(3\left(x^{2}+2\right)-3 x(1-x) ?\) (i) \(6+3 x\) (ii) \(-3 x+6 x^{2}+6\) (iii) \(3 x^{2}+6-3 x-3 x^{2}\) (iv) \(3 x^{2}+6-3 x\)
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