ORCCA Dividing by a Monomial

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Section 1.1 Dividing by a Monomial

Objectives: PCC Course Content and Outcome Guide

MTH 65 CCOG 1.c

We learned how to add and subtract polynomials in [cross-reference to target(s) "section-adding-and-subtracting-polynomials" missing or not unique]. Then in [cross-reference to target(s) "section-introduction-to-exponent-rules" missing or not unique], we learned how to multiply monomials together (but not yet how to multiply general polynomials together). In this section we learn how to divide a general polynomial by a monomial.

Figure 1.1.1. Alternative Video Lesson

Subsection 1.1.1 Quotient of Powers Rule

When we multiply the same base raised to powers, we add the exponents, as in \(2^{2}\cdot2^{3}=2^{5}\text{.}\) What happens when we divide the same base raised to powers?

Example 1.1.2.

Simplify \(\frac{x^5}{x^2}\) by first writing out what each power means.

Without knowing a rule for simplifying this quotient of powers, we can write the expressions without exponents and simplify.

\begin{align*} \frac{x^5}{x^2} \amp= \frac{x \cdot x \cdot x \cdot x \cdot x}{x \cdot x}\\ \amp= \frac{\cancel{x} \cdot \cancel{x} \cdot x \cdot x \cdot x}{\cancel{x} \cdot \cancel{x} \cdot 1}\\ \amp= \frac{x \cdot x \cdot x}{1}\\ \amp= x^3 \end{align*}

Notice that the difference of the exponents of the numerator and the denominator (\(5\) and \(2\text{,}\) respectively) is \(3\text{,}\) which is the exponent of the simplified expression.

When we divide as we've just done, we end up canceling factors from the numerator and denominator one-for-one. These common factors cancel to give us factors of \(1\text{.}\) The general rule for this is:

Fact 1.1.3. Quotient of Powers Rule.

For any non-zero real number \(a\) and integers \(m\) and \(n\) where \(m\gt n\text{,}\)

\begin{equation*} \frac{a^{m}}{a^{n}} = a^{m-n} \end{equation*}

This rule says that when you're dividing two expressions that have the same base, you can simplify the quotient by subtracting the exponents. In Example 1.1.2, this means that we can directly compute \(\frac{x^5}{x^2}\text{:}\)

\begin{align*} \frac{x^5}{x^2} \amp= x^{5-2}\\ \amp=x^3 \end{align*}

Now we can update the list of exponent rules from [cross-reference to target(s) "section-introduction-to-exponent-rules" missing or not unique].

List 1.1.4. Summary of the Rules of Exponents (Thus Far)

If \(a\) and \(b\) are real numbers, and \(m\) and \(n\) are positive integers, then we have the following rules:

Product Rule: \(\displaystyle a^{m} \cdot a^{n} = a^{m+n}\)
Power to a Power Rule: \(\displaystyle (a^{m})^{n} = a^{m\cdot n}\)
Product to a Power Rule: \(\displaystyle (ab)^{m} = a^{m} \cdot b^{m}\)
Quotient of Powers Rule: \(\frac{a^m}{a^n} = a^{m-n}\) (when \(m\gt n\))

Subsection 1.1.2 Dividing a Polynomial by a Monomial

Recall that dividing by a number \(c\) is the same as multiplying by the reciprocal \(\frac{1}{c}\text{.}\) For example, whether you divide \(8\) by \(2\) or multiply \(8\) by \(\frac{1}{2}\text{,}\) the result is \(4\) either way. In symbols,

\begin{equation*} \frac{8}{2}=\frac{1}{2}\cdot8\qquad\text{(both work out to }4\text{)} \end{equation*}

If we apply this idea to a polynomial being divided by a monomial, say with \(\frac{a+b}{c}\text{,}\) we can see that the distributive law works for this kind of division as well as with multiplication:

\begin{align*} \frac{a+b}{c} \amp= \frac{1}{c}\cdot(a+b)\\ \amp= \frac{1}{c}\cdot a + \frac{1}{c}\cdot b\\ \amp= \frac{a}{c} + \frac{b}{c} \end{align*}

In the end, the \(c\) has been “distributed” into the \(a\) and the \(b\text{.}\) Once we recognize that division by a monomial is distributive, we are left with individual monomial pairs that we can divide.

Example 1.1.5.

Simplify \(\dfrac{2x^3+4x^2-10x}{2}\text{.}\)

We recognize that the \(2\) we're dividing by can be divided into each and every term of the numerator. Once we recognize that, we will simply perform those divisions.

\begin{align*} \frac{2x^3+4x^2-10x}{2}\amp=\frac{2x^3}{2}+\frac{4x^2}{2}+\frac{-10x}{2}\\ \amp=x^3+2x^2-5x \end{align*}

Example 1.1.6.

Simplify \(\dfrac{15x^4-9x^3+12x^2}{3x^2}\text{.}\)

We recognize that each term in the numerator can be divided by \(3x^2\text{.}\) To actually carry out that division we'll need to use the Quotient of Powers Rule. This is going to cause a change in each coefficient and exponent.

\begin{align*} \frac{15x^4-9x^3+12x^2}{3x^2}\amp=\frac{15x^4}{3x^2}+\frac{-9x^3}{3x^2}+\frac{12x^2}{3x^2}\\ \amp=5x^2-3x+4 \end{align*}

Remark 1.1.7.

Once you become comfortable with this process, you might leave out the step where we wrote out the distribution. You will do the distribution in your head and this will become a one-step exercise. Here's how Example 1.1.6 would be visualized:

\begin{equation*} \frac{15x^4-9x^3+12x^2}{3x^2}=\boxed{\phantom{5}}x^{\boxed{\phantom{2}}} - \boxed{\phantom{3}}x^{\boxed{\phantom{1}}} + \boxed{\phantom{4}}x^{\boxed{\phantom{0}}} \end{equation*}

And when calculated, we'd get:

\begin{equation*} \frac{15x^4-9x^3+12x^2}{3x^2}=5x^2-3x+4 \end{equation*}

(With the last term, note that \(\frac{x^2}{x^2}\) reduces to \(1\text{.}\))

Example 1.1.8.

Simplify \(\dfrac{20x^3y^4+30x^2y^3-5x^2y^2}{-5xy^2}\text{.}\)

\begin{align*} \frac{20x^3y^4+30x^2y^3-5x^2y^2}{-5xy^2}\amp= \frac{20x^3y^4}{-5xy^2}+\frac{30x^2y^3}{-5xy^2}+\frac{-5x^2y^2}{-5xy^2}\\ \amp= -4x^2y^2 -6xy+x \end{align*}

Checkpoint 1.1.9.

Identify a variable you might use to represent each quantity. Then identify what units would be most appropriate.

Let be the age of a student, measured in .
Let be the amount of time passed since a driver left Portland, Oregon, bound for Boise, Idaho, measured in .
Let be the area of a two-bedroom apartment, measured in .

The unknown quantity is age, which we generally measure in years. So we could say:

“Let \(a\) be the age of a student, measured in years.”
The amount of time passed is the unknown quantity. Since this is a drive from Portland to Boise, it would make sense to measure this in hours. So we could say:

“Let \(t\) be the amount of time passed since a driver left Portland, Oregon, bound for Boise, Idaho, measured in hours.”
The unknown quantity is area. Apartment area is usually measured in square feet. So we’ll say:

“Let \(A\) be the area of a two-bedroom apartment, measured in \(\text{ft}^2\text{.}\)”

Example 1.1.10.

The density of an object, \(\rho\) (pronounced “rho”), can be calculated by the formula

\begin{equation*} \rho=\frac{m}{V} \end{equation*}

where \(m\) is the object's mass, and \(V\) is its volume. The mass of a certain cancerous growth can be modeled by \(4t^3-6t^2+8t\) grams, where \(t\) is the number of days since the growth began. If its volume is \(2t\) cubic centimeters, find the growth's density.

We have:

\begin{align*} \rho\amp=\frac{m}{V}\\ \amp=\frac{4t^3-6t^2+8t}{2t}\,\frac{\text{g}}{\text{cm}^3}\\ \amp=\frac{4t^3}{2t}-\frac{6t^2}{2t}+\frac{8t}{2t}\quad\frac{\text{g}}{\text{cm}^3}\\ \amp=2t^2-3t+4\quad\frac{\text{g}}{\text{cm}^3} \end{align*}

The growth's density can be modeled by \(2t^2-3t+4\) ^g⁄_cm³.

Reading Questions 1.1.3 Reading Questions

1.

How is dividing a polynomial by a monomial similar to distributing multiplication over a polynomial? For example, how is the process of simplifying \(\frac{15x^3+5x^2+10x}{5x}\) similar to simplifying \(5x\left(15x^3+5x^2+10x\right)\text{?}\)

Exercises 1.1.4 Exercises

Quotient of Powers Rule

1.

If a temperature is \(50{^\circ}\mathrm{F}\text{,}\) what is that temperature measured in Celsius?
If a temperature is \(-20{^\circ}\mathrm{F}\text{,}\) what is that temperature measured in Celsius?

\(10\ {\rm degC}\)

\(-28.8889\ {\rm degC}\)

\(\begin{aligned}[t] \frac{5}{9}(F - 32)\amp=\frac{5}{9}(\substitute{50} - 32)\\ \amp=\frac{5}{9}(18)\\ \amp=\frac{5}{1}(2)\\ \amp=10 \end{aligned}\)

So \(50{^\circ}\mathrm{F}\) is equavalent to \(10{^\circ}\mathrm{C}\text{.}\)
\(\begin{aligned}[t] \frac{5}{9}(F - 32)\amp=\frac{5}{9}(\substitute{-20} - 32)\\ \amp=\frac{5}{9}(-52)\\ \amp=-\frac{260}{9}\\ \amp\approx-28.89 \end{aligned}\)

So \(-20{^\circ}\mathrm{F}\) is equavalent to about \(-28.89{^\circ}\mathrm{C}\text{.}\)

2.

We can use the expression \(\frac{p}{100}(220-a)\) to represent a person’s target heart rate when their target rate is \(p\%\) of their maximum heart rate, and they are \(a\) years old.

Determine the target heart rate at the \(53\%\) level for moderate exercise for someone who is \(56\) years old.

At the \(53\%\) level, the target heart rate for moderate exercise for someone who is \(56\) years old is beats per minute.

\(86.92\)

The expression is \(\frac{p}{100}(220-a)\text{,}\) and we must sustitute in \(53\) for \(p\) and \(56\) for \(a\text{.}\)

\begin{equation*} \begin{aligned} \frac{p}{100}(220-a)\amp=\frac{\substitute{53}}{100}(220-\substitute{56})\\ \amp=\frac{53}{100}(164)\\ \amp=\frac{53}{25}(41)\\ \amp=86.92 \end{aligned} \end{equation*}

So at the \(53\%\) level, the target heart rate for moderate exercise for someone who is \(56\) years old is \(86.92\) beats per minute.

3.

From January, 2011, to October, 2016, an expression estimating the average rent of a one-bedroom apartment in Portland, Oregon, is given by \(10.173x+974.78\text{,}\) where \(x\) is the number of months since January, 2011.

According to this model, what was the average rent of a one-bedroom apartment in Portland in January, 2011?
According to this model, what was the average rent of a one-bedroom apartment in Portland in January, 2016?

\(\$974.78\)

\(\$1{,}585.16\)

This model uses \(x\) as the number of months since January, 2011. So for January, 2011, \(x\) is \(0\text{:}\)

\begin{equation*} \begin{aligned} 10.173x+974.78 \amp=10.173(\substitute{0})+974.78\\ \amp \approx 974.78 \end{aligned} \end{equation*}

According to this model, the average monthly rent for a one-bedroom apartment in Portland, Oregon, in January, 2011, was \(\$974.78\text{.}\)
The date we are given is January, 2016, which is \(5\) years after January, 2011. Recall that \(x\) is the number of months since January, 2011. So we need to use \(x=5\cdot12=60\text{:}\)

\begin{equation*} \begin{aligned} 10.173x+974.78 \amp=10.173(\substitute{60})+974.78\\ \amp \approx 1585.16 \end{aligned} \end{equation*}

According to this model, the average monthly rent for a one-bedroom apartment in Portland, Oregon, in January, 2016, was \(\$1585.16\text{.}\)

4.

While camping, the height inside a tent when you are \(d\) feet from the north side of the tent is given by the formula \(h=-2\abs{d-3}+6\text{,}\) where \(h\) is in feet.

When you are \(5\) ft from the north side, the height is .
When you are \(2.5\) ft from the north side, the height is .

\(2\ {\rm ft}\)

\(5\ {\rm ft}\)

When \(d=5\text{,}\) we have:

\begin{equation*} \begin{aligned} h\amp= -2\abs{d-3}+6\\ \amp= -2\abs{\substitute{5}-3}+6\amp\amp\text{Review order of operations in }\text{A.5}\text{.}\\ \amp= -2\abs{2}+6\amp\amp\text{Review absolute value in }\text{A.3}\text{.}\\ \amp= -2(2)+6\\ \amp= -4+6\\ \amp=2 \end{aligned} \end{equation*}

So when you are \(5\) ft from the north side, the height of the tent is \(2\) ft.
When \(d=2.5\text{,}\) we have:

\begin{equation*} \begin{aligned} h\amp= -2\abs{d-3}+6\\ \amp= -2\abs{\substitute{2.5}-3}+6\\ \amp= -2\abs{-0.5}+6\\ \amp= -2(0.5)+6\\ \amp=-1+6\\ \amp=5 \end{aligned} \end{equation*}

So when you are \(2.5\) ft from the north side, the height of the tent is \(5\) ft.

5.

If we borrow \(L\) dollars for a home mortgage loan at an annual interest rate \(r\text{,}\) and intend to pay off the loan after \(n\) months, then the amount we should pay each month \(M\text{,}\) in dollars, is given by the formula

\begin{equation*} M=\frac{rL\left(1+\frac{r}{12}\right)^{n}}{12\left(\left(1+\frac{r}{12}\right)^{n}-1\right)} \end{equation*}

If we borrow \(\$200{,}000\) at an interest rate of \(6\%\) with the intent to pay off the loan in \(30\) years, what should our monthly payment be? (Using a calculator is appropriate here.)

\(\$1{,}199.10\)

We must use \(L=200000\text{.}\) Because the interest rate is a percentage, \(r=0.06\) (not \(6\)). The variable \(n\) is supposed to be a number of months, but we will pay off the loan in \(30\) years. Therefore we take \(n=360\text{.}\)

\begin{equation*} \begin{aligned} M=\frac{rL\left(1+\frac{r}{12}\right)^{n}}{12\left(\left(1+\frac{r}{12}\right)^{n}-1\right)} \amp=\frac{(\substitute{0.06})(\substitute{200000})\left(1+\frac{\substitute{0.06}}{12}\right)^{\substitute{360}}}{12\left(\left(1+\frac{\substitute{0.06}}{12}\right)^{\substitute{360}}-1\right)}\\ \amp=\frac{(0.06)(200000)(1+0.005)^{360}}{12\left((1+0.005)^{360}-1\right)}\\ \amp\approx\frac{(0.06)(200000)(6.022575\ldots)}{12\left(6.022575\ldots-1\right)}\\ \amp\approx\frac{(0.06)(200000)(6.022575\ldots)}{12(5.022575\ldots)}\\ \amp\approx\frac{72270.90\ldots}{60.2709\ldots}\\ \amp\approx1199.10 \end{aligned} \end{equation*}

Our monthly payment should be \({\$1{,}199.10}\text{.}\)

6.

Evaluate and simplify the following expressions for \(x=-5\) and \(y=-2\text{:}\)

\(x^3y^2=\)
\((-2x)^3=\)
\(-3x^2y=\)

\(-500\)

\(1000\)

\(150\)

You may review multiplying negative numbers in Section A.1.

\(\displaystyle \begin{aligned}[t] x^3y^2 \amp= (\substitute{-5})^3(\substitute{-2})^2\\ \amp=(-125)(4)\\ \amp=-500 \end{aligned}\)
\(\displaystyle \begin{aligned}[t] (-2x)^3\amp= (-2(\substitute{-5}))^3\\ \amp=(10)^3\\ \amp=1000 \end{aligned}\)
\(\displaystyle \begin{aligned}[t] -3x^2y\amp=-3(\substitute{-5})^2(\substitute{-2})\\ \amp=-3(25)(-2)\\ \amp=150 \end{aligned}\)

7.

In Figure 1.1.7, when you change the value of \(F\text{,}\) why do some values of \(F\) cause there to be more steps in the calculation than other values of \(F\text{?}\)

8.

Identify a variable you might use to represent each quantity. Then identify what units would be most appropriate.

Let be the depth of a swimming pool, measured in .
Let be the weight of a dog, measured in .

\(d\hbox{, }D\hbox{, }x\hbox{, or }y\)

\(\text{feet}\hbox{ or }\text{meters}\)

\(w\hbox{, }W\hbox{, }x\hbox{, or }y\)

\(\text{pounds}\hbox{ or }\text{kilograms}\)

The unknown quantity is depth, which starts with “d”. We generally measure depth in feet for a swimming pool. (Meters is another reasonable unit.) So we could define this variable as:

“Let \(d\) be the depth of a swimming pool, measured in feet.”
The weight of the dog is the unknown quantity, and “weight” starts with “w”. We generally measure the weight of a dog in pounds. (Kilograms is another reasonable unit.) So we could define this variable as:

“Let \(w\) be be the weight of a dog, measured in pounds.”

9.

Identify a variable you might use to represent each quantity. Then identify what units would be most appropriate.

Let be the amount of time a person sleeps each night, measured in .
Let be the surface area of a patio, measured in .

\(t\hbox{, }T\hbox{, }x\hbox{, or }y\)

\(\text{hours}\)

\(a\hbox{, }A\hbox{, }s\hbox{, }S\hbox{, }x\hbox{, or }y\)

\({\verb!ft^2!}\hbox{ or }{\verb!m^2!}\)

The unknown quantity is time, and “time” starts with a “t”. The amount of time a person sleeps would typically be measured in hours. So we could define this variable as:

“Let \(t\) be the amount of time a person sleeps each night, measured in hours.”
The unknown quantity is area, and “area” starts with an “a”. Patio area is usually measured in square feet. (Square meters is another reasonable unit.) So we could define this variable as:

“Let \(A\) be the area of a patio, measured in \(\text{ft}^2\text{.}\)”

10.

Evaluate \({x-4}\) for \(x = 6\text{.}\)

\(2\)

We evaluate \({x-4}\) by replacing \(x\) with \(6\) in the formula.

\(\begin{aligned} {x-4} \amp = 6 - 4\\ \amp = {2} \end{aligned}\)

11.

Evaluate \({x+9}\) for \(x = 8\text{.}\)

\(17\)

We evaluate \({x+9}\) by replacing \(x\) with \(8\) in the formula.

\(\begin{aligned} {x+9} \amp = 8+9\\ \amp = {17} \end{aligned}\)

12.

Evaluate \({2-x}\) for \(x = 10\text{.}\)

\(-8\)

We evaluate \({2-x}\) by replacing \(x\) with \(10\) in the formula.

\(\begin{aligned} {2-x} \amp = 2-10\\ \amp = {-8} \end{aligned}\)

13.

Evaluate \({-5-x}\) for \(x = -8\text{.}\)

\(3\)

We evaluate \({-5-x}\) by replacing \(x\) with \(-8\) in the formula.

\(\begin{aligned} {-5-x} \amp = -5-(-8)\\ \amp = -5 + 8\\ \amp = {3} \end{aligned}\)

14.

Evaluate \({8x+2}\) for \(x = -6\text{.}\)

\(-46\)

We evaluate \({8x+2}\) by replacing \(x\) with \(-6\) in the formula.

\(\begin{aligned} {8x+2} \amp = 8(-6)+2\\ \amp = {-46} \end{aligned}\)

15.

Evaluate \({x-10}\) for \(x = -4\text{.}\)

\(-14\)

We evaluate \({x-10}\) by replacing \(x\) with \(-4\) in the formula.

\(\begin{aligned} {x-10} \amp = 1(-4) - 10\\ \amp = {-14} \end{aligned}\)

16.

Evaluate \({-5c}\) for \(c = -6\text{.}\)

\(30\)

We evaluate \({-5c}\) by replacing \(c\) with \(-6\) in the formula.

\(\begin{aligned} {-5c} \amp = -5(-6)\\ \amp = {30} \end{aligned}\)

17.

Evaluate \({9A}\) for \(A = 7\text{.}\)

\(63\)

We evaluate \({9A}\) by replacing \(A\) with \(7\) in the formula.

\(\begin{aligned} {9A} \amp = 9(7)\\ \amp = {63} \end{aligned}\)

18.

Evaluate the expression \({r^{2}}\text{:}\)

For \(r=6\text{.}\)
For \(r=-4\text{.}\)

\(36\)

\(16\)

We evaluate \({r^{2}}\) by replacing \(r\) with \(6\) in the formula.

\(\displaystyle{\begin{aligned} {r^{2}} \amp = (6)^2 \\ \amp = {36} \end{aligned}}\)
We evaluate \({r^{2}}\) by replacing \(r\) with \(-4\) in the formula.

\(\displaystyle{\begin{aligned} {r^{2}} \amp = (-4)^2 \\ \amp = {16} \end{aligned}}\)

19.

Evaluate the expression \({t^{2}}\text{:}\)

For \(t=3\text{.}\)
For \(t=-7\text{.}\)

\(9\)

\(49\)

We evaluate \({t^{2}}\) by replacing \(t\) with \(3\) in the formula.

\(\displaystyle{\begin{aligned} {t^{2}} \amp = (3)^2 \\ \amp = {9} \end{aligned}}\)
We evaluate \({t^{2}}\) by replacing \(t\) with \(-7\) in the formula.

\(\displaystyle{\begin{aligned} {t^{2}} \amp = (-7)^2 \\ \amp = {49} \end{aligned}}\)

20.

Evaluate the expression \({-t^{2}}\text{:}\)

For \(t=5\text{.}\)
For \(t=-2\text{.}\)

\(-25\)

\(-4\)

We evaluate \({-t^{2}}\) by replacing \(t\) with \(5\) in the formula.

\(\displaystyle{\begin{aligned} {-t^{2}} \amp = -(5)^2 \\ \amp = -(5)(5) \\ \amp = {-25} \end{aligned}}\)
We evaluate \({-t^{2}}\) by replacing \(t\) with \(-2\) in the formula.

\(\displaystyle{\begin{aligned} {-t^{2}} \amp = -(-2)^2 \\ \amp = -(-2)(-2) \\ \amp = {-4} \end{aligned}}\)

Dividing Polynomials by Monomials

21.

Evaluate the expression \({-t^{2}}\text{:}\)

For \(t=3\text{.}\)
For \(t=-4\text{.}\)

\(-9\)

\(-16\)

We evaluate \({-t^{2}}\) by replacing \(t\) with \(3\) in the formula.

\(\displaystyle{\begin{aligned} {-t^{2}} \amp = -(3)^2 \\ \amp = -(3)(3) \\ \amp = {-9} \end{aligned}}\)
We evaluate \({-t^{2}}\) by replacing \(t\) with \(-4\) in the formula.

\(\displaystyle{\begin{aligned} {-t^{2}} \amp = -(-4)^2 \\ \amp = -(-4)(-4) \\ \amp = {-16} \end{aligned}}\)

22.

Evaluate the expression \({x^{3}}\text{:}\)

For \(x=2\text{.}\)
For \(x=-5\text{.}\)

\(8\)

\(-125\)

We evaluate \({x^{3}}\) by replacing \(x\) with \(2\) in the formula.

\(\displaystyle{\begin{aligned} {x^{3}} \amp = (2)^3 \\ \amp = (2)(2)(2) \\ \amp = {8} \end{aligned}}\)
We evaluate \({x^{3}}\) by replacing \(x\) with \(-5\) in the formula.

\(\displaystyle{\begin{aligned} {x^{3}} \amp = (-5)^3 \\ \amp = (-5)(-5)(-5) \\ \amp = {-125} \end{aligned}}\)

23.

Evaluate the expression \({x^{3}}\text{:}\)

For \(x=5\text{.}\)
For \(x=-3\text{.}\)

\(125\)

\(-27\)

We evaluate \({x^{3}}\) by replacing \(x\) with \(5\) in the formula.

\(\displaystyle{\begin{aligned} {x^{3}} \amp = (5)^3 \\ \amp = (5)(5)(5) \\ \amp = {125} \end{aligned}}\)
We evaluate \({x^{3}}\) by replacing \(x\) with \(-3\) in the formula.

\(\displaystyle{\begin{aligned} {x^{3}} \amp = (-3)^3 \\ \amp = (-3)(-3)(-3) \\ \amp = {-27} \end{aligned}}\)

24.

Evaluate \({5x^{2}}\) when \(x=2\text{.}\)
Evaluate \({\left(5x\right)^{2}}\) when \(x=2\text{.}\)

\(20\)

\(100\)

We evaluate \({5x^{2}}\) by replacing \(x\) with \(2\) in the formula.

\(\displaystyle{\begin{aligned} {5x^{2}} \amp = 5(2)^2 \\ \amp = 5(4) \\ \amp = {20} \end{aligned}}\)
We evaluate \({\left(5x\right)^{2}}\) by replacing \(x\) with \(2\) in the formula.

\(\displaystyle{\begin{aligned} {\left(5x\right)^{2}} \amp = (5\cdot2)^2 \\ \amp = (10)^2 \\ \amp = {100} \end{aligned}}\)

25.

Evaluate \({3x^{2}}\) when \(x=2\text{.}\)
Evaluate \({\left(3x\right)^{2}}\) when \(x=2\text{.}\)

\(12\)

\(36\)

We evaluate \({3x^{2}}\) by replacing \(x\) with \(2\) in the formula.

\(\displaystyle{\begin{aligned} {3x^{2}} \amp = 3(2)^2 \\ \amp = 3(4) \\ \amp = {12} \end{aligned}}\)
We evaluate \({\left(3x\right)^{2}}\) by replacing \(x\) with \(2\) in the formula.

\(\displaystyle{\begin{aligned} {\left(3x\right)^{2}} \amp = (3\cdot2)^2 \\ \amp = (6)^2 \\ \amp = {36} \end{aligned}}\)

26.

Evaluate \({-4\!\left(r+4\right)}\) for \(r = 5\text{.}\)

\(-36\)

We evaluate \({-4\!\left(r+4\right)}\) by replacing \(r\) with \(5\) in the formula.

\(\begin{aligned} {-4\!\left(r+4\right)} \amp = -4(5+4)\\ \amp = -4(9) \\ \amp = {-36} \end{aligned}\)

27.

Evaluate \({-10\!\left(r+1\right)}\) for \(r = -3\text{.}\)

\(20\)

We evaluate \({-10\!\left(r+1\right)}\) by replacing \(r\) with \(-3\) in the formula.

\(\begin{aligned} {-10\!\left(r+1\right)} \amp = -10(-3+1)\\ \amp = -10(-2) \\ \amp = {20} \end{aligned}\)

28.

Evaluate \(\displaystyle{{\frac{5t-3}{3t}}}\) for \(t=-10\text{.}\)

\({\textstyle\frac{53}{30}}\)

We evaluate \({\frac{5t-3}{3t}}\) by replacing \(t\) with \(-10\) in the formula.

\(\begin{aligned} {\frac{5t-3}{3t}} \amp = {\frac{5\cdot (-10)-3}{3\cdot (-10)}} \\ \amp = \frac{-53}{-30}\\[0.5pc] \amp = \frac{53}{30} \end{aligned}\)

29.

Evaluate \(\displaystyle{{\frac{9t-6}{3t}}}\) for \(t=4\text{.}\)

\({\textstyle\frac{5}{2}}\)

We evaluate \({\frac{9t-6}{3t}}\) by replacing \(t\) with \(4\) in the formula.

\(\begin{aligned} {\frac{9t-6}{3t}} \amp = {\frac{9\cdot 4-6}{3\cdot 4}} \\ \amp = \frac{30}{12}\\[0.5pc] \amp = \frac{5}{2} \end{aligned}\)

30.

Evaluate \({3C-2c}\) for \(C = -5\) and \(c = -6\text{.}\)

\(-3\)

We evaluate \({3C-2c}\) by replacing \(C\) with \(-5\) and \(c\) with \(-6\) in the formula.

\(\begin{aligned} {3C-2c} \amp = 3(-5) - 2(-6) \\ \amp = -15 + 12 \\ \amp = {-3} \end{aligned}\)

31.

Evaluate \({-2a+4C}\) for \(a = 4\) and \(C = 9\text{.}\)

\(28\)

We evaluate \({-2a+4C}\) by replacing \(a\) with \(4\) and \(C\) with \(9\) in the formula.

\(\begin{aligned} {-2a+4C} \amp = -2(4) + 4(9) \\ \amp = -8 + 36 \\ \amp = {28} \end{aligned}\)

32.

Evaluate \(\displaystyle{{\frac{-3}{r}-\frac{2}{A}}}\) for \(r = 2\) and \(A = -3\text{.}\)

\(-{\textstyle\frac{5}{6}}\)

We evaluate \({\frac{-3}{r}-\frac{2}{A}}\) by replacing \(r\) with \(2\) and \(A\) with \(-3\) in the formula.

\(\begin{aligned} {\frac{-3}{r}-\frac{2}{A}} \amp = \frac{-3}{2} - \frac{2}{-3} \\ \amp = -\frac{3}{2} + \frac{2}{3} \\ \amp = -\frac{9}{6} + \frac{4}{6} \\ \amp = {-{\textstyle\frac{5}{6}}} \end{aligned}\)

33.

Evaluate \(\displaystyle{{\frac{-3}{a}-\frac{2}{b}}}\) for \(a = 5\) and \(b = -7\text{.}\)

\(-{\textstyle\frac{11}{35}}\)

We evaluate \({\frac{-3}{a}-\frac{2}{b}}\) by replacing \(a\) with \(5\) and \(b\) with \(-7\) in the formula.

\(\begin{aligned} {\frac{-3}{a}-\frac{2}{b}} \amp = \frac{-3}{5} - \frac{2}{-7} \\ \amp = -\frac{3}{5} + \frac{2}{7} \\ \amp = -\frac{21}{35} + \frac{10}{35} \\ \amp = {-{\textstyle\frac{11}{35}}} \end{aligned}\)

34.

Evaluate \(\displaystyle{{\frac{-7b+5C+1}{-10b-C}}}\) for \(b=9\) and \(C=-8\text{.}\)

\({\textstyle\frac{51}{41}}\)

We evaluate \({\frac{-7b+5C+1}{-10b-C}}\) by replacing \(b\) with \(9\) and \(C\) with \(-8\) in the formula.

\(\begin{aligned} {\frac{-7b+5C+1}{-10b-C}} \amp = {\frac{-7\cdot 9+5\cdot (-8)+1}{-10\cdot 9-(-8)}} \\ \amp = \frac{-102}{-82}\\[0.5pc] \amp = {{\textstyle\frac{51}{41}}} \end{aligned}\)

Application Problems

35.

Evaluate \(\displaystyle{{\frac{-6A+3c-1}{6A+9c}}}\) for \(A=7\) and \(c=-8\text{.}\)

\({\textstyle\frac{67}{30}}\)

We evaluate \({\frac{-6A+3c-1}{6A+9c}}\) by replacing \(A\) with \(7\) and \(c\) with \(-8\) in the formula.

\(\begin{aligned} {\frac{-6A+3c-1}{6A+9c}} \amp = {\frac{-6\cdot 7+3\cdot (-8)-1}{6\cdot 7+9\cdot (-8)}} \\ \amp = \frac{-67}{-30}\\[0.5pc] \amp = {{\textstyle\frac{67}{30}}} \end{aligned}\)

36.

Evaluate \({1-{\textstyle\frac{1}{10}}\!\left(C-5\right)^{2}}\) for \(C=-5\text{.}\)

\(-9\)

37.

Evaluate \({{\textstyle\frac{1}{14}}\!\left(m+6\right)^{2}-4}\) for \(m=8\text{.}\)

\(10\)

38.

Evaluate \({2p^{2}+9p-8}\) for \(p=1\text{.}\)

\(3\)