Synthetic Division and the Remainder Theorem: Free Response

The divisor $x + 4$ is $x - (-4)$, so the corner number is $-4$ and not $4$. Every power from $x^3$ down to the constant is present, so the top row is the coefficient list as it stands, $3,\; 10,\; -5,\; -4$, with no placeholder needed.

Bring down the $3$. Then $3 \times (-4) = -12$ and $10 + (-12) = -2$; then $(-2) \times (-4) = 8$ and $-5 + 8 = 3$; then $3 \times (-4) = -12$ and $-4 + (-12) = -16$.

$\begin{array}{c|rrrr} -4 & 3 & 10 & -5 & -4 \\ & & -12 & 8 & -12 \\ \hline & 3 & -2 & 3 & -16 \end{array}$

The dividend is a cubic and the divisor is linear, so the quotient is a quadratic: the first three bottom entries are its coefficients and the last entry stands alone as the remainder. That reads as $Q(x) = 3x^2 - 2x + 3$ and $r = -16$.

Now the check, which costs one expansion and catches nearly every slip:

$(x + 4)\left(3x^2 - 2x + 3\right) = 3x^3 + 10x^2 - 5x + 12$

Adding $r = -16$ turns the constant $12$ into $-4$ and leaves the other three terms untouched, so the right side is $3x^3 + 10x^2 - 5x - 4$, which is $P(x)$ exactly.

Set the long division out and keep each subtracted product on its own line.

The first stage divides the leading terms, $3x^3 \div x = 3x^2$, multiplies the whole divisor, and subtracts:

$3x^2(x + 4) = 3x^3 + 12x^2, \qquad \left(3x^3 + 10x^2\right) - \left(3x^3 + 12x^2\right) = -2x^2$

The second stage repeats on what is left, $-2x^2 - 5x - 4$. Dividing leading terms gives $-2x$, and

$-2x(x + 4) = -2x^2 - 8x, \qquad (-5x) - (-8x) = 3x$

The third stage works on $3x - 4$. Dividing leading terms gives $3$, and

$3(x + 4) = 3x + 12, \qquad (-4) - 12 = -16$

The quotient is $3x^2 - 2x + 3$ and the remainder is $-16$, matching part A.

Now the correspondence, which is the point of the exercise, and it is a correspondence between NUMBERS: the long division writes terms, the tableau writes their coefficients. The three quotient terms $3x^2$, $-2x$ and $3$ carry the coefficients $3$, $-2$ and $3$, which are exactly the first three entries of the tableau's bottom row, in the same order. The terms that were subtracted into the three later columns, $12x^2$, $-8x$ and $12$, carry the coefficients $12$, $-8$ and $12$, and those are the negatives of the tableau's middle row $-12,\; 8,\; -12$, so each long-division subtraction is the corresponding tableau addition.

What the long division writes and the tableau does not is, first, every power of $x$, and second, the leading term of each subtracted product. That leading term always cancels: the quotient term was chosen for precisely that purpose, by dividing the leading terms. Writing a subtraction whose answer is known in advance to be $0$ records no information, so deleting it costs nothing. Every number that did carry information survives in one of the tableau's two lower rows.

Take one stage of the long division and look at what it does to a single column, keeping the arithmetic on numbers. Let $q$ be the newest quotient coefficient, so the quotient term at that stage is $q$ times some power of $x$. Multiplying that term by the whole divisor $x + 4$ produces two terms: one of the next power up, which cancels the leading term by construction and carries no information, and one of $4q$ times the original power. Only the second reaches the next column, and it is subtracted there:

$\text{next entry} = (\text{coefficient above}) - 4q$

Subtracting $4q$ is the same as adding $(-4)q$, and $-4$ is exactly the number sitting in the corner, because the corner holds the value that makes $x + 4$ zero. So the tableau's addition and the long division's subtraction are the same operation written with the sign attached to a different object. The sign is flipped once, at the moment the corner number is written, and never again. That is why a tableau contains no minus sign of its own even though the method it abbreviates is built out of subtractions, and it is also why writing the divisor's own constant in the corner breaks the correspondence: the one flip would never have happened.

The first step is different because the first stage of the long division is different. To choose the first quotient term it divides $3x^3$ by $x$, and the divisor is monic, so that division is by $1$ and the leading coefficient comes through unchanged:

$\frac{3x^3}{x} = 3x^2$

There is no earlier entry to multiply by the corner, and nothing to add, so the tableau simply brings the $3$ down. Every later coefficient does have an entry before it, which is why the multiply-and-add cycle starts only at the second column.

The $x^2$ term is missing, so the top row needs a placeholder zero in its slot: $2,\; -3,\; 1,\; 0,\; 6,\; -11$, with $3$ in the corner.

Each step is one multiplication and one addition. Bring down the $2$; then $2 \times 3 = 6$ and $-3 + 6 = 3$; then $3 \times 3 = 9$ and $1 + 9 = 10$; then $10 \times 3 = 30$ and $0 + 30 = 30$; then $30 \times 3 = 90$ and $6 + 90 = 96$; finally $96 \times 3 = 288$ and $-11 + 288 = 277$.

$\begin{array}{c|rrrrrr} 3 & 2 & -3 & 1 & 0 & 6 & -11 \\ & & 6 & 9 & 30 & 90 & 288 \\ \hline & 2 & 3 & 10 & 30 & 96 & 277 \end{array}$

The bottom row starts one degree below the dividend, so the quotient is $2x^4 + 3x^3 + 10x^2 + 30x + 96$, and the last entry is the remainder $277$. The Remainder Theorem says the remainder on dividing by $x - 3$ is the number $P(3)$, so that same $277$ is the value:

$P(3) = 277$

The brute-force route confirms it: $2(243) - 3(81) + 27 + 18 - 11 = 486 - 243 + 27 + 18 - 11 = 277$. A cheaper spot check is to test the identity at a convenient input. At $x = 1$ the quotient sums to $2 + 3 + 10 + 30 + 96 = 141$, so the identity predicts $(1 - 3)(141) + 277 = -5$, and $P(1) = 2 - 3 + 1 + 0 + 6 - 11 = -5$.

Count route one in its two stages, because they are different work. Building the powers takes one multiplication each for $c^2, c^3, \ldots, c^n$, which is $n - 1$ of them. Then each of the terms $a_n c^n, a_{n-1}c^{n-1}, \ldots, a_1 c$ needs one multiplication, which is another $n$; the constant term $a_0$ needs none. So

$(n - 1) + n = 2n - 1$

multiplications, together with $n$ additions to total the terms.

Route two is one multiplication and one addition per column after the bring-down, and there are $n$ such columns, so it spends $n$ multiplications and $n$ additions. The additions match, so the whole difference sits in the multiplications:

$(2n - 1) - n = n - 1$

At $n = 5$ route one spends $9$ and route two spends $5$, a saving of $4$. The saving is $n - 1$, which grows without bound, and the ratio $\frac{2n - 1}{n}$ climbs towards $2$, so for a high-degree polynomial the naive route costs close to twice as much.

One honest caveat about the polynomial in part A. The count $2n - 1$ assumes every coefficient has to be multiplied in, and here two of them do not: the missing $x^2$ term makes $a_2 = 0$, and $a_3 = 1$. An opportunistic route one skips both of those products, though it still has to build $c^2$ on the way to $c^3$, so it spends four power builds and three real scalings, which is $7$ rather than $9$. The tableau still spends exactly $5$, because it multiplies bottom-row entries by the corner and none of those entries is $0$ or $1$. Count the trivial products or skip them as you please, but do it on both sides of the comparison; either way the tableau is ahead here, so its advantage is not an artefact of this example.

The bottom row was never two different objects. Its entries $2,\; 3,\; 10,\; 30,\; 96$ are the coefficients of the quotient on dividing $P(x)$ by $x - 3$, and the final entry $277$ is the remainder, which the Remainder Theorem also reports as $P(3)$. So the run produces a division, and the value is the last line of it rather than the whole of it.

What that buys is the identity

$P(x) = (x - 3)\left(2x^4 + 3x^3 + 10x^2 + 30x + 96\right) + 277,$

which holds for every $x$ and is a genuine rewriting of $P$. A bare evaluation returns the single number $277$ and nothing else.

One concrete question the identity settles and the number does not: what is $P(5)$? Substituting into the identity, the quotient at $5$ is $1250 + 375 + 250 + 150 + 96 = 2121$, so

$P(5) = (5 - 3)(2121) + 277 = 4519,$

and no fifth power of $5$ was ever formed. The identity is also a statement about $P$ at every input rather than at one: it can be expanded back to check the division, or evaluated anywhere else you need. A lone value of $P(3)$ serves exactly one input and can be checked against nothing. The intermediate entries are the reason any of that is available.

Expand the right side and collect it by powers of $x$. Multiplying the quadratic by $x$ lifts every coefficient one degree, and multiplying it by $-c$ scales it:

$(x - c)\left(b_2x^2 + b_1x + b_0\right) + r = b_2x^3 + (b_1 - cb_2)x^2 + (b_0 - cb_1)x + (r - cb_0)$

The statement is an identity in $x$, which is stronger than an equation to be solved: it holds for every $x$, so the two sides are the same polynomial and their coefficients must agree power by power. Reading them off gives four equations:

$a_3 = b_2, \quad a_2 = b_1 - cb_2, \quad a_1 = b_0 - cb_1, \quad a_0 = r - cb_0$

Now solve them in the order the unknowns become available, each one using the value found just before it:

$b_2 = a_3, \quad b_1 = a_2 + cb_2, \quad b_0 = a_1 + cb_1, \quad r = a_0 + cb_0$

Match each line to the layout. The first says the leading coefficient of the quotient is the leading coefficient of the dividend, which is the bring-down. Each of the other three takes the entry just found, multiplies it by $c$, and adds the next coefficient of the dividend, which is the multiply-and-add step performed once per remaining column. So the rhythm was not chosen for convenience. Given the divisor $x - c$, these are the only values of $b_2$, $b_1$, $b_0$ and $r$ that can make the identity true.

The cubic was only for concreteness, and nothing above used its degree. In general, write $P(x) = a_nx^n + \cdots + a_0$ and $Q(x) = b_{n-1}x^{n-1} + \cdots + b_0$. In the product $(x - c)Q(x)$, a given power $x^k$ collects $b_{k-1}$ from multiplying $Q$ by $x$ and $-c\,b_k$ from multiplying it by $-c$, so the coefficient of $x^k$ in $(x - c)Q(x) + r$ is $b_{k-1} - c\,b_k$. Matching it against $a_k$ gives

$b_{k-1} = a_k + c\,b_k \qquad \text{for } 1 \le k \le n - 1,$

with the two boundary equations $b_{n-1} = a_n$ from the leading power and $r = a_0 + c\,b_0$ from the constant one. That is the bring-down followed by multiply-and-add in every degree, and the four equations above are exactly this statement at $n = 3$.

The four equations of part A pair off one for one with the coefficients $a_3, a_2, a_1, a_0$: each step consumes exactly one of them, in descending order of degree. The tableau's columns are that pairing made visible. Leaving a column out does not tell the method that a power is absent; it shifts every later coefficient into the slot of a higher power, so each step is handed the wrong $a_k$. The zero is not a formatting habit. It is the value of $a_k$ for that power, and the recurrence needs it.

With the full row, $3x^3 - 8x + 5$ has coefficients $3,\; 0,\; -8,\; 5$, and $x + 3$ puts $-3$ in the corner:

$\begin{array}{c|rrrr} -3 & 3 & 0 & -8 & 5 \\ & & -9 & 27 & -57 \\ \hline & 3 & -9 & 19 & -52 \end{array}$

So the quotient is $3x^2 - 9x + 19$ and the remainder is $-52$. Expanding confirms it, since $(x + 3)\left(3x^2 - 9x + 19\right) = 3x^3 - 8x + 57$ and adding $-52$ gives $3x^3 - 8x + 5$.

With the shortened row $3,\; -8,\; 5$ and the same corner:

$\begin{array}{c|rrr} -3 & 3 & -8 & 5 \\ & & -9 & 51 \\ \hline & 3 & -17 & 56 \end{array}$

That run is not nonsense, and calling it wrong misses what happened. It is a perfectly correct division of the polynomial whose coefficients those three numbers are, namely $3x^2 - 8x + 5$, by $x + 3$:

$(x + 3)(3x - 17) + 56 = 3x^2 - 8x + 5$

So the shortened row answered a question nobody asked. It divided a quadratic that shares its written digits with the cubic and is a different polynomial, which is exactly the damage the missing zero does: the machine cannot see the degree, only the list.

Run the tableau they ran, with $3$ in the corner and the full row $3,\; 0,\; -8,\; 5$. Bring down the $3$; then $3 \times 3 = 9$ and $0 + 9 = 9$; then $9 \times 3 = 27$ and $-8 + 27 = 19$; then $19 \times 3 = 57$ and $5 + 57 = 62$.

$\begin{array}{c|rrrr} 3 & 3 & 0 & -8 & 5 \\ & & 9 & 27 & 57 \\ \hline & 3 & 9 & 19 & 62 \end{array}$

Nothing here is arithmetically wrong, and that is the whole diagnosis. The corner holds the value that makes the divisor zero, so a corner of $3$ belongs to the divisor $x - 3$. Their tableau is a correct division by $x - 3$, and the identity it produces holds:

$(x - 3)\left(3x^2 + 9x + 19\right) + 62 = 3x^3 - 8x + 5$

By the Remainder Theorem their last cell is therefore $P(3)$, and a direct substitution agrees:

$P(3) = 3(27) - 8(3) + 5 = 81 - 24 + 5 = 62$

The number they wanted was $P(-3) = -52$, from part B. So the fault is not in the arithmetic and not in the layout: it is that the tableau silently answered the question for a different divisor, and both answers look equally plausible on the page.

Factor the leading coefficient out of the divisor: $2x + 5 = 2\left(x + \tfrac{5}{2}\right)$. The monic factor is $x + \tfrac{5}{2} = x - \left(-\tfrac{5}{2}\right)$, so the corner number is $-\tfrac{5}{2}$, the value that makes $2x + 5$ zero.

Bring down the $4$; then $4 \times \left(-\tfrac{5}{2}\right) = -10$ and $4 + (-10) = -6$; then $(-6) \times \left(-\tfrac{5}{2}\right) = 15$ and $-7 + 15 = 8$; then $8 \times \left(-\tfrac{5}{2}\right) = -20$ and $14 + (-20) = -6$.

$\begin{array}{c|rrrr} -\tfrac{5}{2} & 4 & 4 & -7 & 14 \\ & & -10 & 15 & -20 \\ \hline & 4 & -6 & 8 & -6 \end{array}$

So far this says $P(x) = \left(x + \tfrac{5}{2}\right)\left(4x^2 - 6x + 8\right) - 6$, which answers a question about the monic divisor, not about $2x + 5$. Convert by writing the monic factor as $\tfrac{1}{2}(2x + 5)$ and letting the $\tfrac{1}{2}$ act on the quadratic beside it:

$P(x) = \tfrac{1}{2}(2x + 5)\left(4x^2 - 6x + 8\right) - 6 = (2x + 5)\left(2x^2 - 3x + 4\right) - 6$

The quotient for the divisor $2x + 5$ is therefore $2x^2 - 3x + 4$ and the remainder is $-6$. Expanding checks it:

$(2x + 5)\left(2x^2 - 3x + 4\right) = 4x^3 + 4x^2 - 7x + 20,$

and adding $-6$ turns the constant $20$ into $14$, returning $P(x)$. As a second check, the Remainder Theorem says the remainder is $P$ evaluated at the number killing the divisor, and $P\left(-\tfrac{5}{2}\right) = -\tfrac{125}{2} + 25 + \tfrac{35}{2} + 14 = -6$.

Test the claim in one line, the way any proposed division can be tested. Their pair asserts that $P(x) = (2x + 5)\left(2x^2 - 3x + 4\right) - 3$, and the product was expanded in part A:

$(2x + 5)\left(2x^2 - 3x + 4\right) - 3 = 4x^3 + 4x^2 - 7x + 20 - 3 = 4x^3 + 4x^2 - 7x + 17$

That is not $P(x)$: the constant term is $17$ where $P$ has $14$, so their identity fails by $3$ at every value of $x$. Substituting one value makes the same point, since at $x = 0$ their right side is $17$ while $P(0) = 14$.

Now the reason, which lives in the structure of the identity rather than in the arithmetic. The monic tableau produced

$P(x) = \left(x + \tfrac{5}{2}\right)\left(4x^2 - 6x + 8\right) + (-6),$

and the conversion replaces $x + \tfrac{5}{2}$ by $\tfrac{1}{2}(2x + 5)$. That $\tfrac{1}{2}$ is now one factor of a product of three things, so it can be moved anywhere inside that product, and the natural home is the quadratic, where it halves each coefficient. What it cannot do is move out of the product and into the term added on beside it. The remainder $-6$ is not a factor of anything; it is a separate summand, and no rearrangement of a product reaches it.

So exactly one of the two pieces changes under the conversion. The quotient is divided by $2$ and the remainder stays as it is.

The divisor $x - \tfrac{b}{a}$ is monic, so it is not the zero polynomial and the division algorithm applies. Dividing $P(x)$ by it gives

$P(x) = \left(x - \tfrac{b}{a}\right)Q_0(x) + r,$

where the remainder is either $0$ or has degree below the divisor's degree $1$. In the first case it is the number $0$; in the second its degree is $0$, which makes it a nonzero constant. Either way it is a number, written $r$ above.

The only step is to rewrite the divisor. Factoring $a$ out of $ax - b$ gives $ax - b = a\left(x - \tfrac{b}{a}\right)$, which is legitimate because $a \neq 0$, and dividing through by $a$ turns it around:

$x - \tfrac{b}{a} = \tfrac{1}{a}(ax - b)$

Substitute that into the identity and change nothing else. Multiplication is associative and commutative, so the constant $\tfrac{1}{a}$ may be regrouped with $Q_0(x)$:

$P(x) = \tfrac{1}{a}(ax - b)\,Q_0(x) + r = (ax - b)\left(\tfrac{1}{a}Q_0(x)\right) + r$

This is an identity of the required shape with divisor $ax - b$, quotient $\tfrac{1}{a}Q_0(x)$ and remainder $r$. Three things deserve checking. First, $\tfrac{1}{a}Q_0(x)$ is a polynomial, since scaling every coefficient by a nonzero constant produces one; it is the zero polynomial exactly when $Q_0$ is, and otherwise it has the same degree, because $\tfrac{1}{a}$ cannot turn a nonzero leading coefficient into $0$. Second, $r$ is the same number as before, so it is again either $0$ or a constant of degree $0$, and both of those clear the degree $1$ of $ax - b$; the degree condition that makes a division finished is met. By the uniqueness of quotient and remainder from the previous lesson, this pair is therefore the answer, not merely an answer.

Third, the value. The Remainder Theorem applied at $c = \tfrac{b}{a}$ says that the remainder on dividing by $x - \tfrac{b}{a}$ is $P\!\left(\tfrac{b}{a}\right)$, and the work above showed that this same $r$ is the remainder for $ax - b$. So

$r = P\!\left(\tfrac{b}{a}\right),$

the polynomial evaluated at the number that makes the divisor zero. Part A is the case $a = 2$, $b = -5$, where $\tfrac{b}{a} = -\tfrac{5}{2}$, the quotient was halved and the remainder $-6$ stayed put.