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Chapter 3: Problem 3

Find each product. $$\left(-2 x^{2}\right)\left(6 x^{3}\right)$$

Short Answer

Expert verified
-12x^5

Step by step solution

01

Multiply the coefficients

First, identify the coefficients in each expression. In \(-2 x^2\), the coefficient is -2, and in \(6 x^3\), the coefficient is 6. Multiply these coefficients: \(-2 imes 6 = -12\).
02

Multiply the variables

Next, multiply the variable parts. The expression \(-2 x^2\) has \(x^2\), and \(6 x^3\) has \(x^3\). Using the laws of exponents, you add the exponents when multiplying like bases: \(x^2 imes x^3 = x^{2+3} = x^5\).
03

Combine the results

Combine the results from Step 1 and Step 2: Multiply the coefficient \(-12\) by the variable part \(x^5\), to get the final product: \(-12 x^5\).

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Key Concepts

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

Coefficients
Coefficients are the numerical part of terms in an algebraic expression. They tell us how many units of the variable are in the expression. In our given exercise, we started with two terms:
  • \(-2x^2\) with a coefficient of \(-2\)
  • \(6x^3\) with a coefficient of \(6\)
To find the product of these terms, we multiply the coefficients together: \(-2 \times 6 = -12\). This is the first and crucial step in polynomial multiplication. By understanding coefficients, you can simplify complex algebraic expressions and work efficiently through polynomial operations. Remember to always keep your negative signs, as these impact the final result significantly.
Laws of Exponents
The laws of exponents are a set of rules that make simplifying expressions with exponents easier. One of the most common rules applies when multiplying two powers with the same base. In such cases, you keep the base and add the exponents together.The generic formula is:
  • \(a^m \times a^n = a^{m+n}\)
For the exercise at hand, we used these laws to multiply the terms \(x^2\) and \(x^3\):
  • \(x^2 \times x^3 = x^{2+3} = x^5\)
This step ensures that the variable and exponent are correctly handled in the expression, helping to maintain mathematical accuracy and integrity. Applying the laws of exponents simplifies expressions and makes data handling more manageable.
Variable Multiplication
Variable multiplication involves multiplying terms that contain the same variables. Each variable is raised to a power represented by an exponent. When multiplying these variables, the exponent laws help to streamline the process. In the given exercise, once the coefficients were multiplied, the variables were the focus:
  • The expression \(-2x^2\) has \(x^2\)
  • The expression \(6x^3\) has \(x^3\)
We then apply the rule of adding exponents when multiplying expressions with the same base:
  • \(x^2 \times x^3 = x^{2+3} = x^5\)
By multiplying variables effectively, we derive the correct resultant expression. Understanding variable multiplication is essential in mathematics, as it allows for the simplification and solution of equations that would otherwise be difficult to manage.

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Most popular questions from this chapter

Solve each equation. You will need to use the factoring techniques that we discussed throughout this chapter. $$15 x^{2}+34 x+15=0$$

Should help you pull together all of the factoring techniques of this chapter. Factor completely each polynomial, and indicate any that are not factorable using integers. $$x^{4}+6 x^{2}+9$$

Factor each trinomial and assume that all variables that appear as exponents represent positive integers. $$x^{2 a}+2 x^{a}-24$$

Consider the following approach to factoring \(12 x^{2}+54 x+60\) $$ \begin{aligned} 12 x^{2}+54 x+60 &=(3 x+6)(4 x+10) \\ &=3(x+2)(2)(2 x+5) \\ &=6(x+2)(2 x+5) \end{aligned} $$ Is this a correct factoring process? Do you have any suggestion for the person using this approach?

Factor each trinomial and assume that all variables that appear as exponents represent positive integers. $$4 x^{2 a}+20 x^{a}+25$$
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