All practice questions

Match each of the following quadratic equations to its correct set of real solutions.

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Answer

The equation

2x(x + 1) = 12 - 3x

matches with the solution set

\{-4, \frac{3}{2}\}

; the equation

3x(x - 1) = 2(x + 4)

matches with the solution set

\{-1, \frac{8}{3}\}

; and the equation

4x(x - 2) = 5

matches with the solution set

\{-\frac{1}{2}, \frac{5}{2}\}

.

Each equation is correctly solved by first distributing, moving all terms to one side to set the equation to zero, factoring the resulting trinomial over the integers, and then applying the zero product property to find the corresponding solution set.

Step-by-Step Solution

1

Rearrange the first equation

2x(x + 1) = 12 - 3x

into standard form

ax^2 + bx + c = 0

.

2x^2 + 5x - 12 = 0

Distributing the

2x

gives

2x^2 + 2x = 12 - 3x

. Adding

3x

and subtracting

12

from both sides moves all terms to one side.

2

Factor the rearranged first equation

2x^2 + 5x - 12 = 0

and solve for

x

.

x = -4

or

x = \frac{3}{2}

Finding two integers that multiply to

-24

and add to

5

gives

8

and

-3

. Splitting the middle term and factoring by grouping yields

(2x - 3)(x + 4) = 0

. Setting each factor to zero gives the solutions.

3

Rearrange the second equation

3x(x - 1) = 2(x + 4)

into standard form

ax^2 + bx + c = 0

.

3x^2 - 5x - 8 = 0

Distributing on both sides gives

3x^2 - 3x = 2x + 8

. Subtracting

2x

and

8

from both sides sets the quadratic expression to zero.

4

Factor the rearranged second equation

3x^2 - 5x - 8 = 0

and solve for

x

.

x = -1

or

x = \frac{8}{3}

Finding two integers that multiply to

-24

and add to

-5

gives

-8

and

3

. Grouping terms gives

(3x - 8)(x + 1) = 0

. Setting the factors to zero gives the solutions.

5

Rearrange the third equation

4x(x - 2) = 5

into standard form

ax^2 + bx + c = 0

.

4x^2 - 8x - 5 = 0

Distributing the

4x

yields

4x^2 - 8x = 5

. Subtracting

5

from both sides sets the equation to zero.

6

Factor the rearranged third equation

4x^2 - 8x - 5 = 0

and solve for

x

.

x = -\frac{1}{2}

or

x = \frac{5}{2}

Finding two integers that multiply to

-20

and add to

-8

gives

-10

and

2

. Grouping terms gives

(2x + 1)(2x - 5) = 0

. Solving each linear factor for

x

provides the solutions.

Key Concept

Solving quadratic equations by rearranging them into standard form, factoring by grouping, and applying the zero product property.

Question 2Question →

A team of plasma physicists conducted an experiment to investigate electrical breakdown phenomena in synthetic atmospheric gas mixtures. In a sealed dielectric chamber, researchers systematically varied the volume ratio of nitrogen ( $N_2$ ) to oxygen ( $O_2$ ) across ten trials while recording the threshold breakdown voltage ( $V_{bd}$ ) required to initiate a spark discharge. Throughout all experimental trials, the spacing between the two planar copper electrodes was fixed at $5.0\text{ mm}$ , the total cell pressure was held constant at $101.3\text{ kPa}$ , and the ambient temperature was maintained at $298\text{ K}$ .

Match each experimental component from this investigation to its correct variable classification.

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Answer

Volume ratio of

N_2

to

O_2

matches Independent Variable (manipulated parameter); Threshold breakdown voltage (

V_{bd}

) matches Dependent Variable (measured outcome); Electrode spacing distance (

5.0\text{ mm}

) matches Controlled Variable (apparatus geometry parameter); Total cell pressure (

101.3\text{ kPa}

) matches Controlled Variable (thermodynamic condition parameter).

In experimental design, the independent variable is the factor intentionally varied by experimenters (the ratio of

N_2

to

O_2

), while the dependent variable is the measured output that changes in response (the threshold breakdown voltage

V_{bd}

). Parameters held uniform throughout all trials to maintain experimental integrity are controlled variables, where electrode spacing standardizes apparatus geometry and chamber pressure standardizes ambient thermodynamic conditions.

Step-by-Step Solution

1

Identify the factor systematically changed by the researchers across experimental trials.

The researchers explicitly varied the volume ratio of

N_2

to

O_2

, identifying it as the independent variable.

The independent variable is the condition purposefully manipulated by the experimenter to observe its effect.

2

Identify the factor measured to evaluate the outcome of the experiment.

The threshold breakdown voltage (

V_{bd}

) is recorded in response to changes in gas composition, making it the dependent variable.

The dependent variable represents the response or yield measured as the experimental output.

3

Identify the parameters held constant during the experiment and differentiate their physical nature.

Electrode spacing (

5.0\text{ mm}

) controls the physical setup/geometry, while total pressure (

101.3\text{ kPa}

) controls the ambient thermodynamic environment.

Controlled variables must remain constant across all trials to prevent confounding influence on the dependent variable.

Key Concept

Distinguishing between independent variables (manipulated inputs), dependent variables (observed outcomes), and controlled variables (standardized conditions) in experimental design.

Estimated Time:2m 0s

Question 3Question →

Simplify each of the algebraic expressions on the left by distributing and combining like terms, then match it with its equivalent simplified expression on the right.

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Show answer & explanation

Answer

The expressions match as follows: the first expression matches

-2x^3 + 6x^2y + 4y^3

; the second expression matches

x^3 - 6x^2y + 6xy^2 + 8y^3

; the third expression matches

0

; and the fourth expression matches

y^2

.

Each expression on the left reduces to its matching counterpart on the right by carefully expanding terms (including cubes, squares, difference of squares, and difference of cubes) and combining like terms while correctly distributing negative signs.

Step-by-Step Solution

1

Simplify the first expression by expanding each term individually.

E_1 = -x^3 + 6x^2y - 12xy^2 + 8y^3 - x^3 + 4xy^2 + 8xy^2 - 4y^3

Expanding the binomial cube, the difference of squares product, and distributing the monomial allows us to identify like terms.

2

Combine like terms in the first expression.

E_1 = -2x^3 + 6x^2y + 4y^3

Combining the

x^3

,

x^2y

,

xy^2

, and

y^3

terms simplifies the expression. The

xy^2

terms sum to zero.

3

Simplify the second expression by expanding each term individually.

E_2 = 2x^3 - 4x^2y + 2xy^2 - x^3 + 8y^3 - 2x^2y + 4xy^2

Using algebraic expansion rules (including the difference of cubes product) exposes all individual terms.

4

Combine like terms in the second expression.

E_2 = x^3 - 6x^2y + 6xy^2 + 8y^3

Adding coefficients of like variable terms gives the simplified form.

5

Simplify the third expression by expanding the cubic terms.

E_3 = (x^3 + 3x^2y + 3xy^2 + y^3) - (x^3 - 3x^2y + 3xy^2 - y^3) - (6x^2y + 2y^3) = 6x^2y + 2y^3 - 6x^2y - 2y^3

Expanding the binomial cubes and distributing the negative signs shows that all terms cancel.

6

Combine like terms in the third expression.

E_3 = 0

All terms cancel out, leaving a final value of zero.

7

Simplify the fourth expression by expanding.

E_4 = x^4 - 2x^2y + y^2 - x^4 + y^2 - y^2 + 2x^2y

Squaring the binomial, using the difference of squares, and distributing the negative variable simplifies the individual components.

8

Combine like terms in the fourth expression.

E_4 = y^2

The

x^4

and

x^2y

terms cancel, and the

y^2

terms simplify to

y^2

.

Key Concept

Simplifying Expressions and Combining Like Terms

Question 4Question →

Solve each quadratic equation by factoring, and match the equation to its correct solution set.

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Answer

The equation

3x(4x + 5) = 5 - 2x

matches the solution set

\{-\frac{5}{3}, \frac{1}{4}\}

; the equation

x(12x + 1) = 35

matches the solution set

\{-\frac{7}{4}, \frac{5}{3}\}

; and the equation

(2x - 3)(3x + 1) = 5(1 - x)

matches the solution set

\{-1, \frac{4}{3}\}

.

Each of the quadratic equations is solved by first expanding any products, collecting all terms on the left-hand side to establish the standard form

ax^2 + bx + c = 0

, dividing by any common numerical factors, factoring the resulting quadratic expression into two linear binomials, and solving each linear equation for

x

. This correctly pairs the first equation with

\{-\frac{5}{3}, \frac{1}{4}\}

, the second equation with

\{-\frac{7}{4}, \frac{5}{3}\}

, and the third equation with

\{-1, \frac{4}{3}\}

.

Step-by-Step Solution

1

Solve

3x(4x + 5) = 5 - 2x

.

12x^2 + 17x - 5 = 0 \Rightarrow (3x + 5)(4x - 1) = 0 \Rightarrow x = -\frac{5}{3}

or

x = \frac{1}{4}

.

Distribute the term on the left, rearrange the terms to set the equation to zero, find factors of

12 \times (-5) = -60

that sum to

17

(which are

20

and

-3

), factor by grouping, and apply the Zero Product Property.

2

Solve

x(12x + 1) = 35

.

12x^2 + x - 35 = 0 \Rightarrow (3x - 5)(4x + 7) = 0 \Rightarrow x = \frac{5}{3}

or

x = -\frac{7}{4}

.

Expand the left side, subtract

35

from both sides, find factors of

12 \times (-35) = -420

that sum to

1

(which are

21

and

-20

), factor by grouping, and solve for

x

.

3

Solve

(2x - 3)(3x + 1) = 5(1 - x)

.

6x^2 - 2x - 8 = 0 \Rightarrow 3x^2 - x - 4 = 0 \Rightarrow (3x - 4)(x + 1) = 0 \Rightarrow x = \frac{4}{3}

or

x = -1

.

Expand both sides, move all terms to the left, divide the quadratic equation by

2

to simplify, factor the trinomial, and solve for the roots.

Key Concept

Rearranging non-standard quadratic equations into the standard form

ax^2 + bx + c = 0

and solving them by factoring over the integers.

Question 5Question →

Match each quadratic equation with its correct solution set by solving the equation by factoring.

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Answer

The equation

x(x - 1) = 12

matches the solution set

\{-3, 4\}

; the equation

2x^2 + 5x = 3

matches the solution set

\{-3, \frac{1}{2}\}

; the equation

3x^2 + 8 = 10x

matches the solution set

\{\frac{4}{3}, 2\}

; and the equation

2x^2 - 24 = 8x

matches the solution set

\{-2, 6\}

.

Each quadratic equation is correctly matched to its solutions by first rewriting the equation in standard form

ax^2 + bx + c = 0

, factoring the trinomial over the integers, and then applying the zero product property to find the roots.

Step-by-Step Solution

1

Set each quadratic equation to standard form

ax^2 + bx + c = 0

by expanding terms and moving all terms to one side.

The equations become:
1)

x^2 - x - 12 = 0

2)

2x^2 + 5x - 3 = 0

3)

3x^2 - 10x + 8 = 0

4)

2x^2 - 8x - 24 = 0

Before a quadratic equation can be solved by factoring, it must be set equal to zero so that the zero product property can be applied.

2

Factor each quadratic expression completely over the integers.

The factored expressions are:
1)

(x - 4)(x + 3) = 0

2)

(2x - 1)(x + 3) = 0

3)

(3x - 4)(x - 2) = 0

4)

2(x - 6)(x + 2) = 0

Factoring rewrites the quadratic expressions as products of linear factors.

3

Apply the zero product property by setting each linear factor equal to zero and solving for

x

.

The solution sets are:
1)

x = 4

or

x = -3

, yielding

\{-3, 4\}

2)

x = \frac{1}{2}

or

x = -3

, yielding

\{-3, \frac{1}{2}\}

3)

x = \frac{4}{3}

or

x = 2

, yielding

\{\frac{4}{3}, 2\}

4)

x = 6

or

x = -2

, yielding

\{-2, 6\}

If the product of two or more algebraic factors is zero, then at least one of the individual factors must equal zero.

Key Concept

Solving Quadratic Equations by Factoring

Alternative Method

You can verify the solution sets by substituting the values of the roots back into the original equations to check if they yield a true statement, or by using the quadratic formula

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

as an alternative algebraic method.

Estimated Time:1m 30s

Question 6Question →

For all real numbers $x$ and $y$ , match each algebraic expression on the left with its simplified equivalent expression on the right.

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Show answer & explanation

Answer

The expression

2(x - 3y) + 4y

simplifies to

2x - 2y

, the expression

-2(x - 3y) + 8y

simplifies to

-2x + 14y

, and the expression

2(x + 3y) - 4y

simplifies to

2x + 2y

.

Each expression on the left-hand side is expanded by applying the distributive property and then simplified by combining the terms involving

y

. This correctly matches

2(x - 3y) + 4y

to

2x - 2y

,

-2(x - 3y) + 8y

to

-2x + 14y

, and

2(x + 3y) - 4y

to

2x + 2y

.

Step-by-Step Solution

1

Simplify the expression

2(x - 3y) + 4y

.

2x - 2y

Distribute

2

to both terms inside the parentheses to get

2x - 6y

, then combine the like terms

-6y

and

4y

to get

-2y

.

2

Simplify the expression

-2(x - 3y) + 8y

.

-2x + 14y

Distribute

-2

to both terms inside the parentheses to get

-2x + 6y

, then combine the like terms

6y

and

8y

to get

14y

.

3

Simplify the expression

2(x + 3y) - 4y

.

2x + 2y

Distribute

2

to both terms inside the parentheses to get

2x + 6y

, then combine the like terms

6y

and

-4y

to get

2y

.

Key Concept

Simplifying algebraic expressions by distributing coefficients and combining like terms.

Estimated Time:1m 0s

Question 7Question →

Match each quadratic equation with its correct set of real solutions.

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Show answer & explanation

Answer

The equation

x^2 - 5x + 6 = 0

matches with the solutions

x = 2

and

x = 3

. The equation

x^2 + 5x + 6 = 0

matches with the solutions

x = -3

and

x = -2

. The equation

x^2 - x - 6 = 0

matches with the solutions

x = -2

and

x = 3

.

Each equation is solved by factoring the quadratic trinomial into two binomials, then applying the zero product property to find the values of

x

that make each factor zero. Specifically,

x^2 - 5x + 6 = 0

factors into

(x - 2)(x - 3) = 0

, yielding solutions

x = 2

and

x = 3

. The equation

x^2 + 5x + 6 = 0

factors into

(x + 2)(x + 3) = 0

, yielding solutions

x = -2

and

x = -3

. Finally,

x^2 - x - 6 = 0

factors into

(x - 3)(x + 2) = 0

, yielding solutions

x = 3

and

x = -2

.

Step-by-Step Solution

1

Factor the quadratic equation

x^2 - 5x + 6 = 0

.

(x - 2)(x - 3) = 0

Identify two integers that multiply to

6

and add up to

-5

. These integers are

-2

and

-3

.

2

Solve for

x

by setting each linear factor equal to zero:

x - 2 = 0

and

x - 3 = 0

.

x = 2

and

x = 3

Applying the zero product property means if the product of two numbers is zero, at least one of them must be zero.

3

Factor the quadratic equation

x^2 + 5x + 6 = 0

.

(x + 2)(x + 3) = 0

Identify two integers that multiply to

6

and add up to

5

. These integers are

2

and

3

.

4

Solve for

x

by setting each linear factor equal to zero:

x + 2 = 0

and

x + 3 = 0

.

x = -2

and

x = -3

Applying the zero product property gives the solutions as the negations of the terms inside the binomials.

5

Factor the quadratic equation

x^2 - x - 6 = 0

.

(x - 3)(x + 2) = 0

Identify two integers that multiply to

-6

and add up to

-1

. These integers are

-3

and

2

.

6

Solve for

x

by setting each linear factor equal to zero:

x - 3 = 0

and

x + 2 = 0

.

x = 3

and

x = -2

Setting the linear factors to zero yields the roots of the equation.

Key Concept

Solving Quadratic Equations by Factoring

Estimated Time:1m 30s

Question 8Question →

Match each algebraic expression on the left with its equivalent simplified form on the right. Assume all variables represent real numbers.

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Show answer & explanation

Answer

The expression

2(3a^2 - 4b) - 3(a^2 - 2b)

matches

3a^2 - 2b

;

a(3a - b) - 2b(a - b)

matches

3a^2 - 3ab + 2b^2

;

(a + b)(3a - 2b) - b^2

matches

3a^2 + ab - 3b^2

; and

4a^2 - (a - b)(a + 2b) - 2b^2

matches

3a^2 - ab

.

Each expression is correctly simplified by distributing coefficients, expanding binomial products, and collecting like terms.

Step-by-Step Solution

1

Simplify the expression

2(3a^2 - 4b) - 3(a^2 - 2b)

3a^2 - 2b

Distribute the coefficients to remove the parentheses:

6a^2 - 8b - 3a^2 + 6b

. Group the

a^2

terms and the

b

terms, and then combine:

(6 - 3)a^2 + (-8 + 6)b = 3a^2 - 2b

.

2

Simplify the expression

a(3a - b) - 2b(a - b)

3a^2 - 3ab + 2b^2

Distribute the variables

a

and

-2b

:

3a^2 - ab - 2ab + 2b^2

. Combine the like terms

-ab

and

-2ab

to get

-3ab

, resulting in

3a^2 - 3ab + 2b^2

.

3

Simplify the expression

(a + b)(3a - 2b) - b^2

3a^2 + ab - 3b^2

Multiply the binomial factors using the distributive property:

(a + b)(3a - 2b) = 3a^2 - 2ab + 3ab - 2b^2 = 3a^2 + ab - 2b^2

. Subtract the remaining

b^2

term:

3a^2 + ab - 2b^2 - b^2 = 3a^2 + ab - 3b^2

.

4

Simplify the expression

4a^2 - (a - b)(a + 2b) - 2b^2

3a^2 - ab

Expand

(a - b)(a + 2b) = a^2 + ab - 2b^2

. Subtract this product from

4a^2

by distributing the negative sign:

4a^2 - a^2 - ab + 2b^2

. Finally, subtract the last term

2b^2

:

3a^2 - ab + 2b^2 - 2b^2 = 3a^2 - ab

.

Key Concept

Simplifying algebraic expressions by distributing factors and combining like terms

Estimated Time:2m 0s

Question 9Question →

For each algebraic expression on the left, match it to its completely simplified equivalent expression on the right by distributing terms and combining like terms.

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Show answer & explanation

Answer

The expressions match as follows: the first simplifies to

-3x^2y + 3y^3

; the second simplifies to

12x^2y + 6xy^2 - 2y^3

; the third simplifies to

13xy^2 - 4y^3

; and the fourth simplifies to

x^3 - 3x^2y - 6xy^2 + 2y^3

.

Each expression is expanded fully by distributing multiplication and powers, then simplified by combining terms that have the exact same variable bases and exponents.

Step-by-Step Solution

1

Simplify the first expression by distributing coefficients and combining like terms.

-3x^2y + 3y^3

Expanding the expression gives

2x^3 - 6x^2y - 3x^2y + 3y^3 - 2x^3 + 6x^2y

. Grouping the like terms:

(2 - 2)x^3 + (-6 - 3 + 6)x^2y + 3y^3

, which simplifies to

-3x^2y + 3y^3

.

2

Simplify the second expression using the binomial cube formula

(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3

.

12x^2y + 6xy^2 - 2y^3

Expanding

(2x - y)^3

yields

8x^3 - 12x^2y + 6xy^2 - y^3

. Subtracting the remaining terms gives

8x^3 - 12x^2y + 6xy^2 - y^3 - 8x^3 + 24x^2y - y^3

. Combining like terms yields

12x^2y + 6xy^2 - 2y^3

.

3

Simplify the third expression by squaring the binomials and distributing.

13xy^2 - 4y^3

First expand the squares:

(2x - 3y)^2 = 4x^2 - 12xy + 9y^2

and

(x - 2y)^2 = x^2 - 4xy + 4y^2

. Distributing the variables yields

4x^3 - 12x^2y + 9xy^2 - x^2y + 4xy^2 - 4y^3 - 4x^3 + 13x^2y

. Combining like terms results in

13xy^2 - 4y^3

.

4

Simplify the fourth expression by expanding the cubic term and distributing signs.

x^3 - 3x^2y - 6xy^2 + 2y^3

Expand

(x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3

. Distribute all negative signs to get

2x^3 - 6x^2y - x^3 + 3x^2y - 3xy^2 + y^3 - 3xy^2 + y^3

. Combining like terms results in

x^3 - 3x^2y - 6xy^2 + 2y^3

.

Key Concept

Simplifying Expressions and Combining Like Terms

Question 10Question →

Match each algebraic expression on the left with its fully simplified equivalent expression on the right. All variables represent real numbers.

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Show answer & explanation

Answer

The first expression

3a(a^2 - 2ab) - (a^3 - 5a^2b)

matches

2a^3 - a^2b

. The second expression

a^2(2a - b) - 2a(a^2 - ab)

matches

a^2b

. The third expression

-(a^3 + 3a^2b) + 3a^2(a + b)

matches

2a^3

.

Each expression on the left is simplified by distributing the coefficients and combining the like terms. The first expression simplifies to

2a^3 - a^2b

. The second expression simplifies to

a^2b

. The third expression simplifies to

2a^3

. These match the corresponding simplified expressions on the right.

Step-by-Step Solution

1

Simplify the first expression

3a(a^2 - 2ab) - (a^3 - 5a^2b)

2a^3 - a^2b

Multiply

3a

by both terms in the first parentheses to get

3a^3 - 6a^2b

. Distribute the negative sign to both terms in the second parentheses to get

-a^3 + 5a^2b

. Combine the

a^3

terms to get

2a^3

and the

a^2b

terms to get

-a^2b

.

2

Simplify the second expression

a^2(2a - b) - 2a(a^2 - ab)

a^2b

Multiply

a^2

by both terms in the first parentheses to get

2a^3 - a^2b

. Multiply

-2a

by both terms in the second parentheses to get

-2a^3 + 2a^2b

. Combine the

a^3

terms to get

0

and the

a^2b

terms to get

a^2b

.

3

Simplify the third expression

-(a^3 + 3a^2b) + 3a^2(a + b)

2a^3

Distribute the negative sign to get

-a^3 - 3a^2b

. Multiply

3a^2

by both terms in the second parentheses to get

3a^3 + 3a^2b

. Combine the

a^3

terms to get

2a^3

and the

a^2b

terms to get

0

.

Key Concept

Simplifying Expressions and Combining Like Terms

Estimated Time:1m 30s

Question 11Question →

Match each algebraic expression on the left with its fully simplified equivalent expression on the right.

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Show answer & explanation

Answer

The expression

--(a^2 - 2ab) + a(a - 3b)

simplifies to

-ab

; the expression

2(a^2b - ab) - ab(2a - 3)

simplifies to

ab

; and the expression

a^2(a - b) - a(a^2 - ab)

simplifies to

0

.

Each expression is correctly simplified by distributing the external factors across parenthetical terms and combining the resulting like terms.

Step-by-Step Solution

1

Simplify the first expression by distributing coefficients and combining like terms.

--a^2 + 2ab + a^2 - 3b = -ab

Distributing the negative sign yields

--a^2 + 2ab

and distributing

a

yields

a^2 - 3ab

. Adding them eliminates the

a^2

terms, leaving

-ab

.

2

Simplify the second expression by distributing coefficients and combining like terms.

2a^2b - 2ab - 2a^2b + 3ab = ab

Distributing

2

yields

2a^2b - 2ab

and distributing

-ab

yields

-2a^2b + 3ab

. Adding them eliminates the

2a^2b

terms, leaving

ab

.

3

Simplify the third expression by distributing coefficients and combining like terms.

a^3 - a^2b - a^3 + a^2b = 0

Distributing

a^2

yields

a^3 - a^2b

and distributing

-a

yields

-a^3 + a^2b

. Adding them eliminates all terms, leaving

0

.

Key Concept

Simplifying algebraic expressions containing multiple variables and higher powers by applying the distributive property and combining like terms.

Question 12Question →

Match each algebraic expression on the left with its fully simplified equivalent expression on the right for all real values of $m$ and $n$ .

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches are: (1)

m(m^2 - 3mn) - 2n(m^2 - n^2) - (m^3 - mn^2)

matches with

-5m^2n + mn^2 + 2n^3

; (2)

(m - n)^3 - m(m^2 - 3n^2) + n^3

matches with

-3m^2n + 6mn^2

; (3)

3mn(m - n) - 2m(n^2 - mn) - (m^2n - 5mn^2)

matches with

4m^2n

; and (4)

m^2(2m - n) - n(m^2 - n^2) - (m^3 + n^3)

matches with

m^3 - 2m^2n

.

Each expression is simplified by systematically expanding parenthetical groups and collecting like terms with identical variable powers.

Step-by-Step Solution

1

Simplify the expression

m(m^2 - 3mn) - 2n(m^2 - n^2) - (m^3 - mn^2)

-5m^2n + mn^2 + 2n^3

Distribute each term across the parenthetical expressions:

m^3 - 3m^2n - 2m^2n + 2n^3 - m^3 + mn^2

. Group the like terms:

(m^3 - m^3) + (-3m^2n - 2m^2n) + mn^2 + 2n^3

, which simplifies to

-5m^2n + mn^2 + 2n^3

.

2

Simplify the expression

(m - n)^3 - m(m^2 - 3n^2) + n^3

-3m^2n + 6mn^2

Use the binomial expansion formula to expand

(m - n)^3 = m^3 - 3m^2n + 3mn^2 - n^3

. Distribute the

-m

term to obtain

-m^3 + 3mn^2

. Sum all the expressions and combine like terms:

(m^3 - m^3) - 3m^2n + (3mn^2 + 3mn^2) + (-n^3 + n^3) = -3m^2n + 6mn^2

.

3

Simplify the expression

3mn(m - n) - 2m(n^2 - mn) - (m^2n - 5mn^2)

4m^2n

Expand the terms by distributing the outer coefficients to get

3m^2n - 3mn^2 - 2mn^2 + 2m^2n - m^2n + 5mn^2

. Grouping similar variables yields

(3 + 2 - 1)m^2n + (-3 - 2 + 5)mn^2 = 4m^2n + 0 = 4m^2n

.

4

Simplify the expression

m^2(2m - n) - n(m^2 - n^2) - (m^3 + n^3)

m^3 - 2m^2n

Expand the terms by distributing the multiplication:

2m^3 - m^2n - m^2n + n^3 - m^3 - n^3

. Grouping like terms yields

(2m^3 - m^3) + (-m^2n - m^2n) + (n^3 - n^3) = m^3 - 2m^2n

.

Key Concept

Simplifying multivariable expressions by distributing terms (including negative signs) and combining like terms.

Question 13Question →

For each of the given quadratic equations, solve for $x$ by factoring. Match each quadratic equation on the left to its correct solution set on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The equation

2x^2 + 5x = 3

matches the solution set

\left\{-3, \frac{1}{2}\right\}

; the equation

3x^2 - 10x + 8 = 0

matches the solution set

\left\{\frac{4}{3}, 2\right\}

; and the equation

x(x + 2) = 15

matches the solution set

\left\{-5, 3\right\}

.

Each equation matches its corresponding solution set through distributing terms if necessary, rewriting the equation in standard quadratic form

ax^2 + bx + c = 0

, factoring the quadratic trinomial over the integers, and then using the zero product property to solve for

x

.

Step-by-Step Solution

1

For the equation

2x^2 + 5x = 3

, rewrite in standard form

ax^2 + bx + c = 0

by subtracting 3 from both sides to get

2x^2 + 5x - 3 = 0

.

The equation is rewritten as

2x^2 + 5x - 3 = 0

.

Before factoring a quadratic equation, all terms must be moved to one side so the other side is equal to zero.

2

Factor the trinomial

2x^2 + 5x - 3

by grouping. Find two integers that multiply to

2 \times (-3) = -6

and add to

5

. These integers are

6

and

-1

. Rewrite the middle term and factor by grouping:

2x^2 + 6x - x - 3 = 2x(x + 3) - 1(x + 3) = (2x - 1)(x + 3) = 0

.

The equation becomes

(2x - 1)(x + 3) = 0

.

Factoring allows us to apply the zero product property to find the solutions.

3

Set each factor of

(2x - 1)(x + 3) = 0

to zero and solve for

x

:

2x - 1 = 0 \Rightarrow x = \frac{1}{2}

and

x + 3 = 0 \Rightarrow x = -3

.

The solutions are

x = \frac{1}{2}

and

x = -3

, forming the solution set

\left\{-3, \frac{1}{2}\right\}

.

By the zero product property, if the product of two factors is zero, at least one of the factors must be zero.

4

For the equation

3x^2 - 10x + 8 = 0

, factor the trinomial by finding two integers that multiply to

3 \times 8 = 24

and add to

-10

. These integers are

-6

and

-4

. Rewrite the middle term and factor by grouping:

3x^2 - 6x - 4x + 8 = 3x(x - 2) - 4(x - 2) = (3x - 4)(x - 2) = 0

.

The equation becomes

(3x - 4)(x - 2) = 0

.

The equation is already in standard form, so we can directly proceed with factoring.

5

Set each factor of

(3x - 4)(x - 2) = 0

to zero and solve for

x

:

3x - 4 = 0 \Rightarrow x = \frac{4}{3}

and

x - 2 = 0 \Rightarrow x = 2

.

The solutions are

x = \frac{4}{3}

and

x = 2

, forming the solution set

\left\{\frac{4}{3}, 2\right\}

.

Solving each linear factor yields the roots of the quadratic equation.

6

For the equation

x(x + 2) = 15

, first distribute

x

to get

x^2 + 2x = 15

, then subtract 15 from both sides to write in standard form:

x^2 + 2x - 15 = 0

.

The equation is rewritten as

x^2 + 2x - 15 = 0

.

Distributing and moving terms sets the quadratic to zero, which is necessary for factoring.

7

Factor the quadratic

x^2 + 2x - 15 = 0

by finding two integers that multiply to

-15

and add to

2

. These integers are

5

and

-3

, yielding

(x + 5)(x - 3) = 0

.

The equation becomes

(x + 5)(x - 3) = 0

.

Factoring a quadratic trinomial with a leading coefficient of 1 involves finding numbers that sum to the linear coefficient and multiply to the constant term.

8

Set each factor of

(x + 5)(x - 3) = 0

to zero and solve for

x

:

x + 5 = 0 \Rightarrow x = -5

and

x - 3 = 0 \Rightarrow x = 3

.

The solutions are

x = -5

and

x = 3

, forming the solution set

\left\{-5, 3\right\}

.

Solving the resulting linear equations gives the roots of the original quadratic equation.

Key Concept

Solving Quadratic Equations by Factoring

Alternative Method

Instead of factoring, the solutions to these quadratic equations can be verified by substituting the values in the solution sets back into the original equations, or by applying the quadratic formula:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

after rewriting them in standard form.

Estimated Time:2m 0s

Question 14Question →

Match each unsimplified algebraic expression on the left with its equivalent simplified form on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings match

3x(x - 2y) - 2x(x - 3y)

to

x^2

;

2(x^2 - xy) - 3(xy - y^2)

to

2x^2 - 5xy + 3y^2

;

x^2(x - 2y) - x(x^2 - xy)

to

-x^2y

; and

(x + y)^2 - (x - y)^2

to

4xy

.

Each unsimplified expression is correctly matched to its simplified equivalent by expanding parenthetical terms (taking care to distribute negative signs) and combining like terms.

Step-by-Step Solution

1

Distribute and combine like terms for the first expression

3x(x - 2y) - 2x(x - 3y)

.

The expression simplifies to

x^2

.

First distribute the coefficients to get

3x^2 - 6xy - 2x^2 + 6xy

. Then combine

3x^2 - 2x^2 = x^2

and

-6xy + 6xy = 0

.

2

Distribute and combine like terms for the second expression

2(x^2 - xy) - 3(xy - y^2)

.

The expression simplifies to

2x^2 - 5xy + 3y^2

.

Distribute the coefficients to get

2x^2 - 2xy - 3xy + 3y^2

. Note that distributing the negative sign of

-3

to

-y^2

results in

+3y^2

. Then combine

-2xy - 3xy = -5xy

.

3

Distribute and combine like terms for the third expression

x^2(x - 2y) - x(x^2 - xy)

.

The expression simplifies to

-x^2y

.

Distribute to get

x^3 - 2x^2y - x^3 + x^2y

. Note that distributing

-x

to

-xy

gives

+x^2y

. The

x^3 - x^3

terms cancel, leaving

-2x^2y + x^2y = -x^2y

.

4

Expand and simplify the fourth expression

(x + y)^2 - (x - y)^2

.

The expression simplifies to

4xy

.

Expand both squared binomials:

(x^2 + 2xy + y^2) - (x^2 - 2xy + y^2)

. Distribute the negative sign to get

x^2 + 2xy + y^2 - x^2 + 2xy - y^2

. Combine terms to cancel

x^2

and

y^2

, leaving

4xy

.

Key Concept

Simplifying Expressions and Combining Like Terms

Estimated Time:1m 30s

Question 15Question →

For each quadratic equation on the left, solve for $x$ by factoring and match it to its correct solution set on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The equation

2x^2 + 5x = 3

matches the solution set

\{-3, \frac{1}{2}\}

; the equation

3x^2 - 10x = -8

matches the solution set

\{\frac{4}{3}, 2\}

; and the equation

x(x - 4) = 12

matches the solution set

\{-2, 6\}

.

Each of the quadratic equations can be solved by first rearranging the terms to set the equation equal to zero. After rewriting them in the standard form

ax^2 + bx + c = 0

, they can be factored into a product of linear binomials. Setting each factor equal to zero and solving for

x

yields the solutions. Specifically:

2x^2 + 5x = 3

simplifies to

2x^2 + 5x - 3 = 0

, which factors as

(2x - 1)(x + 3) = 0

and gives the solution set

\{-3, \frac{1}{2}\}

.

3x^2 - 10x = -8

simplifies to

3x^2 - 10x + 8 = 0

, which factors as

(3x - 4)(x - 2) = 0

and gives the solution set

\{\frac{4}{3}, 2\}

.

x(x - 4) = 12

simplifies to

x^2 - 4x - 12 = 0

, which factors as

(x - 6)(x + 2) = 0

and gives the solution set

\{-2, 6\}

.

Step-by-Step Solution

1

Rearrange the equation

2x^2 + 5x = 3

into the standard form

ax^2 + bx + c = 0

.

2x^2 + 5x - 3 = 0

To apply factoring and the zero product property, the quadratic expression must equal zero.

2

Factor the trinomial

2x^2 + 5x - 3

.

(2x - 1)(x + 3) = 0

Since the constant term is negative and the middle coefficient is positive, the factors must have opposite signs.

3

Apply the zero product property to find the solutions for the first equation.

x = \frac{1}{2}

and

x = -3

, yielding the set

\{-3, \frac{1}{2}\}

Setting the individual linear factors

2x - 1

and

x + 3

to zero gives the solutions.

4

Rearrange the equation

3x^2 - 10x = -8

into standard form.

3x^2 - 10x + 8 = 0

Add

8

to both sides to set the right side to zero.

5

Factor the trinomial

3x^2 - 10x + 8

.

(3x - 4)(x - 2) = 0

The constant term is positive and the middle coefficient is negative, meaning both constant terms in the binomial factors must be negative.

6

Solve for

x

using the zero product property.

x = \frac{4}{3}

and

x = 2

, yielding the set

\{\frac{4}{3}, 2\}

Setting

3x - 4 = 0

gives

x = \frac{4}{3}

, and setting

x - 2 = 0

gives

x = 2

.

7

Expand and rearrange the equation

x(x - 4) = 12

into standard form.

x^2 - 4x - 12 = 0

Distribute the

x

on the left side to get

x^2 - 4x

and subtract

12

from both sides to set the equation to zero.

8

Factor the trinomial

x^2 - 4x - 12

.

(x - 6)(x + 2) = 0

Find two numbers that multiply to

-12

and add to

-4

. Those numbers are

-6

and

+2

.

9

Solve for

x

using the zero product property.

x = 6

and

x = -2

, yielding the set

\{-2, 6\}

Setting

x - 6 = 0

gives

x = 6

, and setting

x + 2 = 0

gives

x = -2

.

Key Concept

Solving quadratic equations by rewriting them in standard form, factoring the trinomials, and using the zero product property.

Question 16Question →

Match each of the unsimplified algebraic expressions on the left with its equivalent simplified form on the right. (Assume all variables represent real numbers.)

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches pair the first expression with

6r^2 - 5rs + 6s^2

, the second expression with

2r^2 - 2rs + s^2

, the third expression with

3s^2 - 4rs

, and the fourth expression with

6r^2 - 2r^2s + 5s^2

.

Each expression is correctly simplified by performing distribution first (carefully tracking negative signs and binomial expansion rules) and then combining terms that share the exact same variable powers.

Step-by-Step Solution

1

Simplify the expression

3r(2r - s) - 2s(r - 3s)

.

6r^2 - 5rs + 6s^2

Distribute

3r

to get

6r^2 - 3rs

. Then distribute

-2s

to get

-2rs + 6s^2

(noting that a negative times a negative is positive). Combine the like terms

-3rs

and

-2rs

to get

-5rs

.

2

Simplify the expression

(2r - s)^2 - 2(r^2 - rs)

.

2r^2 - 2rs + s^2

Square the binomial

(2r - s)^2

using the formula

(a - b)^2 = a^2 - 2ab + b^2

to get

4r^2 - 4rs + s^2

. Distribute

-2

to get

-2r^2 + 2rs

. Combine

4r^2 - 2r^2

to get

2r^2

, and

-4rs + 2rs

to get

-2rs

.

3

Simplify the expression

\frac{1}{2}(4r^2 - 6rs) - (r - s)(2r + 3s)

.

3s^2 - 4rs

Distribute

\frac{1}{2}

to get

2r^2 - 3rs

. Expand the binomial product to get

2r^2 + 3rs - 2rs - 3s^2 = 2r^2 + rs - 3s^2

. Subtract this entire expression:

(2r^2 - 3rs) - (2r^2 + rs - 3s^2) = 2r^2 - 3rs - 2r^2 - rs + 3s^2

. Combine like terms to get

-4rs + 3s^2

.

4

Simplify the expression

r^2(6 - s) - s(r^2 - 5s)

.

6r^2 - 2r^2s + 5s^2

Distribute

r^2

to get

6r^2 - r^2s

. Distribute

-s

to get

-sr^2 + 5s^2

. Combine the like terms

-r^2s

and

-sr^2

(since multiplication is commutative) to get

-2r^2s

.

Key Concept

Simplifying expressions by applying the distributive property, expanding products of binomials, and combining like terms.

Question 17Question →

The following paragraphs are from a natural science essay about deep-sea ecosystems:

[Paragraph 1] In 1977, researchers aboard the research submersible Alvin made a discovery that transformed biology. Deep on the ocean floor near the Galápagos Rift, they found hydrothermal vents spewing superheated, mineral-rich water. Surrounding these vents were thriving communities of giant tube worms, clams, and crabs, existing in complete darkness. This finding shattered the long-held scientific consensus that all biological communities on Earth require sunlight as their primary source of energy.

[Paragraph 2] At the heart of this dark ecosystem are specialized bacteria that have bypassed the need for solar radiation entirely. Rather than relying on photosynthesis, these microbes perform chemosynthesis. They metabolize hydrogen sulfide—a compound toxic to most land-based organisms—flowing from the vents and convert it into organic matter. This process forms the base of the food web, nourishing the larger animals that live clustered around the vent openings.

[Paragraph 3] The realization that life can flourish in such extreme, sunless environments has profound implications for astrobiology. Scientists now look to the icy moons of the outer solar system, such as Jupiter's Europa and Saturn's Enceladus, with renewed optimism. If liquid oceans exist beneath their frozen crusts, heated by tidal forces, hydrothermal vent communities could theoretically survive there, completely isolated from any star's light.

Match each paragraph from the passage to its primary paragraph-level main idea.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Paragraph 1 matches the introduction of a finding that challenged assumptions about solar energy; Paragraph 2 matches the explanation of how organisms generate energy without sunlight; Paragraph 3 matches the implications for life on icy moons in the outer solar system.

The correct pairings align each paragraph with its specific focus: Paragraph 1 introduces the discovery challenging solar dependency, Paragraph 2 explains the chemical process of chemosynthesis, and Paragraph 3 outlines the implications for life on outer solar system moons.

Step-by-Step Solution

1

Analyze Paragraph 1 to identify its main focus.

Paragraph 1 describes the discovery of hydrothermal vents in 1977 and how it shattered the consensus that all life requires sunlight.

Understanding the paragraph's primary purpose helps match it to the correct statement.

2

Analyze Paragraph 2 to identify its main focus.

Paragraph 2 explains chemosynthesis, the chemical process bacteria use to generate food from hydrogen sulfide without sunlight.

This establishes the biological mechanism described in the second section.

3

Analyze Paragraph 3 to identify its main focus.

Paragraph 3 discusses astrobiology, icy moons like Europa and Enceladus, and the potential for life in extraterrestrial oceans.

This connects the local discovery to broader celestial search efforts.

Key Concept

Determining Paragraph-Level Main Ideas

Question 18Question →

Match each angle measure in degrees on the left to its equivalent angle measure in radians on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The degree measures

30^\circ

,

45^\circ

,

60^\circ

, and

90^\circ

correspond to

\frac{\pi}{6}

,

\frac{\pi}{4}

,

\frac{\pi}{3}

, and

\frac{\pi}{2}

radians, respectively.

Each degree measure matches its correct radian value by multiplying the degree measure by

\frac{\pi}{180^\circ}

and simplifying the fraction.

Step-by-Step Solution

1

Apply the degree-to-radian conversion formula.

Multiply each degree measure by the conversion factor

\frac{\pi}{180^\circ}

.

A full circle has

360^\circ

or

2\pi

radians, meaning

180^\circ = \pi

radians. Therefore, the conversion factor from degrees to radians is

\frac{\pi}{180^\circ}

.

2

Simplify the resulting fractions.

30^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{6}

,

45^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{4}

,

60^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{3}

, and

90^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{2}

.

Reducing the fractions by dividing the numerator and denominator by their greatest common factor gives the simplified radian values.

Key Concept

Converting degree measures to radian measures on the unit circle

Estimated Time:45s

Question 19Question →

For an angle $\theta$ in standard position, match each description of its terminal side on the left with the corresponding coordinates of its intersection point on the unit circle on the right.

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Answer

The terminal side in Quadrant II with reference angle

60^\circ

matches

(-\frac{1}{2}, \frac{\sqrt{3}}{2})

; in Quadrant III with reference angle

45^\circ

matches

(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2})

; in Quadrant IV with reference angle

30^\circ

matches

(\frac{\sqrt{3}}{2}, -\frac{1}{2})

; in Quadrant III with reference angle

30^\circ

matches

(-\frac{\sqrt{3}}{2}, -\frac{1}{2})

.

Each terminal side is matched correctly to its coordinates by applying the quadrant signs to the trigonometric values of the reference angles. Quadrant II corresponds to

(-, +)

, Quadrant III corresponds to

(-, -)

, and Quadrant IV corresponds to

(+, -)

. Using standard unit circle coordinates, a

60^\circ

reference angle gives magnitudes of

(\frac{1}{2}, \frac{\sqrt{3}}{2})

, a

45^\circ

reference angle gives

(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})

, and a

30^\circ

reference angle gives

(\frac{\sqrt{3}}{2}, \frac{1}{2})

.

Step-by-Step Solution

1

Determine the signs of the

x

- and

y

-coordinates based on the quadrant of the terminal side.

Quadrant II points have

(-, +)

coordinates; Quadrant III points have

(-, -)

coordinates; Quadrant IV points have

(+, -)

coordinates.

On the unit circle,

x = \cos\theta

and

y = \sin\theta

. Cosine is negative in Quadrants II and III, while sine is negative in Quadrants III and IV.

2

Find the absolute values of the coordinates using the reference angle.

A

30^\circ

reference angle corresponds to coordinates of magnitude

(\frac{\sqrt{3}}{2}, \frac{1}{2})

; a

45^\circ

reference angle corresponds to

(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})

; a

60^\circ

reference angle corresponds to

(\frac{1}{2}, \frac{\sqrt{3}}{2})

.

The reference angle determines the basic trigonometric values

\cos\theta_{\text{ref}}

and

\sin\theta_{\text{ref}}

.

3

Combine the quadrant signs and coordinate magnitudes to find the unique point.

Quadrant II with

60^\circ

reference angle is

(-\frac{1}{2}, \frac{\sqrt{3}}{2})

. Quadrant III with

45^\circ

reference angle is

(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2})

. Quadrant IV with

30^\circ

reference angle is

(\frac{\sqrt{3}}{2}, -\frac{1}{2})

. Quadrant III with

30^\circ

reference angle is

(-\frac{\sqrt{3}}{2}, -\frac{1}{2})

.

Applying the quadrant signs from Step 1 to the magnitude values from Step 2 yields the exact coordinates on the unit circle.

Key Concept

Coordinates of points on the unit circle are given by

(\cos\theta, \sin\theta)

, where the magnitude is determined by the reference angle and the signs are determined by the quadrant of the angle.

Question 20Question →

For an angle in standard position on the unit circle, match each rotation scenario on the left with its corresponding terminal angle and location on the right.

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Show answer & explanation

Answer

Matching the scenarios: the wheel rotation matches

\frac{3\pi}{2}

radians on the negative

y

-axis; the combined rotation of

\frac{5\pi}{3}

and

-450^\circ

matches

\frac{7\pi}{6}

radians in Quadrant III; the clockwise travel of

\frac{11\pi}{4}

units matches

\frac{5\pi}{4}

radians in Quadrant III; the combined rotation of

-150^\circ

and

\frac{11\pi}{6}

matches

\pi

radians on the negative

x

-axis.

Each scenario is correctly matched by converting all angular values to radians, determining the net rotation direction (positive for counterclockwise, negative for clockwise), and finding the coterminal angle in the range

[0, 2\pi)

to locate the terminal side.

Step-by-Step Solution

1

Convert the rotation from revolutions to radians for the first scenario.

3.75 \text{ revolutions} \times 2\pi \text{ radians/revolution} = 7.5\pi

radians. Subtract

3

full rotations (

6\pi

radians) to find the coterminal angle in

[0, 2\pi)

:

7.5\pi - 6\pi = 1.5\pi = \frac{3\pi}{2}

radians. This lies on the negative

y

-axis.

One full revolution corresponds to

2\pi

radians, and subtracting multiples of

2\pi

yields the coterminal position.

2

Calculate the net angle in radians for the second scenario.

Convert

450^\circ

to radians:

-450^\circ \times \frac{\pi}{180^\circ} = -\frac{5\pi}{2}

radians (negative due to clockwise direction). Net angle is

\frac{5\pi}{3} - \frac{5\pi}{2} = -\frac{5\pi}{6}

radians. Find the positive coterminal angle:

-\frac{5\pi}{6} + 2\pi = \frac{7\pi}{6}

radians. This lies in Quadrant III.

Converting all angles to radians with correct sign conventions allows addition to find the net angle.

3

Relate arc length to angle measure on the unit circle for the third scenario.

On a circle with

r = 1

, the arc length

s = \frac{11\pi}{4}

corresponds to a rotation of

\frac{11\pi}{4}

radians. Clockwise direction makes it

-\frac{11\pi}{4}

radians. Find the coterminal angle in

[0, 2\pi)

:

-\frac{11\pi}{4} + 4\pi = \frac{5\pi}{4}

radians. This lies in Quadrant III.

The arc length formula

s = r\theta

simplifies to

s = \theta

on the unit circle, and clockwise motion represents a negative angle.

4

Compute the net angle in radians for the fourth scenario.

Convert

-150^\circ

to radians:

-150^\circ \times \frac{\pi}{180^\circ} = -\frac{5\pi}{6}

radians. Net angle is

-\frac{5\pi}{6} + \frac{11\pi}{6} = \frac{6\pi}{6} = \pi

radians. This lies on the negative

x

-axis.

Converting degrees to radians enables direct fraction addition to determine the final terminal position.

Key Concept

Calculating coterminal angles and conversions between degrees, radians, and revolutions on the unit circle.

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