The Language of Algebra: Chapter Test

20 multiple-choice questions and 10 core practice problems, drawn from across the chapter and mixed together.

Multiple choice

20 questions, 100 points in total, 5 points each. Answer in any order and change your mind as often as you like. When you submit, your answers lock and every question shows its worked solution.

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Question 1 of 20

1

Simplify $9b - 4 - 3b + 11$ .

Answer choices for question 1 A $12b + 7$ B $6b + 7$ C $6b - 15$ D $13$
2

Evaluate $121^{1/2}$ .

Answer choices for question 2 A $60.5$ B $\dfrac{1}{11}$ C $242$ D $11$
3

Factor $18x^2 + 30x$ completely.

Answer choices for question 3 A $6x(3x + 5)$ B $6(3x^2 + 5x)$ C $3x(6x + 10)$ D $2x(9x + 15)$
4

Which equation is true for every value of $n$ ?

Answer choices for question 4 A $n + n = n^2$ B $5(n - 2) = 5n - 2$ C $7(n + 2) = 7n + 14$ D $\dfrac{n + 6}{6} = n + 1$
5

For which value of $x$ is $\dfrac{2x - 1}{5x + 20}$ undefined?

Answer choices for question 5 A $x = 4$ B $x = \dfrac{1}{2}$ C $x = -20$ D $x = -4$
6

Simplify $\sqrt{180}$ .

Answer choices for question 6 A $6\sqrt{30}$ B $6\sqrt{5}$ C $36\sqrt{5}$ D $90$
7

Expand $-5a(2a - 3)$ .

Answer choices for question 7 A $-10a^2 + 15a$ B $-10a^2 - 15a$ C $-10a^2 - 3$ D $-10a^2 + 15$
8

Simplify $4(2y - 5) - 3(y - 6)$ .

Answer choices for question 8 A $5y - 38$ B $11y - 2$ C $5y - 2$ D $5y - 26$
9

Evaluate $64^{-2/3}$ .

Answer choices for question 9 A $-16$ B $\dfrac{1}{16}$ C $\dfrac{1}{512}$ D $\dfrac{1}{4096}$
10

Write $\dfrac{4}{x} + \dfrac{3}{5}$ as a single fraction.

Answer choices for question 10 A $\dfrac{7}{5x}$ B $\dfrac{7}{x + 5}$ C $\dfrac{4 + 3x}{5x}$ D $\dfrac{3x + 20}{5x}$
11

A student rewrites $9 + 5m$ as $5m + 9$ , and then rewrites the term $5m$ as $m \cdot 5$ . Which laws license those two steps, in that order?

Answer choices for question 11 A the associative law of addition, then the commutative law of multiplication B the distributive law, then the commutative law of multiplication C the commutative law of addition, then the commutative law of multiplication D the commutative law of addition, then the associative law of multiplication
12

Multiply $(3x - 4)(2x + 5)$ .

Answer choices for question 12 A $6x^2 + 7x - 20$ B $6x^2 - 20$ C $6x^2 + 23x - 20$ D $6x^2 - 7x - 20$
13

Which of these is $45x^3 - 27x^2$ in fully factored form?

Answer choices for question 13 A $3x^2(15x - 9)$ B $9x^2(5x - 3)$ C $9x(5x^2 - 3x)$ D $9(5x^3 - 3x^2)$
14

Which expression takes the same value as $(5x - 2) - (2x - 7) + 3$ at every value of $x$ ?

Answer choices for question 14 A $3x - 6$ B $7x + 8$ C $3x + 5$ D $3x + 8$
15

Simplify $\sqrt{98} - 16^{3/4}\sqrt{2}$ .

Answer choices for question 15 A $5\sqrt{2}$ B $-5\sqrt{2}$ C $-\sqrt{2}$ D $7\sqrt{2} - 8$
16

Expand and completely factor $(x + 6)(x + 2) - (x^2 + 4x)$ .

Answer choices for question 16 A $4(x + 3)$ B $4x + 12$ C $2(2x + 6)$ D $4(2x + 3)$
17

Write $\dfrac{5x + 2}{4x} - \dfrac{x - 6}{2x}$ as a single fraction in lowest terms.

Answer choices for question 17 A $\dfrac{3x - 10}{4x}$ B $\dfrac{3x + 14}{4x}$ C $\dfrac{4x + 8}{4x}$ D $\dfrac{3x + 14}{2x}$
18

Exactly one of these equations fails for some choice of numbers. Which one?

Answer choices for question 18 A $a(bc) = (ab)c$ B $(a + b) + c = c + (a + b)$ C $a(b - c) = ab - ac$ D $a + bc = (a + b)(a + c)$
19

Evaluate $\left(\dfrac{27}{64}\right)^{-2/3}$ .

Answer choices for question 19 A $\dfrac{16}{9}$ B $\dfrac{9}{16}$ C $-\dfrac{16}{9}$ D $\dfrac{4}{3}$
20

Simplify $\sqrt{63} + \sqrt[3]{56} - \sqrt{28}$ .

Answer choices for question 20 A $3\sqrt{7}$ B $\sqrt{91}$ C $\sqrt{7} + 2\sqrt[3]{7}$ D $5\sqrt{7} + 2\sqrt[3]{7}$

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Where you stand

Decide whether two expressions are equivalent, meaning they agree at every input, and refute a claim made about every number with a single value. Review: From Arithmetic to Algebra
Name the law (commutative, associative, identity, inverse, distributive) that licenses a rewriting, and recognize a rewriting no law licenses. Review: From Arithmetic to Algebra
Move between a radical and a fractional exponent, and evaluate $a^{m/n}$ , including a negative exponent, by taking the root of index $n$ before the power $m$ . Review: Fractional Exponents and Radicals
Simplify a radical by pulling out the largest factor of the radicand that is a perfect power for the index, and combine radical terms only when index and radicand both match. Review: Fractional Exponents and Radicals
Combine like terms, and add or subtract whole expressions, sending a leading minus sign onto every term of the expression being subtracted. Review: Arithmetic with Expressions
Expand a product, from one factor across every term of a bracket to every term of one expression against every term of another, then gather like terms. Review: Expanding and Factoring Expressions , Arithmetic with Expressions
Pull out the greatest common factor, the greatest common divisor of the coefficients times each common variable at its lowest power, and judge whether a factorization is complete. Review: Expanding and Factoring Expressions
Simplify, multiply, divide, add and subtract algebraic fractions, canceling only a factor of the whole numerator and the whole denominator, never a term. Review: Algebraic Fractions
Find every value of the variable that makes a denominator zero and report it as a restriction on the expression. Review: Algebraic Fractions

Core practice

10 problems from across the chapter. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A root in a denominator, in a second notation

For $x > 0$ , write $\dfrac{1}{\sqrt[5]{x^2}}$ as a single power of $x$ . Then find the exact value of this expression at $x = 243$ .
1. Hint 1
  A radical and a fractional exponent are two names for the same number, and a negative exponent is how a power records a reciprocal.
2. Hint 2
  Rewrite $\sqrt[5]{x^2}$ first: the index $5$ becomes the denominator of the exponent and the power $2$ under the root becomes its numerator. Moving that power out of the denominator makes its exponent negative.
3. Hint 3
  To evaluate, find the number whose fifth power is $243$ , and take that root before you square.
Answer

$\dfrac{1}{\sqrt[5]{x^2}} = x^{-2/5}$ , and its value at $x = 243$ is $\dfrac{1}{9}$ .

Full solution

The index $5$ of the root becomes the denominator of the exponent, and the power $2$ under the root becomes its numerator.
So
$\sqrt[5]{x^2} = x^{2/5}.$

A power in a denominator moves to the numerator with its exponent made negative, because $a^{-m/n}$ is the reciprocal of $a^{m/n}$ for $a > 0$ , and here $x > 0$ .
$\begin{aligned} \frac{1}{\sqrt[5]{x^2}} &= \frac{1}{x^{2/5}} \\ &= x^{-2/5}. \end{aligned}$

At $x = 243$ , take the fifth root first.
Since $3^5 = 243$ , the fifth root of $243$ is $3$ .
$\begin{aligned} 243^{-2/5} &= \frac{1}{\left(\sqrt[5]{243}\right)^2} \\ &= \frac{1}{3^2} \\ &= \frac{1}{9}. \end{aligned}$

The original radical form gives the same value by the longer route of squaring first: $243^2$ is $59049$ , which is $9^5$ , so $\sqrt[5]{243^2}$ is $9$ and the expression is $\tfrac{1}{9}$ .

Answer

$\dfrac{1}{\sqrt[5]{x^2}} = x^{-2/5}$ , and its value at $x = 243$ is $\dfrac{1}{9}$ .

Key idea

The index of a root becomes the denominator of an exponent, a reciprocal makes the exponent negative, and taking the root before the power keeps the numbers small.
Problem 1: Mark complete
Problem 2 Three terms and what they share

Factor $24x^5 - 40x^4 + 56x^2$ completely, and check your factorization by expanding it.
1. Hint 1
  A factor that every term contains can be written once, in front of a bracket, and the greatest common factor is fully out when the terms left inside share no factor other than $1$ .
2. Hint 2
  Find the greatest common divisor of $24$ , $40$ and $56$ , and the lowest power of $x$ that appears in all three terms. Their product is the largest factor every term contains.
3. Hint 3
  Divide each term by that factor to see what stays inside the bracket, then multiply back out and compare with the original.
Answer

$8x^2(3x^3 - 5x^2 + 7)$ ; expanding it gives back $24x^5 - 40x^4 + 56x^2$ .

Full solution

The largest number that divides $24$ , $40$ and $56$ is $8$ .
The powers of $x$ are $x^5$ , $x^4$ and $x^2$ , and the lowest of them, $x^2$ , is contained in all three terms.
So the greatest common factor is $8x^2$ .

Divide each term by $8x^2$ to find what remains inside the bracket.
$\begin{aligned} 24x^5 &= 8x^2 \cdot 3x^3 \\ -40x^4 &= 8x^2 \cdot (-5x^2) \\ 56x^2 &= 8x^2 \cdot 7 \end{aligned}$

The distributive law, read from right to left, then writes the common factor once, in front.
$\begin{aligned} &24x^5 - 40x^4 + 56x^2 \\ &= 8x^2(3x^3 - 5x^2 + 7). \end{aligned}$

Inside the bracket, the coefficients $3$ , $-5$ and $7$ have no common divisor other than $1$ , and the term $7$ has no $x$ , so the three terms share no factor: the greatest common factor has come out in full.

Expanding checks it: $8x^2 \cdot 3x^3$ is $24x^5$ , $8x^2 \cdot (-5x^2)$ is $-40x^4$ , and $8x^2 \cdot 7$ is $56x^2$ , which are the three original terms.

Answer

$8x^2(3x^3 - 5x^2 + 7)$ ; expanding it gives back $24x^5 - 40x^4 + 56x^2$ .

Key idea

The greatest common factor is the greatest common divisor of the coefficients times each shared variable at its lowest power, and expanding the result back checks the factoring.
Problem 2: Mark complete
Problem 3 A two-term factor times a three-term factor

Expand and simplify $(2x - 5)(x^2 - 3x + 4)$ , and check your result by substituting $x = 2$ into both forms.
1. Hint 1
  Every term of the first factor has to multiply every term of the second, so count how many products there will be before you start.
2. Hint 2
  Multiply $2x$ by each of the three terms of the second factor, then $-5$ by each of them, keeping every sign with its term. That gives six products.
3. Hint 3
  Terms combine only when their variable parts match. Collect the $x^2$ terms together and the $x$ terms together; the $x^3$ term and the constant stand alone.
Answer

$2x^3 - 11x^2 + 23x - 20$ ; at $x = 2$ both forms equal $-2$ .

Full solution

Each of the two terms of $2x - 5$ multiplies all three terms of $x^2 - 3x + 4$ , so there are $2 \times 3 = 6$ products.
Distributing $2x$ gives $2x^3 - 6x^2 + 8x$ , and distributing $-5$ gives $-5x^2 + 15x - 20$ , where $15x$ is positive because $(-5)(-3x)$ is a product of two negatives.

Add the six products and collect like terms.
The $x^2$ terms give $-6x^2 - 5x^2 = -11x^2$ , and the $x$ terms give $8x + 15x = 23x$ .
$\begin{aligned} &(2x - 5)(x^2 - 3x + 4) \\ &= 2x^3 - 11x^2 + 23x - 20. \end{aligned}$

Check at $x = 2$ .
The original is $(4 - 5)(4 - 6 + 4)$ , which is
$(-1) \times 2 = -2.$
The expansion gives
$16 - 44 + 46 - 20 = -2.$
The two forms agree.

Agreement at one value is good evidence rather than proof.
The proof is the expansion itself, since every step used the distributive law, which holds for every number.

Answer

$2x^3 - 11x^2 + 23x - 20$ ; at $x = 2$ both forms equal $-2$ .

Key idea

To multiply two expressions, multiply every term of one by every term of the other, then combine the like terms.
Problem 3: Mark complete
Problem 4 A difference of scaled expressions

Simplify $4(3a^2 - 2a) - 5(a^2 - 4a) - (a^2 + 6a)$ , and write the result in fully factored form.
1. Hint 1
  A minus sign in front of a bracket belongs to the factor outside it, so it reaches every term inside, not just the first one.
2. Hint 2
  Distribute $4$ , then $-5$ , then $-1$ across their brackets, writing each product with its sign. Then collect the $a^2$ terms and the $a$ terms separately.
3. Hint 3
  Once two terms remain, ask what their coefficients share and what power of $a$ both of them contain.
Answer

It simplifies to $6a^2 + 6a$ , which in fully factored form is $6a(a + 1)$ .

Full solution

Distribute each factor across its whole bracket.
The $-5$ reaches both terms of $a^2 - 4a$ , and the lone minus sign is a factor of $-1$ that reaches both terms of $a^2 + 6a$ .
$\begin{aligned} 4(3a^2 - 2a) &= 12a^2 - 8a \\ -5(a^2 - 4a) &= -5a^2 + 20a \\ -(a^2 + 6a) &= -a^2 - 6a \end{aligned}$

Collect the like terms.
The $a^2$ terms have coefficients $12$ , $-5$ and $-1$ , which add to $6$ , and the $a$ terms have coefficients $-8$ , $20$ and $-6$ , which also add to $6$ .
So the whole expression simplifies to $6a^2 + 6a$ .

Both terms contain the factor $6$ and one factor of $a$ , the lowest power of $a$ present, so the greatest common factor is $6a$ .
$6a^2 + 6a = 6a(a + 1).$

Inside the bracket, $a$ and $1$ share no factor, so the factorization is complete.
Expanding confirms it, since $6a \cdot a$ is $6a^2$ and $6a \cdot 1$ is $6a$ .

A check at $a = 2$ : the original expression is $4(8) - 5(-4) - 16$ , which is $32 + 20 - 16 = 36$ , and $6a(a + 1)$ gives
$12 \times 3 = 36.$

Answer

It simplifies to $6a^2 + 6a$ , which in fully factored form is $6a(a + 1)$ .

Key idea

Distribute every factor, a lone minus sign included, across its whole bracket before collecting like terms, and then look for what the remaining terms share.
Problem 4: Mark complete
Problem 5 A quotient of two fractions, and the values it forbids

Write $\dfrac{4x^2 + 8x}{x - 5} \div \dfrac{5x + 10}{2x - 10}$ as a single fraction in lowest terms, and state every value of $x$ at which the original expression is undefined.
1. Hint 1
  A fraction has no value where its denominator is zero, and a division has no value where the quantity being divided by is zero. Both kinds of value are ruled out here.
2. Hint 2
  Division by a fraction is multiplication by its reciprocal. Before multiplying, write $4x^2 + 8x$ , $5x + 10$ and $2x - 10$ as products, so that a factor shared by a top and a bottom can be seen.
3. Hint 3
  Cancel only factors that stand on both the top and the bottom. For the excluded values, set each denominator of the original equal to zero, and do the same for the numerator of the fraction you divide by.
Answer

$\dfrac{8x}{5}$ ; the original expression is undefined at $x = 5$ and at $x = -2$ , so the result holds for every other $x$ .

Full solution

Find the excluded values first.
The denominators $x - 5$ and $2x - 10$ are both zero at $x = 5$ .
The fraction being divided by, $\dfrac{5x + 10}{2x - 10}$ , is zero where its numerator $5x + 10$ is zero, at $x = -2$ , and division by zero has no meaning.
So the original expression is undefined at $x = 5$ and at $x = -2$ .

Write each part as a product by taking out its greatest common factor: $4x^2 + 8x = 4x(x + 2)$ , $5x + 10 = 5(x + 2)$ and
$2x - 10 = 2(x - 5).$

Dividing by a fraction is multiplying by its reciprocal, so turn the divisor over and multiply the tops and the bottoms.
The factors $x + 2$ and $x - 5$ then stand on both the top and the bottom, and neither is zero at an allowed value, so they cancel.
$\begin{aligned} &\frac{4x(x + 2)}{x - 5} \cdot \frac{2(x - 5)}{5(x + 2)} \\ &= \frac{8x(x + 2)(x - 5)}{5(x + 2)(x - 5)} \\ &= \frac{8x}{5}. \end{aligned}$

Check at $x = 1$ , which is allowed.
The first fraction is $\dfrac{12}{-4}$ , which is $-3$ , and the divisor is $\dfrac{15}{-8}$ , so the original is $-3 \cdot \left(-\dfrac{8}{15}\right)$ , which is $\dfrac{24}{15}$ or $\dfrac{8}{5}$ .
The simplified form gives $\dfrac{8(1)}{5}$ , the same value.

The fraction $\dfrac{8x}{5}$ has a value at $x = 5$ and at $x = -2$ , but the original does not, so the two are equal only for $x \neq 5$ and $x \neq -2$ .
The restrictions belong to the original expression and stay with the answer.

Answer

$\dfrac{8x}{5}$ ; the original expression is undefined at $x = 5$ and at $x = -2$ , so the result holds for every other $x$ .

Key idea

When dividing algebraic fractions, the excluded values come from every denominator and from the numerator of the divisor, and they stay excluded after the canceling.
Problem 5: Mark complete
Problem 6 A comet, an asteroid and a three-halves power

Kepler's third law says that an object orbiting the Sun at an average distance of $a$ astronomical units takes $T = a^{3/2}$ years to complete one orbit.

Halley's comet orbits at an average distance of about $18$ astronomical units, and a certain asteroid orbits at an average distance of $2$ astronomical units. Taking both distances as exact, find how many more years the comet takes to complete one orbit than the asteroid does. Give the difference exactly, in simplest radical form, and then to the nearest tenth of a year.
1. Hint 1
  In $a^{3/2}$ the denominator $2$ asks for a square root and the numerator $3$ for a cube. Neither $18$ nor $2$ is a perfect square, so each period will carry a radical.
2. Hint 2
  Take the square root first, and simplify it by splitting off the largest perfect-square factor of the radicand. Then cube it, using $\sqrt{2} \cdot \sqrt{2} = 2$ .
3. Hint 3
  Two radical terms can be combined when they have the same index and the same radicand. Write each period in simplest form and see whether they qualify.
Answer

The comet takes $52\sqrt{2}$ more years, which is about $73.5$ years.

Full solution

The exponent $\tfrac{3}{2}$ means a square root followed by a cube, and taking the root first keeps the numbers small.
For the comet, $18 = 9 \times 2$ with $9$ a perfect square, so $\sqrt{18}$ simplifies to $3\sqrt{2}$ .

The cube of $\sqrt{2}$ is $\sqrt{2} \cdot \sqrt{2} \cdot \sqrt{2}$ , which is $2\sqrt{2}$ , and the cube of $3$ is $27$ .
So cubing $3\sqrt{2}$ gives the comet's period in years.
$\begin{aligned} 18^{3/2} &= \left(3\sqrt{2}\right)^3 \\ &= 27 \cdot 2\sqrt{2} \\ &= 54\sqrt{2}. \end{aligned}$

For the asteroid, $\sqrt{2}$ is already in simplest form, and its cube is $2\sqrt{2}$ , so the asteroid's period is $2\sqrt{2}$ years.

Both periods are whole numbers times $\sqrt{2}$ , so they are like radicals, and the distributive law subtracts their coefficients.
$\begin{aligned} 54\sqrt{2} - 2\sqrt{2} &= (54 - 2)\sqrt{2} \\ &= 52\sqrt{2}. \end{aligned}$

With $\sqrt{2} \approx 1.41421$ , the difference is $52 \times 1.41421$ , about $73.54$ years, which is $73.5$ years to the nearest tenth.
As a check, the two periods are about $76.37$ years and about $2.83$ years, which differ by about $73.54$ years.

Answer

The comet takes $52\sqrt{2}$ more years, which is about $73.5$ years.

Key idea

To evaluate a power like $a^{3/2}$ exactly, take the root first and simplify it; results that share an index and a radicand then combine as like terms.
Problem 6: Mark complete
Problem 7 A longer, narrower car park

A rectangular car park is $3x + 5$ meters long and $x + 6$ meters wide, where $x > 0$ . It is redesigned to be $2$ meters longer and $1$ meter narrower.

Write the change in area, the redesigned car park's area minus the original car park's area, as a simplified expression in $x$ , in square meters. Then decide whether the redesigned car park has more or less area than the original when $x = 7$ , and by how much.
1. Hint 1
  Each car park's area is its length times its width, and the change is one whole area taken away from the other, so the minus sign has to reach every term of the area being taken away.
2. Hint 2
  The redesigned car park is $3x + 7$ meters long and $x + 5$ meters wide. Expand both areas, multiplying every term of one factor by every term of the other.
3. Hint 3
  After subtracting, collect the $x^2$ terms, the $x$ terms and the constants separately. A negative change means the redesigned car park has less area.
Answer

The change is $5 - x$ square meters (equivalently $-x + 5$ ). When $x = 7$ it is $-2$ : the redesigned car park has $336$ square meters, $2$ square meters less than the original $338$ .

Full solution

The original car park is $3x + 5$ meters by $x + 6$ meters.
Two meters longer makes the redesigned car park $3x + 7$ meters long, and one meter narrower makes it $x + 5$ meters wide, which is positive since $x > 0$ .

Expand each area, multiplying every term by every term.
The redesigned area $(3x + 7)(x + 5)$ has the four products $3x^2$ , $15x$ , $7x$ and $35$ , so it is $3x^2 + 22x + 35$ .
The original area $(3x + 5)(x + 6)$ has the four products $3x^2$ , $18x$ , $5x$ and $30$ , so it is $3x^2 + 23x + 30$ .

Subtract the whole original area, so that the minus sign reaches all three of its terms and turns them into $-3x^2$ , $-23x$ and $-30$ .
Collecting like terms, $3x^2 - 3x^2$ is $0$ , $22x - 23x$ is $-x$ , and $35 - 30$ is $5$ , so the change in area is $5 - x$ square meters.

When $x = 7$ the change is $5 - 7$ , which is $-2$ , so the redesigned car park has $2$ square meters less area than the original.

Directly, at $x = 7$ the original car park is $26$ meters by $13$ meters, with area $338$ square meters, and the redesigned car park is $28$ meters by $12$ meters, with area $336$ square meters.
The difference is $-2$ square meters, which matches.

The expression also shows when the redesign gains area: $5 - x$ is positive when $x$ is less than $5$ , zero at $x = 5$ , and negative when $x$ is greater than $5$ .
Adding length and removing width can move the area either way.

Answer

The change is $5 - x$ square meters (equivalently $-x + 5$ ). When $x = 7$ it is $-2$ : the redesigned car park has $336$ square meters, $2$ square meters less than the original $338$ .

Key idea

Subtract a whole expanded area term by term, sending the minus sign onto every term, and read the sign of the result to see whether the area grew or shrank.
Problem 7: Mark complete
Problem 8 Crossing out the $x^2$

A student simplifies $\dfrac{x^2 + 6x}{x^2 + 2x}$ by crossing out the $x^2$ on the top and the $x^2$ on the bottom, which leaves $\dfrac{6x}{2x}$ , and then writes the answer $3$ .

Decide whether the student's answer equals the original fraction at every value of $x$ where the fraction is defined, and support your decision. Then write the fraction in simplest form, and state every value of $x$ that must be excluded.
1. Hint 1
  A claim that two expressions agree at every allowed value is overturned by a single allowed value where they disagree.
2. Hint 2
  Canceling divides the whole top and the whole bottom by the same factor. Ask whether $x^2$ is a factor of all of $x^2 + 6x$ or a term added to $6x$ , and test the student's answer at a small allowed value such as $x = 1$ .
3. Hint 3
  For the correct simplification, write the top and the bottom as products by taking out what their terms share, and find where the original bottom is zero before you cancel anything.
Answer

It does not: at $x = 1$ the original is $\tfrac{7}{3}$ , not $3$ (any allowed $x$ shows it). The correct form is $\dfrac{x + 6}{x + 2}$ , for $x \neq 0$ and $x \neq -2$ .

Full solution

Test the student's answer at an allowed value, say $x = 1$ .
The original fraction gives $\dfrac{1 + 6}{1 + 2}$ , which is $\dfrac{7}{3}$ , while the student's answer is $3$ .
One allowed value where the two disagree is enough to show they are not equal at every allowed value, so the student's answer is wrong.

The crossing out is where it went wrong.
Canceling divides the whole top and the whole bottom by a factor they share, and $x^2$ is not a factor of the whole top: it is a term, added to $6x$ .
The same is true on the bottom, so removing the two $x^2$ terms changes the value of the fraction.

Find the excluded values from the original bottom first.
Both of its terms contain $x$ , so it factors as $x(x + 2)$ , which is zero at $x = 0$ and at $x = -2$ .
Both values are excluded.

To simplify, write the top as $x(x + 6)$ in the same way.
The factor $x$ is then shared by the whole top and the whole bottom, and it is not zero at any allowed value, so it cancels.
$\begin{aligned} \frac{x^2 + 6x}{x^2 + 2x} &= \frac{x(x + 6)}{x(x + 2)} \\ &= \frac{x + 6}{x + 2}. \end{aligned}$

Nothing more cancels, because $x + 6$ and $x + 2$ share no factor: the $6$ and the $2$ are terms, not factors.
The simplified form has a value at $x = 0$ , but the original fraction does not, so both restrictions stay with the answer.

At $x = 1$ the correct form gives $\tfrac{7}{3}$ , matching the original.
It could equal $3$ only where $x + 6$ equals $3(x + 2)$ , which is $3x + 6$ , and those differ by $2x$ , which is $0$ only at the excluded value $x = 0$ .
So the student's answer is wrong at every allowed value, not just at $x = 1$ .

Answer

It does not: at $x = 1$ the original is $\tfrac{7}{3}$ , not $3$ (any allowed $x$ shows it). The correct form is $\dfrac{x + 6}{x + 2}$ , for $x \neq 0$ and $x \neq -2$ .

Key idea

Only a factor of the whole top and the whole bottom cancels, never a term, and the excluded values come from the original fraction, because a canceled factor can hide one.
Problem 8: Mark complete
Problem 9 Three rewritings and a classmate's claim

A student rewrites $x \cdot 7 + 7$ in three steps.
$\begin{aligned} &x \cdot 7 + 7 \\ &= 7x + 7 \\ &= 7x + 7 \cdot 1 \\ &= 7(x + 1) \end{aligned}$
Name the law of arithmetic that licenses each of the three steps. A classmate then claims that $7(2x) = (7 \cdot 2)(7x)$ for every number $x$ , because the $7$ must reach every part of what it multiplies. Decide whether that claim is true for every $x$ , and support your decision.
1. Hint 1
  Each law is a statement true for every number, and a step is licensed when it is one instance of a law. A claim about every number fails as soon as one value makes its two sides differ.
2. Hint 2
  For each of the three steps, look at what changed: two factors swapped places, a number was written as itself times $1$ , or a factor that two terms share was written once in front of their sum.
3. Hint 3
  Try a nonzero value, such as $x = 1$ , in both sides of the classmate's claim. Then notice that $2x$ is a product of the two numbers $2$ and $x$ , with no sum anywhere in it.
Answer

Step 1: commutative law of multiplication; step 2: identity law of multiplication, read right to left; step 3: distributive law, read right to left. The claim is false: at $x = 1$ its sides are $14$ and $98$ .

Full solution

The first step swaps the order of the two factors in $x \cdot 7$ and leaves the added $7$ alone.
That is the commutative law of multiplication, $ab = ba$ , with $a = x$ and $b = 7$ .

The second step writes the term $7$ as $7 \cdot 1$ .
That is the identity law of multiplication, $a \cdot 1 = a$ with $a = 7$ , read from right to left: multiplying by $1$ leaves a number unchanged, so $7$ and $7 \cdot 1$ are the same number.

The third step writes the factor $7$ , which both terms now show, once in front of their sum.
That is the distributive law, $a(b + c) = ab + ac$ with $a = 7$ , $b = x$ and $c = 1$ , read from right to left.

Now the classmate's claim.
At $x = 1$ the left side is $7(2)$ , which is $14$ , and the right side is $(7 \cdot 2)(7)$ , which is $98$ .
The two sides differ, so the claim fails at $x = 1$ , and a claim about every number is false once a single value breaks it.
At $x = 0$ both sides happen to be $0$ , so that value would not have shown anything.

No law licenses the classmate's step.
The distributive law spreads a factor across the terms of a sum, and $2x$ is not a sum: it is the product of $2$ and $x$ , so the $7$ multiplies it once.
Sending the $7$ to both factors multiplies by $7$ twice, which makes the right side $7$ times the left side at every $x$ .

Answer

Step 1: commutative law of multiplication; step 2: identity law of multiplication, read right to left; step 3: distributive law, read right to left. The claim is false: at $x = 1$ its sides are $14$ and $98$ .

Key idea

A factor spreads across the terms of a sum, not across the factors of a product, and one value where the two sides differ refutes a claim about every number.
Problem 9: Mark complete
Problem 10 A common factor in two letters

Jamal factors $20a^3b^2 - 30a^2b^3 + 10a^2b^2$ and writes $10ab(2a^2b - 3ab^2 + ab)$ .

Decide whether Jamal's factorization is complete, and if it is not, complete it. Then explain why no factor common to all three terms can contain $a^3$ or $b^3$ .
1. Hint 1
  A factor common to all three terms has to divide each of them, and the greatest common factor is fully out when the terms left inside the bracket share no factor other than $1$ .
2. Hint 2
  Look at the three terms inside Jamal's bracket, $2a^2b$ , $-3ab^2$ and $ab$ , and find what all of them still contain.
3. Hint 3
  For the last part, compare the power of $a$ in each of the three original terms, and then the power of $b$ . A power of a letter can be shared only if every term contains at least that many factors of it.
Answer

Jamal's factorization is not complete. The complete factorization is $10a^2b^2(2a - 3b + 1)$ , and the lowest powers of $a$ and $b$ in the three terms are $a^2$ and $b^2$ .

Full solution

The greatest common divisor of $20$ , $30$ and $10$ is $10$ .
The powers of $a$ in the three terms are $a^3$ , $a^2$ and $a^2$ , and the powers of $b$ are $b^2$ , $b^3$ and $b^2$ .
The lowest of each is contained in every term, so the greatest common factor is $10a^2b^2$ .

Jamal took out $10ab$ : the right number, but only one factor of each letter.
The terms inside his bracket, $2a^2b$ , $-3ab^2$ and $ab$ , all still contain $a$ and $b$ , so they share the factor $ab$ , and his factorization is not complete.

Taking that shared $ab$ out as well finishes the job.
$\begin{aligned} &10ab(2a^2b - 3ab^2 + ab) \\ &= 10ab \cdot ab(2a - 3b + 1) \\ &= 10a^2b^2(2a - 3b + 1). \end{aligned}$

Check it term by term: $10a^2b^2 \cdot 2a$ is $20a^3b^2$ , $10a^2b^2 \cdot (-3b)$ is $-30a^2b^3$ , and $10a^2b^2 \cdot 1$ is $10a^2b^2$ .
The terms left inside, $2a$ , $-3b$ and $1$ , share no factor other than $1$ , so the greatest common factor has now come out in full.

No common factor can contain $a^3$ , because a common factor has to divide every term, and $-30a^2b^3$ and $10a^2b^2$ each contain only two factors of $a$ .
In the same way, $20a^3b^2$ and $10a^2b^2$ contain only two factors of $b$ , so no common factor can contain $b^3$ .
That is why each letter in the greatest common factor appears at its lowest power.

Answer

Jamal's factorization is not complete. The complete factorization is $10a^2b^2(2a - 3b + 1)$ , and the lowest powers of $a$ and $b$ in the three terms are $a^2$ and $b^2$ .

Key idea

Each letter in the greatest common factor appears at its lowest power among the terms, and that factor is fully out only when the terms left inside share no factor other than $1$ .
Problem 10: Mark complete

Star problems Ten optional challenges to stretch your reasoning Advanced. This problem set goes beyond core Algebra I. You can skip it. → Continue to Chapter 2: Linear Equations →

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