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Chapter 3: Problem 3

For Problems \(1-36\), find each product. $$ \left(-2 x^{2}\right)\left(6 x^{3}\right) $$

Short Answer

Expert verified
The product is \\(-12x^5\\).

Step by step solution

01

Multiply the Coefficients

First, identify the coefficients in each term of the product. The coefficients for the terms \(-2x^2\) and \(6x^3\) are \(-2\) and \(6\), respectively. Multiply these coefficients to get the coefficient of the product: \(-2 \times 6 = -12\).
02

Multiply the Variables

Next, consider the variables \(x\). We have \(x^2\) in the first term and \(x^3\) in the second term. When multiplying variables with the same base, add their exponents: \(x^2 \times x^3 = x^{2+3} = x^5\).
03

Combine the Results

Now combine the results from Step 1 and Step 2. The product of the coefficients \(-12\) and the product of the variables \(x^5\) gives the final product: \(-12x^5\).

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Key Concepts

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

Coefficients
Coefficients are the numbers that appear in front of variables in algebraic expressions. They tell us how much of a variable we have, serving as a kind of multiplier. For example, in the term \(-2x^2\), the coefficient is \(-2\). This means we have \(-2\) times the base variable raised to some power. Similarly, in \(6x^3\), the coefficient is \(6\). When multiplying polynomials, it's important to focus on these coefficients first.
  • Identify coefficients: For terms like \(-2x^2\) and \(6x^3\), the coefficients are \(-2\) and \(6\).
  • Multiply coefficients: Simply multiply these numbers together. Thus, \(-2 \times 6 = -12\).
Understanding coefficients can simplify solving math problems involving polynomials. They lay the foundation before considering variables and their powers.
Exponents
Exponents are the numbers placed above and to the right of a base number, indicating how many times to multiply the base number by itself. In algebraic expressions like \(x^2\) and \(x^3\), the "2" and "3" are exponents. They show the number of times the base \(x\) is used as a factor. Multiplying such terms involves a specific rule:
  • Same base rule: When multiplying terms with the same base, add the exponents together.
  • Example application: For \(x^2\) and \(x^3\), you add the exponents: \(2 + 3 = 5\).
This results in \(x^5\) for the product of \(x^2\) and \(x^3\). Recognizing and using this rule can greatly help when simplifying polynomial expressions.
Algebraic Expressions
Algebraic expressions are combinations of numbers, variables, and operators like addition or subtraction. They form the backbone of many algebra problems. In the context of polynomial multiplication, understanding how to work with these expressions is crucial.
  • Components: Includes terms like \(-2x^2\) and \(6x^3\), which can consist of coefficients, bases, and exponents.
  • Operations: When multiplying expressions, you handle coefficients and variables individually before combining them back.
The resultant product, such as \(-12x^5\) from multiplying \(-2x^2\) and \(6x^3\), is an algebraic expression in itself. This indicates the power of algebra in breaking down complex operations into manageable steps.

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Most popular questions from this chapter

Consider the following two solutions for the equation \((x+3)(x-4)=(x+3)(2 x-1)\) Solution A $$ \begin{aligned} &(x+3)(x-4)=(x+3)(2 x-1) \\ &(x+3)(x-4)-(x+3)(2 x-1)=0 \\ &(x+3)[x-4-(2 x-1)]=0 \\ &(x+3)(x-4-2 x+1)=0 \\ &(x+3)(-x-3)=0 \end{aligned} $$ $$ \begin{aligned} x+3 &=0 & & \text { or } & &-x-3=0 \\ x &=-3 & & \text { or } & &-x=3 \\ x &=-3 & & \text { or } & & x=-3 \end{aligned} $$ The solution set is \(\\{-3\\}\). Solution B $$ \begin{aligned} (x+3)(x-4) &=(x+3)(2 x-1) \\ x^{2}-x-12 &=2 x^{2}+5 x-3 \\ 0 &=x^{2}+6 x+9 \\ 0 &=(x+3)^{2} \\ x+3 &=0 \\ x &=-3 \end{aligned} $$ The solution set is \(\\{-3\\}\). Are both approaches correct? Which approach would you use, and why?

For Problems \(1-54\), solve each equation. You will need to use the factoring techniques that we discussed throughout this chapter. (Objective 1) $$ x^{4}-9 x^{2}=0 $$

How can you determine that \(x^{2}+5 x+12\) is not factorable using integers?

For Problems \(1-54\), solve each equation. You will need to use the factoring techniques that we discussed throughout this chapter. (Objective 1) $$ 3 x^{2}-46 x-32=0 $$

For Problems \(1-54\), solve each equation. You will need to use the factoring techniques that we discussed throughout this chapter. (Objective 1) $$ 9 x^{4}-37 x^{2}+4=0 $$
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