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Expanding and Factoring Expressions: Core practice

10 practice problems for this lesson. 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 One factor, three terms

Expand $-4x(2x^2 - 3x + 5)$ , and check your expansion by evaluating both it and the original product at $x = 2$ .
1. Hint 1
  The factor outside the parentheses has to reach every term inside, all three of them, and its minus sign goes with it into each product.
2. Hint 2
  Work out each product in two pieces: multiply the numbers, signs included, then add the exponents where $x$ meets a power of $x$ .
3. Hint 3
  The middle product is $(-4x)(-3x)$ , and two negative coefficients multiply to a positive one. At $x = 2$ , work out the parentheses first in the original.
Answer

$-8x^3 + 12x^2 - 20x$ ; check: at $x = 2$ both forms equal $-56$ .

Full solution

The factor $-4x$ has to multiply each of the three terms inside the parentheses, and its minus sign goes with it into every product.
In each product, multiply the numbers with their signs, then add the exponents of $x$ :
$\begin{aligned} (-4x)(2x^2) &= -8x^3 \\ (-4x)(-3x) &= 12x^2 \\ (-4x)(5) &= -20x. \end{aligned}$

The middle product is positive because its coefficients, $-4$ and $-3$ , are both negative, and their product is $12$ .
Adding the three products gives the expansion
$-8x^3 + 12x^2 - 20x.$

Check at $x = 2$ .
In the original, the parentheses give $2(4) - 3(2) + 5$ , which is $7$ , and the factor outside is $-4(2)$ , which is $-8$ , so the product is $-56$ .
In the expansion, the three terms at $x = 2$ are $-64$ , $48$ and $-40$ , which also add to $-56$ .

One agreeing value does not prove the two forms equal for every $x$ , but a disagreement would have exposed a slip.
A value such as $x = 2$ is a better test than $x = 1$ , where every power of $x$ equals $1$ and a wrong exponent would go unnoticed.

Answer

$-8x^3 + 12x^2 - 20x$ ; check: at $x = 2$ both forms equal $-56$ .

Key idea

A factor outside the parentheses multiplies every term inside, carrying its sign into each product and adding exponents where powers of $x$ meet.
Problem 1: Mark complete
Problem 2 Two binomials

Multiply $(3x - 7)(2x + 5)$ and simplify, then check your result by evaluating both it and the original product at $x = 3$ .
1. Hint 1
  Each term of the first factor must multiply each term of the second, so the product starts out as four separate pieces.
2. Hint 2
  FOIL is one way to be sure all four are there: First, Outer, Inner, Last. Keep the minus sign with the $7$ in both products it enters.
3. Hint 3
  Two of the four products are $x$ terms with opposite signs. Add their coefficients, signs included, before you write the result.
Answer

$6x^2 + x - 35$ ; check: at $x = 3$ both forms equal $22$ .

Full solution

Every term of $3x - 7$ multiplies every term of $2x + 5$ , which makes four products.
FOIL names them First, Outer, Inner and Last:
$\begin{aligned} (3x)(2x) &= 6x^2 \\ (3x)(5) &= 15x \\ (-7)(2x) &= -14x \\ (-7)(5) &= -35. \end{aligned}$

The outer and inner products, $15x$ and $-14x$ , are like terms.
Their coefficients add to $1$ , so together they make $1x$ , which is written $x$ .
The product is
$6x^2 + x - 35.$

Check at $x = 3$ .
The first factor is $3(3) - 7$ , which is $2$ , and the second is $2(3) + 5$ , which is $11$ , so the original product is $22$ .
The expansion gives $6(9) + 3 - 35$ , which is also $22$ .

Answer

$6x^2 + x - 35$ ; check: at $x = 3$ both forms equal $22$ .

Key idea

Multiplying two binomials makes four products, one for each pairing of terms; FOIL is a way of listing them, and the like terms among them are combined at the end.
Problem 2: Mark complete
Problem 3 The largest shared factor

Pull the greatest common factor out of $18x^4 - 12x^3 + 6x^2$ , and check your factoring by expanding it.
1. Hint 1
  The greatest common factor has a number part and a variable part: the greatest common divisor of the coefficients, and the smallest power of $x$ that appears.
2. Hint 2
  The coefficients $18$ , $12$ and $6$ all divide by $6$ , and every term contains at least $x^2$ . For each term, ask what the common factor must be multiplied by to give that term: that is what goes in the parentheses.
3. Hint 3
  One term is exactly the factor you are pulling out, and the factor times $1$ gives it back. So that term leaves $1$ in the parentheses, not nothing.
Answer

$6x^2(3x^2 - 2x + 1)$ ; expanding it returns $18x^4 - 12x^3 + 6x^2$ .

Full solution

The number part of the greatest common factor is the greatest common divisor of $18$ , $12$ and $6$ , which is $6$ .
The variable part is the smallest power of $x$ that appears, which is $x^2$ , and each of $x^4$ , $x^3$ and $x^2$ has $x^2$ as a factor.
So the greatest common factor is $6x^2$ .

Write each term as $6x^2$ times something:
$\begin{aligned} 18x^4 &= 6x^2 \cdot 3x^2 \\ -12x^3 &= 6x^2 \cdot (-2x) \\ 6x^2 &= 6x^2 \cdot 1. \end{aligned}$

The distributive law read from right to left lifts the shared $6x^2$ out front and leaves the other factors inside:
$6x^2(3x^2 - 2x + 1).$
The last term leaves $1$ behind.
Dropping it would give $6x^2(3x^2 - 2x)$ , which expands to only two terms.

Check by expanding: $6x^2$ times $3x^2$ , $-2x$ and $1$ gives $18x^4$ , $-12x^3$ and $6x^2$ , the three terms of the original.
The terms left inside, $3x^2$ , $-2x$ and $1$ , share no number factor greater than $1$ and no factor of $x$ , so nothing more can be pulled out.

Answer

$6x^2(3x^2 - 2x + 1)$ ; expanding it returns $18x^4 - 12x^3 + 6x^2$ .

Key idea

The greatest common factor is the greatest common divisor of the coefficients times each shared variable at its smallest power, and a term equal to that factor leaves $1$ in the parentheses.
Problem 3: Mark complete
Problem 4 A binomial times a trinomial

Multiply $(4x - 3)(x^2 + 2x - 6)$ and simplify, then check your result by evaluating both it and the original product at $x = -1$ .
1. Hint 1
  FOIL lists the four products of two two-term factors, but the second factor here has three terms. Go back to the rule FOIL names: every term of one factor times every term of the other.
2. Hint 2
  Distribute each term of $4x - 3$ across the whole trinomial: first $4x$ times all three terms, then $-3$ times all three terms. That makes six products.
3. Hint 3
  Two pairs of the six products are like terms, the two $x^2$ terms and the two $x$ terms. At $x = -1$ , remember that $(-1)^2 = 1$ and $(-1)^3 = -1$ .
Answer

$4x^3 + 5x^2 - 30x + 18$ ; check: at $x = -1$ both forms equal $49$ .

Full solution

The second factor has three terms, so FOIL, which lists the four products of two two-term factors, cannot cover this product.
The rule it names still applies: each of the two terms of $4x - 3$ multiplies each of the three terms of $x^2 + 2x - 6$ , which makes $2 \times 3 = 6$ products.

Distributing $4x$ across the trinomial gives $4x^3$ , $8x^2$ and $-24x$ .
Distributing $-3$ gives $-3x^2$ , $-6x$ and $18$ , where the last is positive because $(-3)(-6)$ is a product of two negatives.

Now collect like terms.
The $x^2$ terms and the $x$ terms combine:
$\begin{aligned} 8x^2 - 3x^2 &= 5x^2 \\ -24x - 6x &= -30x. \end{aligned}$
The $x^3$ term and the constant have no partners, so the product is
$4x^3 + 5x^2 - 30x + 18.$

Check at $x = -1$ .
The first factor is $4(-1) - 3$ , which is $-7$ , and the second is $1 - 2 - 6$ , which is also $-7$ , so the original product is $49$ .
In the expansion, $(-1)^3 = -1$ and $(-1)^2 = 1$ , so it gives $-4 + 5 + 30 + 18$ , which is also $49$ .

Answer

$4x^3 + 5x^2 - 30x + 18$ ; check: at $x = -1$ both forms equal $49$ .

Key idea

The number of products is the number of terms in one factor times the number in the other, so a two-term factor times a three-term factor gives $2 \times 3 = 6$ , not FOIL's four.
Problem 4: Mark complete
Problem 5 Two letters, then a common factor

Expand and simplify $(x + 2y)(3x - y) - x(3x + 4y)$ , then factor the result by pulling out its greatest common factor.
1. Hint 1
  Expand each product on its own first. The minus sign in front of $x(3x + 4y)$ belongs to that whole product, so it changes the sign of every term the product makes.
2. Hint 2
  In the first product, $x \cdot (-y)$ and $2y \cdot 3x$ give like terms, because $yx$ and $xy$ are the same product.
3. Hint 3
  After the subtraction, look for terms that cancel. Then ask which letter every remaining term contains, and to what smallest power.
Answer

$xy - 2y^2$ , which factors as $y(x - 2y)$ .

Full solution

Expand the first product, every term times every term.
Multiplying $x$ by each term of $3x - y$ gives $3x^2$ and $-xy$ , and multiplying $2y$ by each term gives $6xy$ and $-2y^2$ .

The middle two are like terms, since $2y \cdot 3x$ is $6yx$ and $yx$ is the same as $xy$ .
Combining $-xy$ and $6xy$ gives $5xy$ , so the first product is $3x^2 + 5xy - 2y^2$ .

The second product expands to $3x^2 + 4xy$ .
It is subtracted as a whole, so both of its terms change sign:
$\begin{aligned} &3x^2 + 5xy - 2y^2 - 3x^2 - 4xy \\ &= xy - 2y^2. \end{aligned}$
The $3x^2$ terms cancel, and $5xy - 4xy$ leaves $xy$ .

Both remaining terms contain $y$ , and the smallest power of $y$ that appears is $y$ itself, in $xy$ .
The coefficients $1$ and $-2$ share no number factor greater than $1$ , and $x$ is missing from $-2y^2$ .
So the greatest common factor is $y$ , and
$xy - 2y^2 = y(x - 2y).$

Check by expanding: $y$ times $x$ is $xy$ and $y$ times $-2y$ is $-2y^2$ .
A spot check at $x = 3$ and $y = 2$ agrees too: the original is $7 \cdot 7 - 3 \cdot 17$ , which is $-2$ , and $y(x - 2y)$ is $2 \cdot (-1)$ , also $-2$ .

Answer

$xy - 2y^2$ , which factors as $y(x - 2y)$ .

Key idea

Expand every product first, then collect like terms; the terms that survive can still share a factor worth pulling out.
Problem 5: Mark complete
Problem 6 Raising the rental price

A bike rental shop rents out $40$ bikes a day at $15$ dollars each. Each time the owner raises the price by $1$ dollar, the shop rents $2$ fewer bikes a day. Let $n$ be the number of one-dollar price rises, a whole number from $0$ to $20$ .

Write the shop's daily takings after $n$ price rises as a product of two expressions in $n$ , then expand and simplify it. Say what the constant term of your expansion tells the owner.
1. Hint 1
  Takings are the price of one rental times the number of rentals. Write each of those two quantities in terms of $n$ before multiplying.
2. Hint 2
  After $n$ rises the price has gone up by $n$ dollars, and the number of bikes rented has gone down by $2$ for each rise. Multiply the two expressions, every term by every term.
3. Hint 3
  The constant term is what the expansion equals when $n = 0$ . Ask what $n = 0$ means for the shop.
Answer

The takings are $(15 + n)(40 - 2n)$ dollars, which expands to $600 + 10n - 2n^2$ (or $-2n^2 + 10n + 600$ ). The constant term $600$ is the takings in dollars with no price rise.

Full solution

Takings are the price of one rental times the number of rentals.
After $n$ one-dollar rises the price is $15 + n$ dollars, and the shop rents $2n$ fewer bikes, so it rents $40 - 2n$ .
The daily takings, in dollars, are
$(15 + n)(40 - 2n).$

Expand, every term times every term:
$\begin{aligned} 15 \cdot 40 &= 600 \\ 15 \cdot (-2n) &= -30n \\ n \cdot 40 &= 40n \\ n \cdot (-2n) &= -2n^2. \end{aligned}$

The two $n$ terms are like terms, and $-30n$ and $40n$ combine to $10n$ .
So the takings are $600 + 10n - 2n^2$ dollars, which can also be written $-2n^2 + 10n + 600$ .

The constant term, $600$ , is what the expansion equals when $n = 0$ , that is, with no price rise.
It is the takings at the original price: $40$ bikes at $15$ dollars each bring in $600$ dollars.

A spot check at $n = 3$ : the product is $18 \cdot 34$ , which is $612$ , and the expansion gives $600 + 30 - 18$ , also $612$ .

Answer

The takings are $(15 + n)(40 - 2n)$ dollars, which expands to $600 + 10n - 2n^2$ (or $-2n^2 + 10n + 600$ ). The constant term $600$ is the takings in dollars with no price rise.

Key idea

The constant term of an expanded expression is its value when the variable is $0$ ; in this model, that is the takings before any price rise.
Problem 6: Mark complete
Problem 7 One bed in place of two

A gardener has two rectangular flower beds, each $3x$ meters wide, where $x$ is positive. One bed is $4x + 5$ meters long and the other is $x + 3$ meters long.

Find the total area of the two beds as a simplified expression in $x$ . Then factor that expression using its greatest common factor, and use the factored form to find how long a single rectangular bed, also $3x$ meters wide, must be to have the same total area.
1. Hint 1
  A rectangle's area is its width times its length, and a product in factored form can be read the same way, as a width times a length.
2. Hint 2
  Multiply $3x$ into every term of each bed's length, then add the two areas and collect like terms. For the greatest common factor, take the greatest common divisor of the two coefficients and the smallest power of $x$ in the two terms.
3. Hint 3
  A bed $3x$ meters wide and $L$ meters long has area $3x \cdot L$ square meters. Set your factored form beside that product and compare the factors.
Answer

The total area is $15x^2 + 24x$ square meters, which factors as $3x(5x + 8)$ . The single bed must be $5x + 8$ meters long.

Full solution

Each bed's area is its width times its length, and the factor $3x$ has to reach every term of each length:
$\begin{aligned} 3x(4x + 5) &= 12x^2 + 15x \\ 3x(x + 3) &= 3x^2 + 9x. \end{aligned}$

Add the two areas and collect like terms: $12x^2$ and $3x^2$ make $15x^2$ , and $15x$ and $9x$ make $24x$ .
So the total area is $15x^2 + 24x$ square meters.

The greatest common divisor of $15$ and $24$ is $3$ , and both terms contain $x$ , whose smallest power there is $x$ itself, so the greatest common factor is $3x$ .
Writing each term as $3x$ times something, $15x^2$ is $3x \cdot 5x$ and $24x$ is $3x \cdot 8$ , so
$\begin{aligned} &15x^2 + 24x \\ &= 3x(5x + 8). \end{aligned}$

Check by expanding: $3x \cdot 5x$ is $15x^2$ and $3x \cdot 8$ is $24x$ , the two terms of the total.
Inside the parentheses, $5x$ and $8$ share no number factor greater than $1$ , and $8$ has no factor of $x$ , so nothing more can be pulled out.

A bed $3x$ meters wide and $L$ meters long has area $3x \cdot L$ square meters.
The factored form writes the total area as $3x$ times $5x + 8$ , so a bed $5x + 8$ meters long has exactly the total area.
No other length does, because $3x$ is not zero when $x$ is positive.

The same length comes straight from the distributive law, since both beds are $3x$ meters wide: $3x(4x + 5) + 3x(x + 3)$ is $3x$ times the sum of the two lengths, and $(4x + 5) + (x + 3)$ is $5x + 8$ .

Check with $x = 2$ .
The beds are $6$ by $13$ and $6$ by $5$ meters, with areas $78$ and $30$ square meters, $108$ in all.
The expanded form gives $15(4) + 24(2)$ , which is $108$ , and a single bed $6$ meters wide and $18$ meters long also covers $108$ square meters.

Answer

The total area is $15x^2 + 24x$ square meters, which factors as $3x(5x + 8)$ . The single bed must be $5x + 8$ meters long.

Key idea

An area in factored form reads as width times length, so pulling a known width out as a factor leaves the matching length inside the parentheses.
Problem 7: Mark complete
Problem 8 A pattern on the calendar

On a monthly calendar laid out in rows of seven days, each date sits directly above the date one week later. Choose any $2 \times 2$ block of four dates from the same month. Multiply the top-right date by the bottom-left date, multiply the top-left date by the bottom-right date, and subtract the second product from the first.

Writing $n$ for the top-left date, express this difference in terms of $n$ and simplify it. Then explain what your result shows about every such block.
1. Hint 1
  Name all four dates in terms of $n$ : one place to the right adds $1$ to a date, and one row down adds $7$ .
2. Hint 2
  The two products are $(n + 1)(n + 7)$ and $n(n + 8)$ . Expand both, every term times every term, and subtract the second as a whole.
3. Hint 3
  Once the difference is simplified, look at whether any $n$ is left in it, and what that means for the choice of block.
Answer

The difference is $(n + 1)(n + 7) - n(n + 8)$ , which simplifies to $7$ : every $2 \times 2$ block gives the same difference, $7$ .

Full solution

One place to the right on the calendar adds $1$ to the date, and one row down adds $7$ .
So with $n$ top-left, the top row holds $n$ and $n + 1$ , and the bottom row holds $n + 7$ and $n + 8$ .

The top-right date times the bottom-left date is $(n + 1)(n + 7)$ , and the top-left date times the bottom-right date is $n(n + 8)$ .
The difference is
$(n + 1)(n + 7) - n(n + 8).$

Expand the first product, every term times every term:
$\begin{aligned} &(n + 1)(n + 7) \\ &= n^2 + 7n + n + 7 \\ &= n^2 + 8n + 7. \end{aligned}$

The second product expands to $n^2 + 8n$ .
It is subtracted as a whole, so both of its terms change sign:
$\begin{aligned} &n^2 + 8n + 7 - n^2 - 8n \\ &= 7. \end{aligned}$

The $n^2$ terms cancel and the $8n$ terms cancel, so no $n$ is left.
The equation $(n + 1)(n + 7) - n(n + 8) = 7$ is true for every number $n$ , so every $2 \times 2$ block on the calendar gives the difference $7$ , whichever date is top-left.

For example, the block with $8$ top-left holds $8$ , $9$ , $15$ and $16$ , and $9 \cdot 15 - 8 \cdot 16$ is $135 - 128$ , which is $7$ .
Trying blocks one at a time shows the pattern only for the blocks tried; the expansion shows it for all of them at once.

Answer

The difference is $(n + 1)(n + 7) - n(n + 8)$ , which simplifies to $7$ : every $2 \times 2$ block gives the same difference, $7$ .

Key idea

Expanding an expression written with a letter can show that a numerical pattern holds for every case at once, not just for the examples tried.
Problem 8: Mark complete
Problem 9 Jordan's four products

Jordan multiplies $(x - 4)(x^2 + 5x + 2)$ using FOIL. He takes the First terms $x \cdot x^2$ , the Outer terms $x \cdot 2$ , the Inner terms $(-4) \cdot x^2$ and the Last terms $(-4) \cdot 2$ , and writes the answer $x^3 - 4x^2 + 2x - 8$ .

Test Jordan's answer against the original product at $x = 2$ , and find the correct product. Then explain which products Jordan missed and why FOIL could not have found them.
1. Hint 1
  FOIL is a name for the four products of two two-term factors. Count the terms in each of Jordan's factors before trusting it.
2. Hint 2
  At $x = 2$ , work out each factor of the original product on its own and multiply them; then work out Jordan's four terms at $x = 2$ and compare.
3. Hint 3
  Every term of $x - 4$ must multiply every term of $x^2 + 5x + 2$ , which makes six products. Look for the two that Jordan never formed: both involve the same term of the trinomial.
Answer

At $x = 2$ the original product is $-32$ but Jordan's answer gives $-12$ . The correct product is $x^3 + x^2 - 18x - 8$ . Jordan missed $5x^2$ and $-20x$ , the products with $5x$ : FOIL names four products, and this multiplication needs six.

Full solution

Test first.
At $x = 2$ the first factor is $2 - 4$ , which is $-2$ , and the second is $4 + 10 + 2$ , which is $16$ , so the original product is $-32$ .
Jordan's answer gives $8 - 16 + 4 - 8$ , which is $-12$ .
The values differ, so Jordan's answer is not equal to the product.

Now multiply correctly.
Every term of $x - 4$ multiplies every term of $x^2 + 5x + 2$ , which makes $2 \times 3 = 6$ products.
Distributing $x$ gives $x^3$ , $5x^2$ and $2x$ , and distributing $-4$ gives $-4x^2$ , $-20x$ and $-8$ .

Collect like terms: $5x^2$ and $-4x^2$ make $x^2$ , and $2x$ and $-20x$ make $-18x$ .
So the product is
$x^3 + x^2 - 18x - 8.$
At $x = 2$ this gives $8 + 4 - 36 - 8$ , which is $-32$ , matching the original product.

Jordan's four products are all correct, but two of the six are missing: $x$ times $5x$ , which is $5x^2$ , and $-4$ times $5x$ , which is $-20x$ .
Both come from the middle term $5x$ of the trinomial.

FOIL names four products, first, outer, inner and last, one for each pairing when both factors have two terms.
With three terms in the second factor there are six pairings, and the words first and last pick out only its end terms, so the middle term $5x$ is never multiplied.
Jordan's answer is in fact what FOIL gives for $(x - 4)(x^2 + 2)$ , the product with the $5x$ deleted.

Answer

At $x = 2$ the original product is $-32$ but Jordan's answer gives $-12$ . The correct product is $x^3 + x^2 - 18x - 8$ . Jordan missed $5x^2$ and $-20x$ , the products with $5x$ : FOIL names four products, and this multiplication needs six.

Key idea

FOIL is a name for the four products of two binomials; when a factor has more terms, every term times every term finds all of the products.
Problem 9: Mark complete
Problem 10 Three attempts at one factoring

Three students factor $16x^3 + 40x^2$ . Ana writes $4x(4x^2 + 10x)$ , Ben writes $8x^2(2x + 5)$ , and Cal writes $8x(2x^2 + 5)$ .

Expand each answer and say which ones are equal to $16x^3 + 40x^2$ . Then say which answer pulls out the greatest common factor, and explain why expanding alone could not settle that second question.
1. Hint 1
  Expand each answer by multiplying the factor outside into every term inside, then compare the result with $16x^3 + 40x^2$ term by term.
2. Hint 2
  For the greatest common factor, find the greatest common divisor of $16$ and $40$ , and the smallest power of $x$ that appears in the two terms.
3. Hint 3
  Two answers expand correctly. Look inside their parentheses: do the terms there still share a number factor greater than $1$ or a factor of $x$ , and would expanding ever reveal that?
Answer

Ana's and Ben's expand to $16x^3 + 40x^2$ ; Cal's gives $16x^3 + 40x$ , so it is not equal. Only Ben's, $8x^2(2x + 5)$ , pulls out the greatest common factor. Expanding tests only equality, which Ana's unfinished factoring also passes.

Full solution

Expand each answer by multiplying the factor outside into every term inside.
Ana's gives $4x \cdot 4x^2 = 16x^3$ and $4x \cdot 10x = 40x^2$ , which is the original.
Ben's gives $8x^2 \cdot 2x = 16x^3$ and $8x^2 \cdot 5 = 40x^2$ , which is the original too.

Cal's gives $8x \cdot 2x^2 = 16x^3$ but $8x \cdot 5 = 40x$ , not $40x^2$ , so Cal's answer is not equal to $16x^3 + 40x^2$ .
Since $40x^2$ is $8x \cdot 5x$ , the second term inside should have been $5x$ , and Cal dropped the $x$ .

The greatest common factor has number part $8$ , the greatest common divisor of $16$ and $40$ , and variable part $x^2$ , the smallest power of $x$ that appears.
So it is $8x^2$ , and Ben's is the answer that pulls it out.
The terms left in Ben's parentheses, $2x$ and $5$ , share no number factor greater than $1$ and no factor of $x$ .

Ana's answer is equal to the original, but it is not finished: the terms inside, $4x^2$ and $10x$ , still share the factor $2x$ .
Pulling that out as well turns $4x(4x^2 + 10x)$ into $4x \cdot 2x(2x + 5)$ , which is Ben's $8x^2(2x + 5)$ .

Expanding answers one question only: is the product equal to the original expression?
Ana's and Ben's are both equal to it, so both pass, and expanding cannot tell them apart.
To see whether the factor outside is the greatest, look inside the parentheses instead: if the terms there still share a number factor greater than $1$ or a factor of $x$ , the factor outside was not the greatest.

Answer

Ana's and Ben's expand to $16x^3 + 40x^2$ ; Cal's gives $16x^3 + 40x$ , so it is not equal. Only Ben's, $8x^2(2x + 5)$ , pulls out the greatest common factor. Expanding tests only equality, which Ana's unfinished factoring also passes.

Key idea

Re-expanding checks that a factoring is correct; to check that the factor pulled out is the greatest, see whether the terms left inside still share a number factor greater than $1$ or a factor of $x$ .
Problem 10: Mark complete

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