Expanding and Factoring Expressions: Free Response

1. Reaching every term, signs included . Foundational, 11 points. Question 1 of 5.

A factor written outside a parenthesis multiplies each term inside, however many terms there are and whatever signs they carry. These parts push that in three directions at once: more than two terms inside, a factor carrying both a sign and a variable, and a factor that is not written as a factor at all.

Part A.
Expand $-4(3x - 7)$ and $6x(2x^2 - x + 4)$ , showing the individual products before you write each final expression.

Write the expression An equation or an expression is enough here. Show how you built it. 4 points
Part B.
Expand and simplify $4x - 2(x^2 - 3x + 6)$ . Say which quantity is multiplying the parenthesis before you distribute anything.

Write the expression An equation or an expression is enough here. Show how you built it. 4 points
Part C.
Before computing anything, a student says the expansion of $6x(2x^2 - x + 4)$ has to come out with two terms, because $6x$ is two things multiplied together. Explain what actually fixes how many products an expansion produces, and give that number for each of the two products in part A.

Explain why it works A sentence or two. Reasons, not steps. 3 points

Hints

One at a time, each one a step further than the last. Take only as many as you need.

Hint 1 of 3
Nothing here calls for a new rule. One law does all three parts, and the only questions are what exactly is playing the part of the outside factor and how many terms it has to reach.
Hint 2 of 3 · Part B
Read the expression as two pieces added together. The second piece is being subtracted and it is a product, so decide which of its two factors owns the minus sign before you multiply anything out.
Hint 3 of 3 · Part C
Test the student's rule on a product whose outside factor is a single number and whose parenthesis holds several terms. If the rule and the multiplication disagree there, you have found the thing the rule never looks at.

Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Worked solution

Part A

Multiply the outside factor against each term inside, one term at a time, and keep the sign that each product earns.

In the first, the factor is the negative number $-4$ , so both products come out with the opposite sign to the terms they came from:

$-4(3x - 7) = (-4)(3x) + (-4)(-7) = -12x + 28.$

In the second, the factor carries a variable, so the exponents add wherever $x$ meets a power of $x$ . The three products are $6x \cdot 2x^2 = 12x^3$ , then $6x \cdot (-x) = -6x^2$ , then $6x \cdot 4 = 24x$ :

$6x(2x^2 - x + 4) = 12x^3 - 6x^2 + 24x.$

Three terms inside produced three products, and none of them was skipped.

Part B

The minus sign belongs to the factor. The expression subtracts twice the parenthesis, so the quantity multiplying it is $-2$ , not $2$ , and every product it makes carries that sign:

$-2(x^2 - 3x + 6) = -2x^2 + 6x - 12.$

The middle product is $(-2)(-3x)$ , a negative times a negative, so it is positive; leaving it negative is the single most common slip here.

Now restore the leading $4x$ and gather the two like terms:

$4x - 2x^2 + 6x - 12 = -2x^2 + 10x - 12.$

The $x^2$ term and the constant have nothing to pair with, so that is as far as the tidying goes.

Part C

Look at what the distributive law actually says. In $a(b + c + d) = ab + ac + ad$ the letter $a$ stands for the whole outside factor, however many symbols it happens to be written with, and the law hands back exactly one product for each term inside the parenthesis:

$a(b + c + d) = ab + ac + ad.$

So the count is decided by the parenthesis, and the outside factor has no say in it. The first product in part A has two terms inside, so it produces two products; the second has three terms inside, so it produces three:

$6x(2x^2 - x + 4) = 6x \cdot 2x^2 + 6x \cdot (-x) + 6x \cdot 4.$

The student's rule looks only at the factor, so applied to $-4(3x - 7)$ , whose outside factor is a single number, it would predict one product where there are plainly two. A rule that never inspects the parenthesis cannot be counting what the law produces.

Gathering like terms afterwards can of course shorten a result, but that is a second step and a separate question; neither expansion in part A has like terms to gather.

In one line

$-4(3x - 7) = -12x + 28$ and $6x(2x^2 - x + 4) = 12x^3 - 6x^2 + 24x$ . Since the quantity multiplying the parenthesis is $-2$ , $4x - 2(x^2 - 3x + 6) = -2x^2 + 10x - 12$ . And the number of products an expansion produces is fixed by the number of terms inside the parenthesis, two for the first product of part A and three for the second.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Multiplies the outside factor against every term inside, so each expansion has one product for each term of its parenthesis. . Worth 2 points.

Gives every product the sign the two factors earn it, and adds exponents wherever a power of $x$ multiplies another power of $x$ . . Worth 2 points.

Part B 4 points

Identifies the quantity multiplying the parenthesis as a signed factor, rather than distributing a positive factor and leaving the subtraction outside it. . Worth 2 points.

Produces one product for each of the three terms inside, each carrying the sign that its two factors give it. . Worth 1 point.

Gathers the like terms afterwards, leaving an expression whose terms are pairwise unlike. . Worth 1 point.

Part C 3 points

Names the feature of a product that fixes how many products an expansion produces, and says why the outside factor's own shape does not affect that count. . Worth 2 points. needs an explanation, not just an answer

Gives a number for each of the two products of part A, rather than answering only in general terms. . Worth 1 point.

Try a similar problem (Optional)

Same idea, different numbers. Work it on paper, then check yourself the same way.

Expand $-5(4x - 3)$ and $3x(x^2 + 2x - 8)$ , then expand and simplify $9x - 4(2x^2 - x + 3)$ .

2. Every term times every term . Foundational, 12 points. Question 2 of 5.

The rectangle below has been cut by one vertical line and one horizontal line, so its width is split into $2x$ and $5$ and its height into $x$ and $3$ . Four cells appear, and this question is about what those four cells are counting, and about how far the picture can be trusted.

The rectangle of the stem: its width is split into

2x

and

5

, its height into

x

and

3

, and the two cuts leave four cells.

Text description of this figure

A rectangle is divided by one vertical line and one horizontal line into four cells. Along the top, the width is split into a segment marked 2x and a segment marked 5. Down the left side, the height is split into a segment marked x and a segment marked 3. The four cells themselves carry no labels.

Part A.
Give the area of each of the four cells, identifying each cell by its own two side lengths, then add the four areas and write the area of the whole rectangle as one expression with its like terms gathered.

Write the expression An equation or an expression is enough here. Show how you built it. 4 points
Part B.
Multiply $(3x - 2)(x^2 + 4x - 1)$ . Before multiplying anything, say how many pairwise products the every-term-by-every-term rule predicts here; then produce them and gather like terms.

Write the expression An equation or an expression is enough here. Show how you built it. 4 points
Part C.
The rectangle in part A is a genuine picture only while every term stands for a positive length, and the first factor in part B contains $-2$ . Decide whether the every-term-by-every-term rule still applies to part B's product, and justify your decision by naming what that rule actually rests on.

Justify your claim State the claim, then give the reason it has to be true. 4 points

Hints

One at a time, each one a step further than the last. Take only as many as you need.

Hint 1 of 3
Two of these parts ask the same thing twice, once with a picture to lean on and once with no picture available at all. Work out what the cells of the rectangle are counting and the second one will look familiar.
Hint 2 of 3 · Part B
Count the pairings first. Each term on the left has to meet each term on the right, so the total is a multiplication of two small numbers, and knowing it in advance is how you can tell when you have finished.
Hint 3 of 3 · Part C
Ask where the rule came from in the first place. If its derivation used only a law that holds for every number, a negative term cannot damage the rule, so it must damage something else, and naming that is part of your answer.

Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Worked solution

Part A

Each cell is itself a rectangle, so its area is the product of its own two side lengths.

The cell with sides $2x$ and $x$ has area $2x^2$ . The cell with sides $2x$ and $3$ has area $6x$ . The cell with sides $5$ and $x$ has area $5x$ . The cell with sides $5$ and $3$ has area $15$ .

Adding the four and gathering the two like terms gives the whole:

$2x^2 + 6x + 5x + 15 = 2x^2 + 11x + 15.$

The same rectangle has width $2x + 5$ and height $x + 3$ , so its area is also the single product $(2x + 5)(x + 3)$ . The four cells are exactly the four ways of pairing a term of one side with a term of the other:

$(2x + 5)(x + 3) = 2x^2 + 11x + 15.$

Part B

Two terms in the first factor and three in the second give $2 \times 3 = 6$ pairings, so six products appear before anything is gathered.

Take the terms of the first factor one at a time, each with its own sign:

$3x(x^2 + 4x - 1) = 3x^3 + 12x^2 - 3x,$

$-2(x^2 + 4x - 1) = -2x^2 - 8x + 2.$

Those are the six products. Adding them, the like $x^2$ terms give $12x^2 - 2x^2 = 10x^2$ and the like $x$ terms give $-3x - 8x = -11x$ , while the $x^3$ term and the constant have nothing to pair with:

$3x^3 + 10x^2 - 11x + 2.$

The constant $+2$ is the product of the two negative terms, one from each factor.

Part C

Separate two things that are easy to run together: what makes the rule true, and what makes it easy to see.

The derivation of the rule never mentions area. Treat the first factor as a single quantity $Q$ , distribute it across the second factor, then distribute each term of the second factor back across $Q$ :

$Q(c + d) = Qc + Qd, \qquad Q = a + b.$

Every step there is the distributive law, and the distributive law is an identity: it holds for every number, positive, negative or zero. So the products it predicts are the same whatever the terms stand for, and part B's $-2$ is no exception.

What the negative term does cost is the drawing. A side of length $-2$ is not a length, so no rectangle can be cut the way part A's rectangle was cut. The area model is an illustration of the rule in the case where every term is positive, and an illustration does not carry the argument.

This is worth being exact about, because the reverse move, treating a picture that works in the easy case as the reason a rule holds in every case, is how a false claim survives longest.

In one line

The four cells have areas $2x^2$ , $6x$ , $5x$ and $15$ , so the rectangle's area is $2x^2 + 11x + 15$ , which is the single product $(2x + 5)(x + 3)$ . The rule predicts six products for $(3x - 2)(x^2 + 4x - 1)$ , and gathering them gives $3x^3 + 10x^2 - 11x + 2$ . The rule survives a negative term because it is derived from the distributive law, an identity for every number; it is only the rectangle that needs every term to be positive.

Another way: Distribute in two stages instead of tracking six pairings

Rather than listing every pairing at once, take the first factor one term at a time and expand two ordinary single-factor products:

$(3x - 2)(x^2 + 4x - 1) = 3x(x^2 + 4x - 1) - 2(x^2 + 4x - 1).$

Each of those is exactly the kind of expansion question 1 was about, and adding the two results gives the same six products. Nothing new happens; the pairings are produced in two organized batches instead of all at once.

When it is worth it When a factor has three terms or more, where a single list of pairings is easy to leave one short. It also keeps the sign of the second term in view, since it is written into the factor $-2$ from the start.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Gives an area for every one of the four cells, each one the product of that cell's own two side lengths. . Worth 2 points.

Adds the four cell areas and gathers the two like terms into a single term. . Worth 1 point.

Reports the four cells as parts of one total, naming the area of the whole rectangle rather than stopping at four separate areas. . Worth 1 point.

Part B 4 points

Predicts the number of products from the number of terms in each factor, before any multiplying is done. . Worth 2 points.

Produces every predicted product with the sign its two terms give it, including where a negative term of one factor meets a negative term of the other. . Worth 2 points.

Part C 4 points

Reaches a verdict on whether the rule applies to part B's product, and supports it by naming what the rule is derived from rather than by appealing to the picture. . Worth 3 points. needs an explanation, not just an answer

Says what a negative term does affect, so that the picture and the rule are not left treated as the same thing. . Worth 1 point.

Try a similar problem (Optional)

Same idea, different numbers. Work it on paper, then check yourself the same way.

Multiply $(4x + 3)(x + 6)$ , and then $(2x - 5)(x^2 - 3x + 2)$ .

3. The gravel around the plot . Application, 12 points. Question 3 of 5.

A rectangular vegetable plot is $x$ meters wide and $x + 4$ meters long. A gravel strip of uniform width $3$ meters is laid all the way around the outside of it, so the gravel forms a frame with the plot in the middle. Every length here is in meters and every area in square meters.

Part A.
Write the width and the length of the whole region, gravel included, in terms of $x$ , and expand their product to give the area of that whole region as a single expression.

Write the expression An equation or an expression is enough here. Show how you built it. 3 points
Part B.
The gravel is the part of the whole region that is not the plot. Write the gravel's area as a single expression with its like terms gathered, and then write that expression in fully factored form.

Carry your own answer forward Take the whole region's area from your own part A expression and the plot's dimensions from the stem. If part A did not come out, carry on with whatever you wrote there: the credit here is for the subtraction and the factoring, not for one particular area appearing.

Write the expression An equation or an expression is enough here. Show how you built it. 4 points
Part C.
This year's plot has $x = 7$ . How much gravel does the gardener have to buy for it?

Carry your own answer forward Evaluate your own expression from part B at the stated width. The credit here is for substituting correctly and reporting an area, not for the expression having come out a particular way.

Solve and show your work Write each step out, and end with the value and its units. 2 points
Part D.
Use the factored form of the gravel area, not the expanded one, to say how that area changes when the plot's width $x$ goes up by one meter. Confirm the same reading from the expanded form, and say why the two forms could not have disagreed.

Carry your own answer forward Argue from whichever pair of forms you produced in part B. The credit here is for reading a change out of a factored form and then confirming it from the expanded one.

Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 3 points

Hints

One at a time, each one a step further than the last. Take only as many as you need.

Hint 1 of 4
Three shapes are in play: the plot, the whole region including the gravel, and the gravel itself. Only two of them are rectangles, which decides which two areas you write down first and which one the algebra has to produce.
Hint 2 of 4 · Part A
Sketch the frame and walk along one edge of it, marking every segment you cross. The strip is met once going in and once coming out, so each dimension of the plot grows twice over.
Hint 3 of 4 · Part B
Subtracting a whole expression means subtracting each of its terms. Once you see what survives that subtraction, ask what its terms have in common, and check whether every one of them really carries the variable.
Hint 4 of 4 · Part D
Compare the area at a width of $x$ with the area at a width of $x + 1$ . In the factored form only one of the two factors moves at all, and it moves by one.

Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Worked solution

Part A

The strip runs along both sides of the plot, so each dimension picks up $3$ meters twice, once at each end, and grows by $6$ rather than by $3$ . The width becomes $x + 6$ and the length becomes $(x + 4) + 6 = x + 10$ .

Multiplying every term of one dimension by every term of the other:

$(x + 6)(x + 10) = x^2 + 10x + 6x + 60 = x^2 + 16x + 60.$

Part B

The plot itself is $x$ by $x + 4$ , so its area is

$x(x + 4) = x^2 + 4x.$

The gravel is what is left when the plot is removed from the whole region, so subtract, letting the minus sign reach both terms of the plot's area:

$(x^2 + 16x + 60) - (x^2 + 4x) = 12x + 60.$

Now factor what is left. The coefficients $12$ and $60$ share the factor $12$ , and only one of the two terms carries an $x$ , so no variable is common and the greatest common factor is the number $12$ by itself:

$12x + 60 = 12(x + 5).$

The terms $x$ and $5$ inside share nothing beyond $1$ , so nothing more can be pulled out.

Part C

Substitute $x = 7$ into the factored form, which is the quicker of the two to evaluate:

$12(7 + 5) = 12 \times 12 = 144.$

So the gardener needs $144$ square meters of gravel. The gathered form agrees, as it must: $12(7) + 60 = 84 + 60 = 144$ .

It is worth checking that against the picture rather than only against the algebra. With $x = 7$ the plot is $7$ by $11$ and the whole region is $13$ by $17$ , so the gravel is $221 - 77 = 144$ square meters.

Part D

Widening the plot by one meter replaces $x$ by $x + 1$ . In the factored form one factor never moves and the other goes up by exactly one, so the whole area goes up by one copy of the fixed factor:

$12(x + 1 + 5) - 12(x + 5) = 12.$

The rise is $12$ square meters, and it is the same rise whatever $x$ was, which is the part worth saying out loud.

The expanded form gives the same reading by a different route: its constant term does not move at all, and its only $x$ term goes up by $12$ when $x$ goes up by $1$ :

$(12(x + 1) + 60) - (12x + 60) = 12.$

The two forms could not have disagreed, because factoring did not change the value at any $x$ ; it only rewrote how that value is built. That is the same guarantee that makes it worth checking a factorization by expanding it.

In one line

The whole region is $(x + 6)$ by $(x + 10)$ , with area $x^2 + 16x + 60$ ; the gravel area is $12x + 60 = 12(x + 5)$ ; at $x = 7$ that is $144$ square meters; and every extra meter of width adds $12$ square meters of gravel, whatever the width was.

Another way: Build the frame out of strips instead of subtracting

The frame can be cut into four rectangles and added directly, with no subtraction anywhere. The two side strips are $3$ wide and as long as the plot, $x + 4$ ; the two end strips are $3$ deep and as wide as the whole region, $x + 6$ :

$2 \cdot 3(x + 4) + 2 \cdot 3(x + 6) = 6x + 24 + 6x + 36.$

Gathering gives $12x + 60$ , the same expression. The cut has to be made carefully: if both pairs of strips are taken as long as the whole region, the four corner squares get counted twice.

When it is worth it When the subtraction route feels like a trick, or when you want a second route to the same expression as a check. It is also the natural method if you are pricing the gravel strip by strip.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Accounts for the strip at both ends of each dimension when writing the outer width and length. . Worth 2 points.

Expands the product of the two outer dimensions and gathers the like terms. . Worth 1 point.

Part B 4 points

Forms the gravel area as a difference of two areas, subtracting every term of the plot's area rather than only its first. . Worth 2 points.

Pulls out the greatest common factor, leaving a bracket whose terms share no factor beyond $1$ . . Worth 1 point.

Reports both the gathered form and the factored form, rather than stopping at one of them. . Worth 1 point.

Part C 2 points

Substitutes the given width into an expression for the gravel area and evaluates it correctly. . Worth 1 point.

Reports the result as an area, with square meters attached to the number. . Worth 1 point.

Part D 3 points

Reads the change out of the factored form and says whether it depends on the starting width, rather than answering only for the width used in part C. . Worth 2 points.

Confirms the same change from the expanded form and says why the two forms have to agree. . Worth 1 point. needs an explanation, not just an answer

Try a similar problem (Optional)

Same idea, different numbers. Work it on paper, then check yourself the same way.

A rectangular patio is $x$ meters wide and $x + 6$ meters long, and a path of uniform width $2$ meters is laid all around it. Write the path's area in fully factored form, and evaluate it for $x = 5$ .

4. A factorization that survived its own check . Reasoning, 10 points. Question 4 of 5.

A student is asked to factor $24x^3 - 36x^2 + 60x$ completely, and writes

$24x^3 - 36x^2 + 60x = 6x(4x^2 - 6x + 10).$

They then check it by expanding the right-hand side, and the check succeeds: every product comes back correctly and the original expression is recovered exactly. The arithmetic is not the problem. The answer is still not the one the instruction asked for.

Part A.
State precisely what the student's answer fails to do, and say what an expansion check does settle about a factorization and what it cannot settle.

Find and correct the error Say which line first goes wrong, why it is wrong, and then do it correctly. 4 points
Part B.
Factor $24x^3 - 36x^2 + 60x$ completely. Show separately how the coefficients decide the numerical part of the factor you take out and how the powers of $x$ decide its variable part.

Write the expression An equation or an expression is enough here. Show how you built it. 3 points
Part C.
Give a test that decides whether a factorization of the shape 'one term times a bracket' is complete, and apply your test both to the student's answer and to your own.

Carry your own answer forward Apply your test to whichever factorization you produced in part B, whether or not it matched the expected one. The credit here is for stating a test that can actually be checked and then using it twice.

Justify your claim State the claim, then give the reason it has to be true. 3 points

Hints

One at a time, each one a step further than the last. Take only as many as you need.

Hint 1 of 3
Do not go hunting for a slip in the arithmetic, because there is not one. Read the instruction again instead, and ask what work the word completely was doing in it.
Hint 2 of 3 · Part B
Handle the numbers and the letters separately. Find the largest number dividing all three coefficients, then the highest power of $x$ dividing all three terms, and multiply those two findings together.
Hint 3 of 3 · Part C
A test worth having can be run on an answer alone, with the original expression covered up. That rules out expanding, so ask what is visible in a finished factorization that could still betray unfinished work.

Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Worked solution

Part A

Two different questions are being run together here, and separating them is the whole of this part.

The first is whether the student's line is true, and it is. Expanding returns the original, so the two forms really are equal, and the check tested exactly that and passed honestly.

The second is whether the answer is what was asked for, namely a complete factorization. Look at what was left inside the bracket:

$4x^2 - 6x + 10 = 2(2x^2 - 3x + 5).$

All three coefficients are even, so a factor of $2$ is still sitting inside, waiting to come out. The student took out a common factor, but not the greatest one.

And the check could never have found this. Taking out a factor that is too small produces a form that is still equal to the original, so equality survives the shortfall untouched. An expansion check tests equality and nothing else: it will catch a wrong sign or a dropped term, and it is blind, by construction, to a factorization that stopped early.

Part B

The greatest common factor has a numerical part and a variable part, and each is found on its own.

For the numerical part, take the greatest common divisor of the three coefficients. Since $24 = 2^3 \cdot 3$ , $36 = 2^2 \cdot 3^2$ and $60 = 2^2 \cdot 3 \cdot 5$ , the primes they all share are $2^2$ and $3$ , so that divisor is $12$ .

For the variable part, take each common variable at the lowest power that appears. The three terms carry $x^3$ , $x^2$ and $x$ , so the lowest power present is $x$ itself.

The greatest common factor is therefore $12x$ , and dividing each term by it gives the bracket:

$24x^3 - 36x^2 + 60x = 12x(2x^2 - 3x + 5).$

Expanding it back returns the original, and the coefficients $2$ , $3$ and $5$ share nothing beyond $1$ , so nothing has been left inside.

Part C

A usable test has to be checkable by someone holding only the factorization, with no original expression to compare against. That rules out expanding, and it points straight at the bracket.

The test: the factor already taken out is the GREATEST common factor, exactly when the terms left inside the bracket share no common factor other than $1$ , counting numbers and variables alike. Since taking out the full greatest common factor is what this chapter means by factoring completely, that settles the instruction.

It works in both directions, which is what makes it a test rather than a hint. If the terms inside do share a factor, that factor can be taken out again, so more factoring was still available. If they share nothing, there is nothing left to take, so the factor already removed was the greatest one.

Applied to the student's answer, the bracket is $4x^2 - 6x + 10$ , whose coefficients $4$ , $6$ and $10$ are all even, so the test fails:

$4x^2 - 6x + 10 = 2(2x^2 - 3x + 5).$

Applied to the complete answer, the bracket is $2x^2 - 3x + 5$ . Its coefficients share no divisor beyond $1$ , and the constant term carries no $x$ , so no variable is common either. The test passes, and it is the passing case that tells you to stop.

In one line

$24x^3 - 36x^2 + 60x = 12x(2x^2 - 3x + 5)$ . The student took out $6x$ , a common factor but not the greatest one, so the bracket $4x^2 - 6x + 10$ still held a factor of $2$ . An expansion check cannot detect that, because a factor that is too small still leaves the two forms equal; whether the greatest common factor has been fully taken out is tested instead by asking whether the terms inside the bracket share anything beyond $1$ .

Another way: Take out what you can see, then look again

Spotting the greatest common factor in one go is not compulsory. Take out any common factor, apply the completeness test to what is left, and repeat if it fails, which turns the student's own work into a valid route:

$6x(4x^2 - 6x + 10) = 6x \cdot 2(2x^2 - 3x + 5) = 12x(2x^2 - 3x + 5).$

The factors taken out multiply together into the greatest common factor, and the process stops exactly when the test passes.

When it is worth it When the coefficients are large or share several primes, so the greatest common divisor is hard to see all at once. The price is that you have to remember to look again, and stopping after one pull is precisely the failure this question is about.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Says what the answer fails to do as a property of the factorization itself, rather than reporting an arithmetic slip in the student's work. . Worth 2 points. needs an explanation, not just an answer

Distinguishes what an expansion check tests from what the instruction asked for, and says what that difference means for the kinds of fault such a check can and cannot detect. . Worth 2 points. needs an explanation, not just an answer

Part B 3 points

Finds the numerical part of the factor from the coefficients and the variable part from the powers, as two separate decisions. . Worth 2 points.

Divides every term by the factor taken out, so the bracket has one term for each term of the original. . Worth 1 point.

Part C 3 points

States a test that can be applied to a factorization on its own, without comparing it against the original expression. . Worth 2 points.

Runs the test on both factorizations and reports what it says about each. . Worth 1 point. needs an explanation, not just an answer

Try a similar problem (Optional)

Same idea, different numbers. Work it on paper, then check yourself the same way.

A student factors $30x^3 + 45x^2 - 60x$ as $5x(6x^2 + 9x - 12)$ and checks it by expanding, correctly. Say what is still wrong with the answer, and give the complete factorization.

5. When the shared factor is a whole bracket . Reasoning, 10 points. Question 5 of 5.

A shared factor need not be a number or a single letter. If the same bracket sits inside every term, it can be lifted out exactly as a $3$ or an $x$ would be, because the distributive law never said what kind of quantity its factor had to be. The last part asks you to make that clause precise.

Part A.
Factor $9(x - 6) + 4x(x - 6)$ by taking out the bracket that both terms carry, then expand your factorization to confirm that it returns the expression you started with.

Write the expression An equation or an expression is enough here. Show how you built it. 3 points
Part B.
Factor $7(x - 4) - x(4 - x)$ completely. The two brackets are not identical as they stand, so say what you do to one of them first and why that is allowed.

Write the expression An equation or an expression is enough here. Show how you built it. 3 points
Part C.
Prove the general fact both earlier parts leaned on: a sum of the form $F \cdot P + F \cdot Q$ equals $F(P + Q)$ whatever quantities $F$ , $P$ and $Q$ stand for, including when $F$ is a bracket with several terms of its own. Then name what plays the part of $F$ in each of the two expressions above.

Complete the derivation Each line should follow from the one above it. Say what lets you take each step. 4 points

Hints

One at a time, each one a step further than the last. Take only as many as you need.

Hint 1 of 3
Ask what the two terms of the first expression genuinely have in common. It is neither a number nor a single letter, and once you have seen it, the reverse reading of the distributive law does the rest of the work.
Hint 2 of 3 · Part B
The two brackets in the second expression are not equal, but they are close. Work out what comes out of $4 - x$ when a factor of $-1$ is taken from it, and keep track of where that $-1$ has to go afterwards.
Hint 3 of 3 · Part C
The law you want to appeal to was stated for numbers. Ask what a bracket denotes once every letter inside it has been given a value, and why that answer is enough to let a whole expression stand in a place reserved for a number.

Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Worked solution

Part A

Read the expression as a sum of two terms. The first term is $9$ times $(x - 6)$ and the second is $4x$ times $(x - 6)$ , so the factor $(x - 6)$ sits in both.

That is the shape $F \cdot P + F \cdot Q$ with $F = x - 6$ , so take $F$ out and leave the other factors behind inside a new bracket:

$9(x - 6) + 4x(x - 6) = (x - 6)(9 + 4x).$

Now the check, which means expanding both forms and comparing the sums. The original expands to

$9x - 54 + 4x^2 - 24x = 4x^2 - 15x - 54,$

and the factorization expands to

$(x - 6)(4x + 9) = 4x^2 + 9x - 24x - 54 = 4x^2 - 15x - 54.$

The two agree, so the factoring changed nothing about the value of the expression, which is all such a check can say and all it needs to.

Part B

The two brackets are $x - 4$ and $4 - x$ . They are not the same quantity, but each is the negative of the other:

$4 - x = -(x - 4).$

That rewrite is allowed because it is an expansion read backwards: distributing $-1$ across $x - 4$ gives $-x + 4$ , which is $4 - x$ .

Substitute it into the second term and let the two minus signs meet:

$-x(4 - x) = -x \cdot (-(x - 4)) = x(x - 4).$

Now both terms carry the same bracket, so it comes out:

$7(x - 4) + x(x - 4) = (x - 4)(7 + x).$

Expanding checks it. The factorization gives $x^2 + 7x - 4x - 28 = x^2 + 3x - 28$ , and the original gives $7x - 28 - 4x + x^2 = x^2 + 3x - 28$ .

Part C

The claim is about every quantity, so the argument cannot lean on what $F$ , $P$ and $Q$ happen to be in the two examples.

Start from the distributive law, which is an identity: for all numbers $a$ , $b$ and $c$ ,

$a(b + c) = ab + ac.$

An equation may be read in either direction, and read from right to left this one says that a factor shared by two terms can be lifted out:

$ab + ac = a(b + c).$

Now put $F$ in the role of $a$ , $P$ in the role of $b$ , and $Q$ in the role of $c$ . The one thing that needs checking is whether $F$ is entitled to stand where $a$ stood, since $a$ ranges over numbers. It is: give every letter inside $F$ a value and the bracket names a number, exactly as $3$ or $x$ does. The law applies to that number, so

$F \cdot P + F \cdot Q = F(P + Q)$

holds at every value of every letter, which is what it means for the two expressions to be equivalent. Nothing anywhere in the argument cared how many terms $F$ was written with, which is the whole point.

That is the licence for both earlier parts. In part A the part of $F$ was played by $(x - 6)$ , with $P = 9$ and $Q = 4x$ . In part B it was played by $(x - 4)$ , once the second bracket had been rewritten so that the same quantity really did appear in both terms.

In one line

$9(x - 6) + 4x(x - 6) = (x - 6)(4x + 9)$ , and $7(x - 4) - x(4 - x) = (x - 4)(x + 7)$ once $4 - x$ is rewritten as $-(x - 4)$ . Both are the distributive law $ab + ac = a(b + c)$ read from right to left with a whole bracket in the role of $a$ , which is legitimate because that law holds for every number and a bracket names a number as soon as its letters are given values.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Treats the repeated bracket as a single factor shared by both terms and lifts it out whole. . Worth 2 points.

Confirms the factorization by expanding, showing that both forms reach the same sum of terms. . Worth 1 point.

Part B 3 points

Rewrites one of the two brackets so that both terms end up carrying the same factor, and adjusts the rest of that term to match, leaving the expression unchanged. . Worth 2 points.

Says why that rewrite leaves the expression unchanged, rather than performing it silently. . Worth 1 point. needs an explanation, not just an answer

Part C 4 points

Derives the general statement from a law already known to hold for every number, rather than from the two worked examples, and addresses the case where the shared factor has several terms of its own. . Worth 3 points. needs an explanation, not just an answer

Names what plays the part of the shared factor in each of the two expressions above. . Worth 1 point.

Try a similar problem (Optional)

Same idea, different numbers. Work it on paper, then check yourself the same way.

Factor $5(x + 8) + 2x(x + 8)$ , and then $3(x - 9) - x(9 - x)$ .

1. Reaching every term, signs included . Foundational, 11 points. Question 1 of 5.

2. Every term times every term . Foundational, 12 points. Question 2 of 5.

3. The gravel around the plot . Application, 12 points. Question 3 of 5.

4. A factorization that survived its own check . Reasoning, 10 points. Question 4 of 5.

5. When the shared factor is a whole bracket . Reasoning, 10 points. Question 5 of 5.

Marking rubrics

1. Reaching every term, signs included 11 points

2. Every term times every term 12 points

3. The gravel around the plot 12 points

4. A factorization that survived its own check 10 points

5. When the shared factor is a whole bracket 10 points