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Chapter 2: Problem 79

In Exercises \(75-86\), simplify the expression. $$ \left(7 y^{2}\right)(-3 y) $$

Short Answer

Expert verified
The simplified expression is \( -21y^{3} \).

Step by step solution

01

Multiplication of Coefficients

Start by multiplying the coefficients (numbers in front of the variables) of the two operands, which are 7 and -3. This gives us -21.
02

Multiplication of Variables

Next, multiply the variable parts together. We have \(y^{2}\) from the first expression and \(y\) from the second expression. By the rule of exponents, when multiplying powers of the same base, you can add the exponents. In this case, 2 + 1 = 3. Hence, \( y^{2} \times y = y^{3} \).
03

Put it all together

Finally, combine the produced coefficient and variable to give the simplified expression. Thus, \( (7y^{2})(-3y) = -21y^{3} \).

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Key Concepts

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

Multiplication of Polynomials
When multiplying polynomials, each term in one polynomial is multiplied by every term in the other polynomial. Polynomials are similar to numerical expressions, but they include variables as well. In this case, we're working with the two terms \(7y^2\) and \(-3y\).
For multiplication:
  • Start by multiplying the coefficients: 7 and -3. The result is \(-21\).
  • Multiply the variable portions, taking into account their exponents: \(y^2\) and \(y\).
  • Combine these results to form a single algebraic expression.
Polynomial multiplication requires attention to both coefficient and variable management. We handle them separately first, then bring them together.
Exponent Rules
Exponents signify how many times a number, known as the base, is multiplied by itself. Here, our key rule for simplifying polynomials is when multiplying like bases, we add the exponents.
Let's look at the \(y^2\) and \(y^1\) present in our exercise:
  • \(y^2\) means \(y \times y\).
  • \(y^1\) means just \(y\).
When multiplying these bases:
  • Add their exponents: \(2 + 1 = 3\). Thus, \(y^2 \times y = y^3\).
Understanding exponent rules is crucial because these rules help us manage and simplify terms effectively. Remember, this simplification makes the process of working with algebraic expressions much more efficient.
Algebraic Expressions
Algebraic expressions are combinations of numbers, variables, and operations. Our focus on these allows for representing complex relationships in a simplified manner.
In this problem, consider the expression \((7y^2)(-3y)\). It is made up of:
  • Numeric coefficients: 7 and -3.
  • Variables with exponents: \(y^2\) and \(y\).
After applying our multiplication and exponent rules, we achieve \(-21y^3\), which is a more compact and simplified form of our original expression.
The primary goal when working with algebraic expressions is simplification 鈥� making expressions easier to work with and understand. Grasping the relationships between the components through the use of multiplication and exponent rules simplifies our analysis and provides clear solutions.

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Most popular questions from this chapter

In Exercises 39-42, write an algebraic equation. Do not solve the equation. During a football game, a running back carried the ball 18 times and his average number of yards per carry was \(4.5\). How many yards did the running back gain for the game?

In Exercises 63-68, translate the verbal phrase into an algebraic expression. Simplify the expression. $$ \text { The square of } x \text { decreased by the product of } x \text { and } 2 x $$

In Exercises 17-26, determine whether an algebraic expression or an algebraic equation is given. $$ \frac{5}{6} x+1 $$

In Exercises 33-38, justify each step of the equation. Then identify any properties of equality used to solve the equation. $$ \begin{aligned} \frac{4}{5} x &=-28 \\ \frac{5}{4}\left(\frac{4}{5} x\right) &=\frac{5}{4}(-28) \\ x &=-35 \end{aligned} $$

In Exercises 27-32, solve the equation. $$ 10+x=3 $$
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