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Chapter 5: Problem 108

Multiply. $$ (2 c)\left(3 c^{2}\right) $$

Short Answer

Expert verified
6c^3

Step by step solution

01

- Multiply the coefficients

Identify the coefficients in each term. Here, the coefficients are 2 and 3. Multiply these coefficients together: \(2 \times 3 = 6\)
02

- Multiply the variables

Identify the variables in each term. Here, the variables are \(c^1\) and \(c^2\). When multiplying variables with the same base, add their exponents: \(c^1 \times c^2 = c^{1+2} = c^3\)
03

- Combine the results

Combine the multiplied coefficients and variables to get the final result: \(6c^3\)

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Key Concepts

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

Coefficients
In algebra, coefficients are the numerical part of terms that include variables. They are the numbers that multiply the variables. For instance, in the term \(3x^2\), the coefficient is 3. Coefficients are essential because they indicate how many times the variable is being multiplied.
In our example, \((2c)(3c^2)\), the coefficients are 2 and 3.
When performing algebraic multiplication, the first step is to multiply these coefficients. Multiplying 2 by 3 gives us 6. This simple multiplication of coefficients sets the stage for combining the rest of the terms.
Exponents
Exponents denote the number of times a variable is multiplied by itself. In the expression \(x^3\), the exponent is 3, meaning \(x\) is multiplied by itself three times: \(x \times x \times x\).
When multiplying terms with the same base variable, we add their exponents. This is because multiplication of exponents corresponds to repeated multiplication of the base.
In our exercise, we have \((2c)(3c^2)\). Notice that \(c\) has an exponent of 1 in the first term and an exponent of 2 in the second term. To multiply the variables, we add their exponents: \(c^1 \times c^2 = c^{1+2} = c^3\). This rule simplifies combining exponential expressions.
Variables
Variables are symbols that stand for unknown values. They are usually represented by letters such as \(x\), \(y\), or in our case, \(c\). Variables form the foundation of algebra since they allow us to write expressions that define relationships and general rules.
In the given problem \((2c)(3c^2)\), our variable is \(c\). Each term contains this variable, making it easier to combine them through multiplication. When variables with the same base are multiplied, their exponents are simply added together.
So, combining \(c^1\) and \(c^2\) gives \(c^{1+2}\) which simplifies to \(c^3\).

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Most popular questions from this chapter

Graph each equation by completing the table of values. \(y=x^{2}-4\) \(\begin{array}{|c|c|}\hline x & {y} \\ \hline-2 & {} \\ \hline-1 & {} \\\ \hline 0 & {} \\ \hline 1 & {} \\ \hline 2 & {} \\ \hline\end{array}\)

Multiply. $$ \frac{1}{2}(4 m-8 n) $$

Perform each division using the "long division" process. $$ \frac{5 y^{4}+5 y^{3}+2 y^{2}-y-8}{y+1} $$

The special product $$ (x+y)(x-y)=x^{2}-y^{2} $$ $$ \text { can be used to perform some multiplication problems. Here are two examples.} $$ $$ \begin{aligned} 51 \times 49 &=(50+1)(50-1) \\ &=50^{2}-1^{2} \\ &=2500-1^{2} \\ &=2499 \end{aligned} \quad | \begin{aligned} 102 \times 98 &=(100+2)(100-2) \\ &=100^{2}-2^{2} \\ &=10,000-4 \\ &=9996 \end{aligned} $$ Once these patterns are recognized, multiplications of this type can be done mentally. Use this method to calculate each product mentally. $$ 20 \frac{1}{2} \times 19 \frac{1}{2} $$

The special product $$ (x+y)(x-y)=x^{2}-y^{2} $$ $$ \text { can be used to perform some multiplication problems. Here are two examples.} $$ $$ \begin{aligned} 51 \times 49 &=(50+1)(50-1) \\ &=50^{2}-1^{2} \\ &=2500-1^{2} \\ &=2499 \end{aligned} \quad | \begin{aligned} 102 \times 98 &=(100+2)(100-2) \\ &=100^{2}-2^{2} \\ &=10,000-4 \\ &=9996 \end{aligned} $$ Once these patterns are recognized, multiplications of this type can be done mentally. Use this method to calculate each product mentally. $$ 101 \times 99 $$
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