Polynomial Long Division: Free Response

1. A forecast, a division, and the reason it ends . Foundational, 10 points. Question 1 of 5.

The division algorithm makes two promises at once: an identity saying dividend equals divisor times quotient plus remainder, and a degree condition saying the work has finished. Some of what the answer will look like is settled by the two degrees and the two leading terms before a single subtraction happens. Work throughout with $P(x) = 2x^4 - 3x^3 + 3x^2 + 15x - 13$ and $D(x) = x^2 - 2x + 4$ .

Part A.
Without dividing, give the degree of the quotient and its leading coefficient, and list every degree the remainder is permitted to have.

Solve and show your work Write each step out, and end with the value and its units. 3 points
Part B.
Now carry the division out. Report the quotient and the remainder, and confirm the pair by expanding.

Write the expression An equation or an expression is enough here. Show how you built it. 3 points
Part C.
Explain why the loop is guaranteed to stop after finitely many passes, and why its stopping test compares degrees rather than comparing the two expressions' values at particular numbers.

Explain why it works A sentence or two. Reasons, not steps. 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 4
Two of the three things asked for in the opening part are settled by the degrees and the leading terms alone, before any subtraction happens. The third is settled by the rule that tells the loop when to halt.
Hint 2 of 4 · Part A
Degrees add when polynomials multiply, so taking the divisor's degree away from the dividend's leaves the quotient's. Then ask what the halting rule permits a finished leftover to look like.
Hint 3 of 4 · Part B
Every pass divides the current leading term by $x^2$ , and the whole product of that term with the divisor comes off, all three of its terms at once. The sign traps sit in the middle column.
Hint 4 of 4 · Part C
Ask what the subtraction in a single pass was engineered to destroy, and what that guarantees about the sequence of degrees. For the second half, weigh a linear expression against a quadratic one at $x = 0$ and again at $x = 100$ .

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

Both facts about the quotient come from the two degrees and the two leading terms, so neither costs a subtraction.

Degrees add when polynomials multiply, and the product $D \cdot Q$ has to carry the dividend's leading term, because the remainder is too low in degree to interfere with it.

$\deg Q = \deg P - \deg D = 4 - 2 = 2.$

The same matching fixes the leading coefficient. The leading term of $D \cdot Q$ is $1$ times the leading coefficient of $Q$ times $x^4$ , and that has to be $2x^4$ .

$\text{leading coefficient of } Q = \frac{2}{1} = 2.$

The remainder is governed by the other promise. The loop halts when the working remainder is the zero polynomial or its degree has fallen below $2$ , so exactly three finishes are legal: a linear remainder, a nonzero constant remainder, or the zero polynomial. All three are finished answers, not failures.

Part B

Both polynomials are already in standard form with every power present, so no placeholder column is needed.

First pass. The two leading terms force the first quotient term, $2x^4 \div x^2 = 2x^2$ , and

$2x^2(x^2 - 2x + 4) = 2x^4 - 4x^3 + 8x^2.$

Subtracting that entire product leaves

$x^3 - 5x^2 + 15x - 13.$

Watch the middle column: subtracting $-4x^3$ adds $4x^3$ to the $-3x^3$ that was there.

Second pass. $x^3 \div x^2 = x$ , and $x(x^2 - 2x + 4) = x^3 - 2x^2 + 4x$ . Subtracting leaves

$-3x^2 + 11x - 13.$

Third pass. $-3x^2 \div x^2 = -3$ , and $-3(x^2 - 2x + 4) = -3x^2 + 6x - 12$ . Subtracting leaves

$5x - 1,$

of degree $1$ , below the divisor's degree $2$ , so the loop stops. Collecting the three recorded terms gives the quotient $2x^2 + x - 3$ , and the last line is the remainder.

Check by expanding:

$(x^2 - 2x + 4)(2x^2 + x - 3) = 2x^4 - 3x^3 + 3x^2 + 10x - 12,$

and adding $5x - 1$ restores $2x^4 - 3x^3 + 3x^2 + 15x - 13$ exactly.

Part C

Look at what one pass does. The quotient term $q$ is chosen so that $q$ times the divisor's leading term equals the working remainder's leading term, so the subtraction destroys that leading term exactly. Every other term of $qD$ is $q$ times a lower term of $D$ , so it sits below that degree, and so does every other term of the working remainder. Nothing at the top degree survives.

$\deg(W - qD) < \deg W.$

So the degrees of the successive working remainders form a strictly decreasing list of whole numbers. Such a list cannot run forever: it is bounded below by $0$ and each step goes down by at least one. After at most $\deg P - \deg D + 1$ passes, which is $3$ for this division, the stopping condition must be met.

The test itself cannot be about numerical size, because a polynomial's size moves with $x$ . Compare a linear expression with this divisor at two inputs:

$x = 0: \quad x + 6 = 6, \quad x^2 - 2x + 4 = 4;$

$x = 10: \quad x + 6 = 16, \quad x^2 - 2x + 4 = 84.$

A rule phrased in terms of values would call the same division finished at one input and unfinished at another, and would never settle anything. Degree does not move with $x$ , so a polynomial either passes the test or fails it once and for all. That is what lets the loop halt, and it is what makes the answer unique.

In one line

Before dividing: the quotient has degree $2$ and leading coefficient $2$ , and the remainder must be linear, a nonzero constant, or the zero polynomial. Dividing gives $2x^4 - 3x^3 + 3x^2 + 15x - 13 = (x^2 - 2x + 4)(2x^2 + x - 3) + 5x - 1$ . The loop had to stop because each pass cancels the working remainder's leading term, so the degrees fall strictly and cannot fall forever, and the halting test is written in degrees because numerical size depends on the input while degree does not.

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

Gets the quotient's degree from the difference of the two degrees and its leading coefficient from the ratio of the two leading coefficients, with no division carried out. . Worth 2 points.

Gives every finish the stopping condition allows the remainder here, not a subset of them. . Worth 1 point.

Part B 3 points

Runs the loop with each quotient term taken from the two leading terms and the entire product subtracted each pass, stopping only once the working remainder falls below the divisor's degree. . Worth 2 points.

Confirms the reported pair by expanding the divisor times the quotient, adding the remainder, and comparing the result with the dividend term by term. . Worth 1 point.

Part C 4 points

Argues that each pass strictly lowers the degree of the working remainder because the quotient term was chosen to cancel its leading term, and concludes from that a strictly decreasing list of whole numbers has to terminate. . Worth 3 points. needs an explanation, not just an answer

Says why the stopping test is stated in degrees rather than in numerical size, naming what changes with the input and what does not. . Worth 1 point.

Try a similar problem (Optional)

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

Forecast the degree and leading coefficient of the quotient for $3x^4 + x^3 + 15x - 9$ divided by $x^2 - x + 3$ , then carry the division out and check your pair by expanding.

2. Gaps on both sides of the bracket . Foundational, 11 points. Question 2 of 5.

A tableau is column bookkeeping, so a power that never appears still needs a column held open for it. Here $P(x) = x^5 - 2x^2 - 18x + 9$ and $D(x) = x^2 - 5$ , and gaps occur on both sides of the division sign.

Part A.
Write both polynomials in the form the tableau needs, then divide $P$ by $D$ and report the quotient and the remainder.

Write the expression An equation or an expression is enough here. Show how you built it. 4 points
Part B.
The termination argument promises only that a pass lowers the working remainder's degree by at least one, not by exactly one. Explain how a single pass can lower it by more than that, and say what such a pass leaves behind in the quotient.

Explain why it works A sentence or two. Reasons, not steps. 3 points
Part C.
A classmate says that dividing a degree- $5$ polynomial by a degree- $2$ polynomial always takes exactly four passes of the loop, one for each power the quotient spans. Decide whether that is right, and justify your verdict.

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 4
Two powers never appear in the dividend, and the divisor is missing a power of its own. Settle what each of those gaps has to become before any column is subtracted from another.
Hint 2 of 4 · Part A
After a subtraction, read off which power the working remainder actually starts with, and take the next quotient term from that leading term rather than from the power you were expecting to see there.
Hint 3 of 4 · Part B
Write down the second coefficient of the product being taken away and set it beside the second coefficient of what it is taken from. Nothing in the construction forces those two numbers to differ.
Hint 4 of 4 · Part C
Count the passes you actually made in the first part, then count the powers the quotient you reported spans. If those two counts differ, the claim is already settled.

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 dividend skips $x^4$ and $x^3$ , and the divisor skips $x$ , so every one of those powers gets an explicit zero coefficient before the first pass.

$x^5 - 2x^2 - 18x + 9 = x^5 + 0x^4 + 0x^3 - 2x^2 - 18x + 9,$

$x^2 - 5 = x^2 + 0x - 5.$

First pass. $x^5 \div x^2 = x^3$ , and $x^3(x^2 - 5) = x^5 - 5x^3$ . That product has nothing at all in the $x^4$ column, so its second term has to be written under $x^3$ and not under $x^4$ . Subtracting leaves

$5x^3 - 2x^2 - 18x + 9.$

The $x^4$ column came out empty, and the $x^3$ column, which was holding $0x^3$ , is where the $+5x^3$ landed. Without that open column there would have been nowhere to put it.

Second pass. $5x^3 \div x^2 = 5x$ , and $5x(x^2 - 5) = 5x^3 - 25x$ . Subtracting leaves

$-2x^2 + 7x + 9.$

Third pass. $-2x^2 \div x^2 = -2$ , and $-2(x^2 - 5) = -2x^2 + 10$ . Subtracting leaves

$7x - 1,$

of degree $1$ , below the divisor's degree, so the loop stops. The three recorded quotient terms are $x^3$ , $5x$ and $-2$ . Checking,

$(x^2 - 5)(x^3 + 5x - 2) = x^5 - 2x^2 - 25x + 10,$

and adding $7x - 1$ rebuilds the dividend.

Part B

Take one pass in general. Let the working remainder be $W$ with leading term $a x^m$ and second term $a' x^{m-1}$ , and let the divisor be $D$ with leading term $b x^n$ and second term $b' x^{n-1}$ . The pass uses $q = \tfrac{a}{b} x^{m-n}$ and subtracts $qD$ .

The leading terms were chosen to cancel, so the $x^m$ column always empties. One column down nothing was arranged at all: the coefficient left there is

$a' - \frac{a}{b}\,b'.$

Nothing forces that number to be nonzero. When it happens to vanish, the $x^{m-1}$ column empties as well and the degree falls by at least $2$ in that single pass.

The quotient records what happened. Each pass writes exactly one term, and the power it writes on is fixed by the leading term the pass started from. If no working remainder ever leads with $x^k$ , no pass ever writes on $x^{k-n}$ , so the quotient carries a zero coefficient there. In the division above the first pass left both the $x^5$ and the $x^4$ columns empty, so the second pass started at $x^3$ instead of $x^4$ , and the quotient reads

$x^3 + 5x - 2 = x^3 + 0x^2 + 5x - 2.$

That zero is not a defect any more than the $0$ in a numeral like $3{,}205$ is.

Part C

Count what a pass produces. Each pass writes exactly one term into the quotient, with coefficient $\tfrac{a}{b}$ , where $a$ is the leading coefficient of the working remainder and $b$ that of the divisor. Neither is zero, so the coefficient a pass writes is never zero. And the passes work at strictly decreasing degrees, so no two of them write on the same power.

That makes the count exact in both directions: every pass contributes one nonzero coefficient of the quotient, and every nonzero coefficient of the quotient came from some pass. So the number of passes is the number of nonzero coefficients, not the number of powers between the quotient's degree and its constant term.

Here the quotient has degree $5 - 2 = 3$ , which allows four coefficients, so four passes is a ceiling. The division in part A reaches it only if all four are nonzero, and it does not: its degrees ran

$5 \to 3 \to 2 \to 1,$

three passes rather than four, because the first pass left two columns empty at once. The classmate's rule would have predicted four and been wrong on the very first example.

In one line

Written for the tableau, the dividend is $x^5 + 0x^4 + 0x^3 - 2x^2 - 18x + 9$ and the divisor $x^2 + 0x - 5$ ; the division gives quotient $x^3 + 5x - 2$ and remainder $7x - 1$ . A pass can lower the degree by more than one, because only the leading column is engineered to empty and the next one may empty by coincidence, and the quotient then carries a zero coefficient on the skipped power. So the classmate is wrong: the number of passes is the number of nonzero coefficients in the quotient, here three rather than four.

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

Opens a column with an explicit zero coefficient for every power the dividend skips, and treats the divisor's own missing power the same way. . Worth 2 points.

Produces a quotient and a remainder whose expansion rebuilds the dividend, with each pass subtracting the entire product and each product's terms written under the powers they carry. . Worth 2 points.

Part B 3 points

Explains that only the leading column is forced to cancel and that the next one down can cancel by coincidence, so the degree may fall by more than one in a single pass. . Worth 2 points. needs an explanation, not just an answer

Says what a drop of more than one leaves behind in the quotient, naming both the coefficient it produces and which powers carry it. . Worth 1 point.

Part C 4 points

States a verdict and supports it by tying the number of passes to the quotient's nonzero coefficients rather than to the range of powers the quotient spans. . Worth 3 points. needs an explanation, not just an answer

Backs the verdict with a specific division, or with an exact description of one, rather than only asserting what could happen in principle. . Worth 1 point.

Try a similar problem (Optional)

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

Divide $x^5 - 3x^3 + 4x^2 - 3x - 5$ by $x^2 - 2$ , writing both polynomials for the tableau first, and say how many passes the loop took and why that count is not the one the quotient's degree suggests.

3. Recovering a height that was never measured . Application, 13 points. Question 3 of 5.

A shop cuts rectangular blocks on a machine whose single dial reading is $x$ centimetres, set large enough that every measurement below comes out positive. Every block in the run stands on the same rectangular base, of area $x^2 + 4x - 3$ square centimetres, and the shop's records list only each block's volume. A height that was never written down has to be recovered from those two figures.

Part A.
The first block has volume $x^4 + 3x^3 - 2x^2 + 23x - 15$ cubic centimetres. Find an expression for its height, and say what your working establishes about the base area and the volume.

Write the expression An equation or an expression is enough here. Show how you built it. 5 points
Part B.
A second block, on the same base, has volume $2x^4 + 5x^3 - 17x^2 + 13x + 4$ cubic centimetres. Divide again, and say what your result reports about a height for this block written as a polynomial in $x$ .

Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 3 points
Part C.
Explain how one division decides, once and for all, whether a given base area can carry a polynomial height for a given volume. Your explanation should say why a single division settles it, rather than leaving open the chance that some other height would have worked.

Justify your claim State the claim, then give the reason it has to be true. 5 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
Volume is base area times height, so the height is the polynomial the base area must be multiplied by. That turns each part of this question into a division with the base area as the divisor.
Hint 2 of 4 · Part A
Run the loop until the working remainder falls below degree $2$ , and pay attention to what the last subtraction leaves. Multiplying your answer back by the base area is the quickest confirmation.
Hint 3 of 4 · Part B
Keep the divisor from the previous part. The informative part of this answer is not the quotient but whatever survives the final subtraction, so report both and then read the survivor.
Hint 4 of 4 · Part C
Suppose a polynomial height did exist. Write down the sentence it would satisfy, notice that the sentence also keeps both of the division algorithm's promises, and then use what those two promises pin down between them.

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

Volume is base area times height, so the height is whatever the base area has to be multiplied by to produce the volume. That is a division, with the base area as the divisor.

First pass. $x^4 \div x^2 = x^2$ , and $x^2(x^2 + 4x - 3) = x^4 + 4x^3 - 3x^2$ . Subtracting leaves

$-x^3 + x^2 + 23x - 15.$

Second pass. $-x^3 \div x^2 = -x$ , and $-x(x^2 + 4x - 3) = -x^3 - 4x^2 + 3x$ . Subtracting leaves

$5x^2 + 20x - 15.$

Third pass. $5x^2 \div x^2 = 5$ , and $5(x^2 + 4x - 3) = 5x^2 + 20x - 15$ . Subtracting leaves $0$ , so the loop ends on the zero polynomial and the division is exact.

$x^4 + 3x^3 - 2x^2 + 23x - 15 = (x^2 + 4x - 3)(x^2 - x + 5).$

The height is $x^2 - x + 5$ centimetres. A zero remainder is what licenses the word factor here: the identity has no leftover term, so the base area divides the volume exactly. The units agree as well, since square centimetres multiplied by centimetres give the cubic centimetres a volume is measured in.

Part B

Same divisor, same loop.

First pass. $2x^2(x^2 + 4x - 3) = 2x^4 + 8x^3 - 6x^2$ , and subtracting leaves

$-3x^3 - 11x^2 + 13x + 4.$

Second pass. $-3x(x^2 + 4x - 3) = -3x^3 - 12x^2 + 9x$ , and subtracting leaves

$x^2 + 4x + 4.$

Third pass. $1 \cdot (x^2 + 4x - 3) = x^2 + 4x - 3$ , and subtracting leaves

$7,$

a constant, of degree $0$ , which is below the divisor's degree $2$ , so the loop has finished.

$2x^4 + 5x^3 - 17x^2 + 13x + 4 = (x^2 + 4x - 3)(2x^2 - 3x + 1) + 7.$

Read that as the shop would. Standing $2x^2 - 3x + 1$ centimetres of height on the given base accounts for all but $7$ cubic centimetres of the recorded volume, at every dial setting. Since the leftover is not zero, the base area is not a factor of this volume, and the height is not any polynomial in $x$ .

Part C

Write $B$ for the base area and $V$ for the volume, and run the division once, producing

$V = B\,H + R, \qquad R = 0 \ \text{ or } \ \deg R < \deg B.$

One direction is immediate. If $R = 0$ then $V = B\,H$ , so the quotient $H$ is a polynomial height and the question is answered yes.

The other direction is where uniqueness does the work. Suppose some polynomial $H^{*}$ were a height, so that $V = B\,H^{*}$ . Write that as

$V = B\,H^{*} + 0.$

That pair satisfies the identity, and its leftover is the zero polynomial, which the stopping condition explicitly allows. So $(H^{*}, 0)$ keeps both promises. But exactly one pair keeps both promises, and the division already produced one, so $H^{*} = H$ and $R = 0$ .

Put the two halves together and the test is a genuine both-ways statement: a polynomial height exists exactly when the division leaves no remainder. That is why one division is enough. A nonzero leftover does not mean this attempt failed and a cleverer one might succeed; it means no polynomial height exists at all, because any that did would have been forced to be the one already on the page.

In one line

The first block's height is $x^2 - x + 5$ centimetres, since $x^4 + 3x^3 - 2x^2 + 23x - 15 = (x^2 + 4x - 3)(x^2 - x + 5)$ with nothing left over. The second gives $(x^2 + 4x - 3)(2x^2 - 3x + 1) + 7$ , so its height is not a polynomial in $x$ . One division settles either case, because the quotient and remainder are unique: a polynomial height would pair with a leftover of zero, so it would have to be the quotient the division already returned, and a polynomial height therefore exists exactly when the division comes out exact.

Another way: Match coefficients instead of dividing

Assume the height is a quadratic, $H(x) = ax^2 + bx + c$ , and expand the product of the base area with it: $(x^2 + 4x - 3)(ax^2 + bx + c)$ has coefficients $a$ , $b + 4a$ , $c + 4b - 3a$ , $4c - 3b$ and $-3c$ from the top down. Setting those against $1, 3, -2, 23, -15$ gives $a = 1$ , then $b = -1$ , then $c = 5$ from the first three, and the last two become checks: $4(5) - 3(-1) = 23$ and $-3(5) = -15$ . Both hold, so the height is $x^2 - x + 5$ and the fit is exact. Had either check failed, no quadratic height would exist.

When it is worth it When the quotient's degree is small and known in advance, or when you want the exactness to show up as a pair of equations that either agree or do not, rather than as a final subtraction.

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 5 points

Sets the height up as the result of dividing the recorded volume by the base area, rather than subtracting the two or multiplying them. . Worth 2 points.

Carries the division through to a working remainder the stopping condition accepts, with every pass subtracting the entire product. . Worth 2 points.

Reports the height with its units, and says what the leftover the division produced establishes about the base area and the volume. . Worth 1 point.

Part B 3 points

Divides the second volume by the same base area and reports the quotient together with the leftover, rather than the quotient alone. . Worth 2 points.

Reads the leftover back into the situation, saying what it establishes about the base area and the volume, and draws the consequence for a height written as a polynomial. . Worth 1 point.

Part C 5 points

Argues from the uniqueness of the quotient-and-remainder pair that any polynomial height would have to coincide with the one the division produced, so a nonzero leftover leaves no room for one. . Worth 3 points. needs an explanation, not just an answer

States the resulting test in both directions, saying what an exact division supplies and what only an exact division permits. . Worth 2 points.

Try a similar problem (Optional)

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

Blocks in a second run stand on a base of area $x^2 - 2x + 5$ square centimetres. Find the height of the block whose volume is $x^4 + x^3 + x^2 + 11x + 10$ cubic centimetres, then decide whether a block of volume $x^4 - x^3 + 2x^2 + 10x - 1$ cubic centimetres can stand on that base with a polynomial height.

4. Two true sentences, one finished division . Reasoning, 12 points. Question 4 of 5.

Both of these hold for every value of $x$ , as expanding either one confirms: $x^4 + 1 = (x^2 + x + 1)(x^2 - x) + (x + 1)$ and $x^4 + 1 = (x^2 + x + 1)(x^2 - x - 1) + (x^2 + 2x + 2)$ . Only one of them reports what long division returns for $x^4 + 1$ divided by $x^2 + x + 1$ .

Part A.
Say which of the two identities reports the quotient and the remainder, and say what disqualifies the other.

Justify your claim State the claim, then give the reason it has to be true. 3 points
Part B.
For each of the two leftovers, decide whether one further copy of the divisor can be taken out of it, and carry out any such step that is available. Do not run the tableau.

Write the expression An equation or an expression is enough here. Show how you built it. 4 points
Part C.
Explain in general why a true identity $P = D\,Q + R$ can fail to report the quotient and the remainder, describe what to do to any such identity to reach the pair that does report them, and say why that repair always finishes.

Explain why it works A sentence or two. Reasons, not steps. 5 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
Expanding settles nothing here, since both lines are correct arithmetic. What separates them is the other promise the division algorithm makes, the one that says the work has finished.
Hint 2 of 4 · Part A
Set the degree of each leftover beside the degree of the divisor, and ask of each whether another copy of the divisor could still be taken out of it.
Hint 3 of 4 · Part B
Write the larger leftover as one copy of the divisor plus whatever is then left, substitute that back in, and look at where the extra copy has to go for the sentence to stay true.
Hint 4 of 4 · Part C
Try adding some multiple of the divisor to the leftover while taking the same multiple away from the quotient, and see what the identity does. Then ask what the degree condition rules out.

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

Both lines are correct as algebra, so expanding them can never separate them. What separates them is the second promise: the leftover must be the zero polynomial or of degree below the divisor's.

$\deg(x + 1) = 1 < 2, \qquad \deg(x^2 + 2x + 2) = 2.$

The first leftover passes that test and the second does not. A leftover of degree $2$ has not finished paying: a copy of the degree- $2$ divisor can still be taken out of it, which is precisely what the loop's stopping condition is watching for.

So the first identity is the division's output, with quotient $x^2 - x$ and remainder $x + 1$ . Notice what makes this a decision rather than a preference. At most one pair can keep both promises, so once the first line is seen to keep them, the second is ruled out without any further inspection. Uniqueness is what turns a degree check into a verdict.

Part B

A leftover of degree $2$ still contains a copy of a degree- $2$ divisor, so split it into that copy plus whatever remains.

$x^2 + 2x + 2 = (x^2 + x + 1) + (x + 1).$

Substitute that into the identity it belongs to and the divisor appears twice:

$(x^2 + x + 1)(x^2 - x - 1) + (x^2 + x + 1) + (x + 1).$

The first two terms share the factor $x^2 + x + 1$ , so collect them:

$(x^2 + x + 1)\bigl[(x^2 - x - 1) + 1\bigr] + (x + 1).$

The bracket is $x^2 - x$ , so the identity has become $x^4 + 1 = (x^2 + x + 1)(x^2 - x) + (x + 1)$ . The extra copy moved out of the leftover and into the quotient, which is exactly what one pass of the loop does, and the new leftover has degree $1$ , so the division is now finished.

The other leftover admits no such step. Taking a copy of the divisor out of $x + 1$ would need a quotient term equal to $x \div x^2$ , which is not a polynomial term at all, so there is nothing to move and nothing to gain.

Part C

The identity by itself is one equation in two unknown polynomials, and one equation like that leaves room to move. Take any pair $(Q, R)$ that satisfies it and any polynomial $S$ , then slide $S$ across from one to the other:

$D(Q + S) + (R - SD) = DQ + R.$

The right side is $P$ again, so the new pair satisfies the identity for every choice of $S$ . There are infinitely many such pairs and nothing in the identity prefers any of them. The degree condition is what cuts that list down, and the uniqueness proof is what shows the cut leaves exactly one survivor.

The repair is one pass of the loop applied to the leftover rather than to the dividend. While the leftover's degree is at least $\deg D$ , divide the leftover's leading term by the divisor's leading term, add that term to the quotient, and subtract that term times the divisor from the leftover. The identity survives each step, by the sliding computation above with $S$ equal to the term being moved.

And it finishes. Each step was built to cancel the leftover's leading term, so it strictly lowers the leftover's degree, and degrees are whole numbers that cannot fall forever. After finitely many steps the leftover is zero or of degree below $\deg D$ ; at that moment both promises hold at once, so the pair on the page is the quotient and the remainder.

In one line

The first identity is the division's output, with quotient $x^2 - x$ and remainder $x + 1$ ; the second is disqualified because its leftover $x^2 + 2x + 2$ has the divisor's own degree. Splitting that leftover as $(x^2 + x + 1) + (x + 1)$ moves one copy of the divisor into the quotient and turns the second identity into the first, while $x + 1$ admits no such step. In general the identity alone has infinitely many solutions, since $(Q + S,\ R - SD)$ satisfies it for every polynomial $S$ ; the degree condition selects one, and repeated peeling reaches it because each peel strictly lowers the leftover's degree.

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

Names the identity that reports the division's output and disqualifies the other by the degree of its leftover, not by rechecking the arithmetic of either line. . Worth 2 points. needs an explanation, not just an answer

States the stopping condition being applied, including why a leftover of the divisor's own degree counts as unfinished. . Worth 1 point.

Part B 4 points

Splits whichever leftover admits it into one copy of the divisor plus what remains, and moves that copy into the quotient rather than discarding it, keeping the identity true throughout. . Worth 3 points.

Says why the remaining leftover admits no such step, on grounds of degree rather than on grounds of the arithmetic being awkward. . Worth 1 point.

Part C 5 points

Shows that the identity on its own has many solutions, for instance by moving a polynomial from the quotient into the leftover and checking that the identity survives. . Worth 2 points. needs an explanation, not just an answer

Describes the repair as repeatedly taking a copy of the divisor out of the leftover and adding it to the quotient, until the leftover's degree falls below the divisor's. . Worth 2 points.

Says why the repair terminates, in terms of what each step does to the degree of the leftover. . Worth 1 point.

Try a similar problem (Optional)

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

The identity $x^4 - 3 = (x^2 - x + 2)(x^2 + x - 2) + (x^2 - 4x + 1)$ holds for every $x$ . Decide whether it reports the quotient and remainder for $x^4 - 3$ divided by $x^2 - x + 2$ , and if it does not, repair it without running the tableau.

5. An answer that passes the wrong checks . Reasoning, 12 points. Question 5 of 5.

A student divided $x^5 + 2x^4 + 2x^3 + 3x^2 - 4x + 5$ by $x^3 + 3x - 1$ and recorded three passes. First pass: quotient term $x^2$ , product $x^5 + 3x^3 - x^2$ , leaving $2x^4 - x^3 + 4x^2 - 4x + 5$ . Second pass: quotient term $2x$ , product $2x^4 + 6x^3 - 2x^2$ , leaving $-7x^3 + 6x^2 - 4x + 5$ . Third pass: quotient term $-7$ , product $-7x^3 - 21x + 7$ , leaving $6x^2 + 17x - 2$ . They reported the quotient $x^2 + 2x - 7$ and the remainder $6x^2 + 17x - 2$ .

Part A.
Find the first line of the student's work that does not follow, say what went wrong there, and give the correct quotient and remainder.

Find and correct the error Say which line first goes wrong, why it is wrong, and then do it correctly. 5 points
Part B.
The student's reported answer passes several checks this lesson recommends: the quotient has the degree and the leading coefficient the bookkeeping predicts, and the leftover's degree is below the divisor's. Explain why those checks let the error through, and name the check that catches it.

Explain why it works A sentence or two. Reasons, not steps. 3 points
Part C.
State exactly what a completed division has to satisfy in order to be the quotient and the remainder, and argue your statement in both directions: that the loop's output satisfies it, and that anything satisfying it is the loop's output.

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 4
Work down the recorded lines one at a time and stop at the first that does not follow. Everything after a bad line inherits its damage, so only the earliest fault needs diagnosing.
Hint 2 of 4 · Part A
Multiply the second quotient term by every term of the divisor yourself, and mind the power the divisor does not have: the honest product has a gap where the recorded line has a term.
Hint 3 of 4 · Part B
Ask of each check whether it is a promise every correct answer keeps, or a promise only a correct answer keeps. Those two are not the same thing, and only the second kind certifies.
Hint 4 of 4 · Part C
Write the two promises side by side, then ask separately whether the loop's output keeps both of them and whether two different pairs could both keep them.

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

Check the passes in order and stop at the first that does not follow.

The first pass is sound: $x^5 \div x^3 = x^2$ , and $x^2(x^3 + 3x - 1) = x^5 + 3x^3 - x^2$ , whose subtraction leaves exactly the line the student wrote.

The second pass is where it breaks. The quotient term $2x$ is right, but the product is not:

$2x(x^3 + 3x - 1) = 2x^4 + 6x^2 - 2x.$

The student wrote $2x^4 + 6x^3 - 2x^2$ , which is that product with its last two terms shifted one column to the left. The divisor has no $x^2$ term, so its product with $2x$ has no $x^3$ term, and skipping that empty column is what pushed both of the remaining terms one place to the left.

Redo from there. Subtracting the correct product from $2x^4 - x^3 + 4x^2 - 4x + 5$ leaves

$-x^3 - 2x^2 - 2x + 5.$

Third pass: $-x^3 \div x^3 = -1$ , and $-1(x^3 + 3x - 1) = -x^3 - 3x + 1$ . Subtracting leaves

$-2x^2 + x + 4,$

of degree $2$ , below the divisor's degree $3$ , so the loop stops. The quotient is $x^2 + 2x - 1$ and the remainder is $-2x^2 + x + 4$ . Multiplying back,

$(x^3 + 3x - 1)(x^2 + 2x - 1) = x^5 + 2x^4 + 2x^3 + 5x^2 - 5x + 1,$

and adding $-2x^2 + x + 4$ restores the dividend.

Part B

Take the checks one at a time. The quotient's degree should be $5 - 3 = 2$ , and $x^2 + 2x - 7$ has degree $2$ . Its leading coefficient should be $1 \div 1 = 1$ , and it is. The leftover's degree should be below $3$ , and it is.

$5 - 3 = 2, \qquad \deg(6x^2 + 17x - 2) = 2 < 3.$

Every one of those is a property the correct answer has, which is why a wrong answer can have them too. They are necessary conditions: failing one proves an error, and that makes them worth running, but passing all of them certifies nothing. None of them so much as looks at the coefficients the passes produced, and this student's fault moved terms between columns without disturbing a single degree, so it slipped under every one.

The check that catches it is the identity itself. Expand the divisor times the reported quotient:

$(x^3 + 3x - 1)(x^2 + 2x - 7) = x^5 + 2x^4 - 4x^3 + 5x^2 - 23x + 7.$

Adding the reported leftover $6x^2 + 17x - 2$ gives $x^5 + 2x^4 - 4x^3 + 11x^2 - 6x + 5$ , which is not the dividend: the $x^3$ terms already disagree, $-4x^3$ against the $2x^3$ that should be there. Multiplying back is the one check that tests the promise the answer is actually making.

Part C

The test is the two promises taken together, and neither half can be dropped.

$P = DQ + R, \qquad R = 0 \ \text{ or } \ \deg R < \deg D.$

Forward direction. The pair the loop produces satisfies both. The identity holds because $P = DS + W$ is true before the first pass, with $S = 0$ and $W = P$ , and a pass adds $q$ to $S$ while subtracting $qD$ from $W$ , which leaves $DS + W$ untouched. The degree condition holds because that is the only circumstance in which the loop stops.

Backward direction. Suppose $P = DQ_1 + R_1$ and $P = DQ_2 + R_2$ , with both leftovers legal. Subtracting one from the other,

$D(Q_1 - Q_2) = R_2 - R_1.$

If $Q_1$ and $Q_2$ differed, the left side would be a product of two nonzero polynomials, of degree $\deg D + \deg(Q_1 - Q_2)$ , hence at least $\deg D$ . The right side is a difference of two leftovers, so it is zero or of degree below $\deg D$ . No polynomial is both, so $Q_1 = Q_2$ , and then $R_1 = R_2$ follows at once.

Both directions earn their keep. Without the forward direction the test might be one nothing satisfies. Without the backward direction, passing it would certify nothing, and the whole point of the test is that it certifies. The student's answer is what happens when one promise is kept and the other is quietly broken.

In one line

The second pass is the first faulty line: $2x(x^3 + 3x - 1) = 2x^4 + 6x^2 - 2x$ , and the recorded version shifted its last two terms one column left, because the empty $x^3$ column that the divisor's missing $x^2$ leaves was skipped rather than kept open. Corrected, the quotient is $x^2 + 2x - 1$ and the remainder is $-2x^2 + x + 4$ . The degree and leading-coefficient checks let the fault through because they are conditions every correct answer meets rather than conditions only a correct answer meets; expanding $DQ + R$ against the dividend is what catches it. A pair is the quotient and remainder exactly when it keeps the identity and the degree condition together, which the loop's output does and which, by uniqueness, nothing else does.

Another way: Let the failed check point at the pass

Instead of reading the lines in order, expand $DQ + R$ for the reported pair first and subtract the dividend. The difference is $-6x^3 + 8x^2 - 2x$ , and its highest power is $x^3$ . Everything above $x^3$ therefore came out right, so the reread can begin at the $x^3$ column: whatever went wrong wrote into that column or below it, and no line that touched only higher columns needs looking at. That narrows the search the longer the division is.

When it is worth it When the work is long and rereading every pass is expensive, or when the recorded work is someone else's and hard to follow line by line.

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 5 points

Locates the first line that does not follow and names the specific fault in that line, rather than only asserting that the line is wrong or that a later line disagrees. . Worth 3 points.

Reworks the division from the faulty line onward and reports a quotient and remainder that rebuild the dividend when expanded. . Worth 2 points.

Part B 3 points

Explains that the degree and leading-coefficient checks are conditions any correct answer meets, so passing them cannot certify an answer even though failing one would condemn it. . Worth 2 points. needs an explanation, not just an answer

Names a check that tests the identity the answer claims rather than only the shape it must have, and says what its result is compared against. . Worth 1 point.

Part C 4 points

States both promises together as the test, rather than either one on its own. . Worth 2 points. needs an explanation, not just an answer

Argues the test in both directions, using uniqueness for the direction that says nothing other than the loop's output can pass it. . Worth 2 points.

Try a similar problem (Optional)

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

Divide $x^5 - x^4 + 2x^3 + 5x^2 - 9x + 2$ by $x^3 - 2x$ , then state the two-part test your answer has to pass and check it against both parts.

1. A forecast, a division, and the reason it ends . Foundational, 10 points. Question 1 of 5.

2. Gaps on both sides of the bracket . Foundational, 11 points. Question 2 of 5.

3. Recovering a height that was never measured . Application, 13 points. Question 3 of 5.

4. Two true sentences, one finished division . Reasoning, 12 points. Question 4 of 5.

5. An answer that passes the wrong checks . Reasoning, 12 points. Question 5 of 5.

Marking rubrics

1. A forecast, a division, and the reason it ends 10 points

2. Gaps on both sides of the bracket 11 points

3. Recovering a height that was never measured 13 points

4. Two true sentences, one finished division 12 points

5. An answer that passes the wrong checks 12 points