Supplies the machinery §3.6 needs. Establishes the division algorithm for polynomials, performs long division and its stripped-down form synthetic division, and proves the Remainder Theorem — that dividing by x minus c leaves f(c) as the remainder, which turns division into a root test and makes degree reduction possible.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 3 — Polynomial and Rational Functions
§3.5 Dividing Polynomials, pp. 362-374
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §3.5 Dividing Polynomials §3.5, pp. 362-374 — the pages these objectives are drawn from
Warm-up
Division looks like a technique in search of a purpose until you meet a cubic that will not factor.
Discussion prompt
You need to factor a cubic and can see by inspection that x equal to 2 is a root. What has that bought you, and what would you do next?
Hint: If 2 is a root, what factor must the cubic contain?
Answer:
A root at 2 means the cubic has x minus 2 as a factor. So the cubic is that factor times something, and that something is a quadratic.
Dividing the cubic by x minus 2 finds that quadratic, and a quadratic can be handled by factoring or by the quadratic formula — techniques you have had since Algebra 1.
So division is a degree-reduction machine. One known root converts a problem with no standard method into one with two. That is what this section is for, and why it sits immediately before the section on finding roots.
Concept
Polynomial division works like whole-number division, with degree playing the role of size. Its purpose here is that dividing by a known factor produces a quotient of lower degree, which is easier to handle.
division algorithm for polynomials — For polynomials f and d with d not zero, there are unique polynomials q and r with f equal to d times q plus r, where r is either zero or has degree lower than d.
\[ f(x)=d(x)\,q(x)+r(x), \qquad \deg r < \deg d \]
The uniqueness matters. Without the degree condition on the remainder there would be many ways to write the identity, and the quotient would not be well defined. The condition is what makes the answer a single answer, exactly as insisting the remainder be smaller than the divisor does for whole numbers.
Figure (svg): The division algorithm written as an identity, showing the dividend equal to the divisor times the quotient plus the remainder, with each part labelled and the degree condition on the remainder noted
OpenStax, Precalculus, §3.5 Dividing Polynomials §3.5, pp. 362-365
Section
Section 1
Concept
Dividing one polynomial by another produces a quotient and a remainder, related by the identity that the dividend equals the divisor times the quotient plus the remainder.
The parallel with whole numbers is exact and worth keeping in mind. Dividing 17 by 5 gives 3 remainder 2, and the remainder must be smaller than 5 or the quotient was too small. Here 'smaller' means lower degree, and everything else transfers.
Figure (svg): The division algorithm written as an identity, showing the dividend equal to the divisor times the quotient plus the remainder, with each part labelled and the degree condition on the remainder noted
OpenStax, Precalculus, §3.5 Dividing Polynomials §3.5, pp. 362-366
Picture it
Four objects, one identity, and one condition that makes the answer unique.
Figure (svg): The division algorithm written as an identity, showing the dividend equal to the divisor times the quotient plus the remainder, with each part labelled and the degree condition on the remainder noted
The degree condition on the remainder is the whole reason the quotient is well defined. Without it, any polynomial could be called a quotient with a suitably chosen remainder.
Worked example
The identity is the definition, so verifying is one multiplication.
\[ \text{Check that } x^3-1 \text{ divided by } x-1 \text{ gives } x^2+x+1 \text{ with remainder } 0. \]
Multiply the divisor by the quotient
Why: Expand the product.
\[ (x - 1) (x ^{2} + x + 1) \]
Expand the first bracket's x term
Why: Distributing x over the quadratic.
\[ x ^{3} + x ^{2} + x \]
Expand the minus one term
Why: Distributing negative 1.
\[ -x ^{2} - x - 1 \]
Collect
Why: The middle terms cancel in pairs.
\[ x ^{3} - 1 \]
Figure (svg): The solution to Worked example check a division by multiplying back shown as a ladder of expressions, one row per legal move
\[ (x-1)(x^2+x+1) = x^3-1 \]
Verify: note what the zero remainder tells you
Why: Since the remainder is zero, x minus 1 is a factor of x cubed minus 1, and 1 is therefore a root. Substituting confirms it: 1 minus 1 is zero. The division has produced a factorisation, which is exactly what it is used for.
OpenStax, Precalculus, §3.5 Dividing Polynomials §3.5, pp. 363-364
Sorting
The remainder must have lower degree than the divisor.
Sort into buckets
For a division by a quadratic divisor, sort each proposed remainder.
Worked example
The degrees are determined before any computation.
\[ \text{A degree } 5 \text{ polynomial is divided by a degree } 2 \text{ one. Describe the result.} \]
Find the quotient's degree
Why: Dividend's degree minus the divisor's.
\[ 5 - 2 = 3 \]
Apply the remainder condition
Why: Lower degree than the divisor.
\[ ^\circ 0\text{ or } 1 \]
Describe the remainder
Why: So it is linear or constant.
Note the special case
Why: A zero remainder means the divisor was a factor.
Figure (svg): The solution to Worked example predict the degrees shown as a ladder of expressions, one row per legal move
\[ \deg q = 3, \quad \deg r \le 1 \]
Verify: check the degrees are consistent
Why: The divisor times the quotient has degree 2 plus 3, which is 5 — matching the dividend, as the identity requires. The remainder's degree of at most 1 is too low to affect the leading behaviour, which is why it cannot disturb that match.
OpenStax, Precalculus, §3.5 Dividing Polynomials §3.5, pp. 365-366
Trap
\[ x^2+3x+5 \div (x+1): \quad \text{quotient } x, \text{ remainder } 2x+5 \]
Divide the leading terms once and stop
Why: The first step gives x, and subtracting leaves 2x plus 5.
The remainder is reported as 2x plus 5.
The remainder still has degree 1, the same as the divisor, so the division is not finished. The condition requires the remainder's degree to be strictly lower.
Continuing gives a quotient of x plus 2 with a remainder of 3, which is a constant and therefore acceptable.
Keep dividing while the remainder's degree is at least the divisor's. Stopping early is not merely inefficient — it produces an answer that fails the definition, and the quotient is wrong as a result.
Prediction
A degree 7 polynomial is divided by a degree 3 one.
Predict first
What is the quotient's degree?
Correct: 4.
Why: Degrees subtract under division, exactly as they add under multiplication, so 7 minus 3 gives 4. Checking: a degree 3 divisor times a degree 4 quotient has degree 7, matching the dividend. Ten would be the degree of their product, which is the opposite operation.
Faded example
Dividing gives a quotient of x plus 4 and a remainder of 7, with divisor x minus 2.
Fill in the blanks
f(x) = (x-2)(x+4) + 7 = x^2 + 2x - 8 + 7 = x^2 + 2x - 1
Why: Multiplying the divisor by the quotient gives x squared plus 2x minus 8, and adding the remainder 7 gives x squared plus 2x minus 1. Reconstructing the dividend this way is the standard check on any polynomial division, and it takes one expansion.
Analogy
Polynomial division mirrors whole-number division exactly.
Match the pairs
Why: Degree plays the role of size throughout. The correspondence is exact, which is why the algorithm has the same name in both settings and why the intuition transfers without adjustment.
Section
Section 2
Concept
Polynomial long division follows the same loop as numerical long division: divide the leading terms, multiply the divisor by that result, subtract, and bring down the next term.
The zero placeholders are not a formality. Dividing a polynomial with no x squared term without inserting a zero for it shifts every subsequent column by one and produces a quotient that is wrong in every coefficient, with no visible sign that anything went astray.
Figure (svg): The division algorithm written as an identity, showing the dividend equal to the divisor times the quotient plus the remainder, with each part labelled and the degree condition on the remainder noted
OpenStax, Precalculus, §3.5 Dividing Polynomials §3.5, pp. 366-369
Picture it
Every long division ends in this identity, whatever the degrees involved.
Figure (svg): The division algorithm written as an identity, showing the dividend equal to the divisor times the quotient plus the remainder, with each part labelled and the degree condition on the remainder noted
The loop repeats until the degree condition is met. Each pass reduces the working polynomial's degree by at least one, so it always terminates.
Worked example
One pass of the loop per degree.
\[ \text{Divide } x^3+2x^2-5x+2 \text{ by } x-1. \]
Divide the leading terms
Why: x cubed over x.
\[ \text{first term } x ^{2} \]
Multiply and subtract
Why: x squared times the divisor, subtracted.
\[ \text{leaves } 3 x ^{2} - 5 x \]
Repeat on what is left
Why: 3x squared over x.
\[ \text{next term } 3 x \]
Continue to the end
Why: Leaves negative 2x plus 2, then negative 2.
\[ \text{quotient } x ^{2} + 3 x - 2 \]
Figure (svg): The solution to Worked example divide with all terms present shown as a ladder of expressions, one row per legal move
\[ x^2+3x-2, \quad \text{remainder } 0 \]
Verify: multiply back
Why: The divisor times the quotient gives x cubed plus 3x squared minus 2x, minus x squared minus 3x plus 2, which collects to x cubed plus 2x squared minus 5x plus 2 — the original. And the zero remainder confirms 1 is a root, which substituting checks: 1 plus 2 minus 5 plus 2 is zero.
OpenStax, Precalculus, §3.5 Dividing Polynomials §3.5, pp. 367-368
Faded example
Prepare x to the fourth minus 3x plus 1 for long division.
Fill in the blanks
x^4 + 0x^3 + 0x^2 - 3x + 1
Why: The cubic and quadratic terms are absent, so both need zero placeholders to hold their columns. Without them the coefficients of the lower terms shift left during the division and every subsequent step operates on the wrong column, producing a quotient that is wrong throughout.
Worked example
The placeholder is what keeps the columns aligned.
\[ \text{Divide } x^3-8 \text{ by } x-2. \]
Insert placeholders
Why: There is no x squared or x term.
\[ x ^{3} + 0 x ^{2} + 0 x - 8 \]
Divide the leading terms
Why: x cubed over x.
\[ \text{first term } x ^{2} \]
Multiply and subtract
Why: Leaves 2x squared plus 0x.
\[ 2 x ^{2} + 0 x \]
Continue with the placeholders in place
Why: Next terms 2x and 4.
\[ \text{quotient } x ^{2} + 2 x + 4 \]
Figure (svg): The solution to Worked example a missing term shown as a ladder of expressions, one row per legal move
\[ x^2+2x+4, \quad \text{remainder } 0 \]
Verify: recognise the pattern
Why: This is the difference of cubes factorisation, and the division has produced it: x cubed minus 8 equals x minus 2 times x squared plus 2x plus 4. Without the placeholders the columns would have collapsed and the quotient would have come out as something like x squared minus 8 over x, which is not even a polynomial.
OpenStax, Precalculus, §3.5 Dividing Polynomials §3.5, pp. 368-369
Error analysis
A student performs one step of a long division.
Annotate
On: \( x^3+2x^2 - (x^3-x^2) = 2x^2 - x^2 = x^2 \)
Write the subtracted polynomial with its signs already flipped and then add. It converts a subtraction into an addition and removes the step where the signs get lost.
Prediction
A polynomial with a missing term is divided without inserting a placeholder.
Predict first
What goes wrong?
Correct: The columns shift and every later coefficient is wrong.
Why: Long division is a column-aligned procedure, and a missing column moves everything after it. The result is a quotient wrong in most of its coefficients, with nothing in the layout to indicate a problem — which is what makes this error so expensive. The division still terminates; it just terminates at the wrong answer.
Ranking
One pass of the long-division loop.
Put in order
Why: The loop divides, multiplies, subtracts and brings down, then repeats. The order is the same as in numerical long division, which is why the layout looks familiar. Subtracting before multiplying, or bringing down before subtracting, both break the alignment the method depends on.
Explain it to yourself
The loop repeats, and it always stops.
Discussion prompt
Explain why polynomial long division cannot go on forever.
Hint: What happens to the degree at each pass?
Answer:
Each pass cancels the current leading term by construction — that is what dividing the leading terms and subtracting achieves. So the working polynomial's degree drops by at least one every time.
Degrees are non-negative whole numbers, so they cannot drop forever. Eventually the working polynomial's degree falls below the divisor's, and at that point the loop's condition fails and the process stops.
That final working polynomial is the remainder, and it satisfies the degree condition automatically — not by choice but because that condition is exactly what made the loop stop. The algorithm's termination and the remainder's defining property are the same fact.
Section
Section 3
Concept
When the divisor is linear, only the coefficients matter, so the whole division can be carried out on a single row of numbers.
The sign of c is the standard trap. Dividing by x plus 3 means dividing by x minus negative 3, so the number written at the left is negative 3, not 3. Getting it wrong produces a completely different and entirely plausible-looking answer.
Figure (svg): A synthetic division laid out as a three-row array: the coefficients along the top, the running products in the middle, and the sums along the bottom, with the last entry circled as the remainder
OpenStax, Precalculus, §3.5 Dividing Polynomials §3.5, pp. 369-372
Picture it
Coefficients on top, products in the middle, sums at the bottom.
Figure (svg): A synthetic division laid out as a three-row array: the coefficients along the top, the running products in the middle, and the sums along the bottom, with the last entry circled as the remainder
The circled entry is the remainder and the rest are the quotient's coefficients, read as a polynomial of one degree lower than the dividend.
Worked example
Bring down, multiply, add, repeat.
\[ \text{Divide } 2x^3-5x^2+3x-4 \text{ by } x-3. \]
Write c and the coefficients
Why: c is 3; coefficients 2, -5, 3, -4.
\[ 3 | 2 - 5 3 - 4 \]
Bring down the first, multiply, add
Why: Two times three is six; negative five plus six is one.
\[ 2,\text{ then } 1 \]
Repeat
Why: One times three is three; three plus three is six.
\[ \text{then } 6 \]
Repeat once more
Why: Six times three is eighteen; negative four plus eighteen is fourteen.
\[ \text{then } 14 \]
Figure (svg): A synthetic division laid out as a three-row array: the coefficients along the top, the running products in the middle, and the sums along the bottom, with the last entry circled as the remainder
\[ 2x^2+x+6, \quad \text{remainder } 14 \]
Verify: check with the Remainder Theorem
Why: The remainder should equal the polynomial at 3. Substituting gives 54 minus 45 plus 9 minus 4, which is 14 — matching. That check costs one substitution and confirms the whole division, which is why the two ideas belong in the same section.
OpenStax, Precalculus, §3.5 Dividing Polynomials §3.5, pp. 370-371
Matching
Rewrite each in the form x minus c.
Match the pairs
Why: A minus in the divisor gives a positive c and a plus gives a negative one, because the standard form subtracts c. The reliable framing is that c is the root being tested: dividing by x plus 5 asks whether negative 5 is a root, so negative 5 is the number to use.
Worked example
The sign of c is the thing to get right.
\[ \text{Divide } x^3+4x^2-x-4 \text{ by } x+4. \]
Rewrite the divisor in standard form
Why: x plus 4 is x minus negative 4.
\[ c = -4 \]
Set up with c equal to negative 4
Why: Coefficients 1, 4, -1, -4.
\[ -4 | 1 4 - 1 - 4 \]
Run the loop
Why: Bring down 1; times -4 is -4; 4 plus -4 is 0.
\[ 1,\text{ then } 0 \]
Continue
Why: Zero times -4 is 0; -1 plus 0 is -1; -1 times -4 is 4; -4 plus 4 is 0.
\[ -1,\text{ then } 0 \]
Figure (svg): The solution to Worked example a divisor with a plus sign shown as a ladder of expressions, one row per legal move
\[ x^2-1, \quad \text{remainder } 0 \]
Verify: confirm the factorisation
Why: The zero remainder says x plus 4 is a factor, so the cubic is x plus 4 times x squared minus 1, which factors further into three linear factors with roots at negative 4, 1 and negative 1. Substituting negative 4 into the original gives negative 64 plus 64 plus 4 minus 4, which is zero — confirming both the root and the sign of c.
OpenStax, Precalculus, §3.5 Dividing Polynomials §3.5, pp. 371-372
Trap
\[ \text{dividing by } x+4 \;\Longrightarrow\; \text{write } 4 \text{ at the left} \]
Read the constant from the divisor
Why: The number 4 appears in x plus 4, so it is used directly.
The synthetic division is run with 4 rather than negative 4.
The layout uses c from the form x minus c. Since x plus 4 is x minus negative 4, the number written is negative 4.
The check is the root: dividing by x plus 4 tests whether negative 4 is a root, and that is the value that should appear at the left.
Ask what root the divisor is testing, and write that. It is the same sign reversal as everywhere else in the course, and framing it as 'which root' rather than 'which constant' removes the ambiguity.
Sorting
It applies only to linear divisors of the form x minus c.
Sort into buckets
Sort each divisor.
Faded example
A synthetic division of a cubic leaves the bottom row 3, 1, negative 2, 5.
Fill in the blanks
\text3 5x^2 + x - 2, \qquad \text___ ___
Why: The last entry is always the remainder, and the entries before it are the quotient's coefficients in descending order. The dividend was a cubic, so the quotient is a quadratic and the first three numbers are its coefficients. Reading all four as the quotient is the standard misreading.
Analogy
Synthetic division is long division with the algebra erased.
Match the pairs
Why: Every synthetic step corresponds to a long-division step. The third pairing is the one that explains the sign convention: writing c rather than negative c is what turns the subtraction into an addition, which is why the layout adds where long division subtracts.
Section
Section 4
Concept
The remainder on dividing a polynomial by x minus c is exactly the polynomial's value at c. So division and evaluation compute the same number.
\[ f(x)=(x-c)q(x)+r \;\Longrightarrow\; f(c)=r \]
The proof is worth seeing because it is so short. The identity holds for every input, so it holds at c in particular, and at c the factor x minus c is zero — so the divisor's whole contribution disappears and the remainder is all that survives.
Figure (svg): A graph with a point marked at an input, alongside a synthetic division whose remainder equals the output at that input, connecting the two
OpenStax, Precalculus, §3.5 Dividing Polynomials §3.5, pp. 372-374
Picture it
The graph's height at 3 and the division's remainder are the same 14.
Figure (svg): A graph with a point marked at an input, alongside a synthetic division whose remainder equals the output at that input, connecting the two
That coincidence is not one: substituting c into the division identity makes the divisor vanish, so the remainder is forced to equal the output there.
Worked example
Synthetic division computes an output.
\[ \text{Use the Remainder Theorem to find } f(2) \text{ for } f(x)=x^4-3x^2+2x-5. \]
Write the coefficients with placeholders
Why: There is no cubic term.
\[ 1, 0, -3, 2, -5 \]
Run synthetic division with c equal to 2
Why: Bring down 1; times 2 is 2; 0 plus 2 is 2.
\[ 1, 2 \]
Continue
Why: Two times 2 is 4; -3 plus 4 is 1; 1 times 2 is 2; 2 plus 2 is 4.
\[ 1, 4 \]
Finish
Why: Four times 2 is 8; -5 plus 8 is 3.
\[ \text{remainder } 3 \]
Figure (svg): A graph with a point marked at an input, alongside a synthetic division whose remainder equals the output at that input, connecting the two
\[ f(2)=3 \]
Verify: substitute directly
Why: Sixteen minus twelve plus four minus five is 3 — the same answer. For a fourth-degree polynomial the two routes are comparable in effort, but for higher degrees the synthetic route involves only multiplications and additions, with no powers to compute, which is why it is what a computer would do.
OpenStax, Precalculus, §3.5 Dividing Polynomials §3.5, pp. 372-373
Prediction
A polynomial has f(5) equal to 0.
Predict first
What is the remainder when it is divided by x minus 5?
Correct: Zero, so x - 5 is a factor.
Why: The Remainder Theorem says the remainder equals the value at 5, which is zero. A zero remainder means the divisor divides exactly, so x minus 5 is a factor — this is the Factor Theorem, and it is the Remainder Theorem in the special case that matters most.
Worked example
A zero remainder is the whole test.
\[ \text{Is } x=-2 \text{ a root of } f(x)=x^3+3x^2-4? \]
Write the coefficients with placeholders
Why: There is no x term.
\[ 1, 3, 0, -4 \]
Run synthetic division with c equal to negative 2
Why: Bring down 1; times -2 is -2; 3 plus -2 is 1.
\[ 1, 1 \]
Continue
Why: One times -2 is -2; 0 plus -2 is -2.
\[ -2 \]
Finish
Why: Negative 2 times -2 is 4; -4 plus 4 is 0.
\[ \text{remainder } 0 \]
Figure (svg): The solution to Worked example test whether a number is a root shown as a ladder of expressions, one row per legal move
\[ \text{Remainder } 0, \text{ so } -2 \text{ is a root and } x+2 \text{ is a factor.} \]
Verify: read off the bonus
Why: The division has also produced the quotient x squared plus x minus 2, which factors as x plus 2 times x minus 1. So the cubic is x plus 2 all squared, times x minus 1 — the root at negative 2 has multiplicity 2, which §3.4 says makes the graph bounce there. One division answered the root question and delivered the full factorisation.
OpenStax, Precalculus, §3.5 Dividing Polynomials §3.5, pp. 373-374
Error analysis
A student checks whether x plus 3 is a factor of a polynomial.
Annotate
On: \( \text{synthetic division with } c=3 \;\Longrightarrow\; \text{remainder } 12 \;\Longrightarrow\; x+3 \text{ is not a factor} \)
Name the root you are testing before setting up. Dividing by x plus 3 tests negative 3, and stating that out loud makes the sign impossible to get wrong.
Faded example
Synthetic division of a polynomial by x minus 4 leaves a remainder of negative 7.
Fill in the blanks
f(4) = -7
Why: The remainder on dividing by x minus 4 is the polynomial's value at 4, so f(4) is negative 7. The division has evaluated the polynomial without any powers being computed, which is the practical value of the theorem beyond its use as a root test.
Sorting
A zero remainder and a nonzero one carry different information.
Sort into buckets
Sort each conclusion for a division by x minus c.
Socratic
The Remainder Theorem has a very short proof.
Discussion prompt
Prove it, starting from the division identity.
Hint: The identity holds for every input. Try one particular input.
Answer:
The division identity says f of x equals x minus c, times the quotient, plus the remainder, and it holds for every input because it is an identity between polynomials.
So it holds at x equal to c in particular. There the factor x minus c is zero, so the entire first term vanishes no matter what the quotient is.
What survives is f of c equals the remainder. That is the theorem, in one substitution. The brevity is not a trick — it is because the identity was already doing all the work, and the theorem is just the observation that one particular input makes half of it disappear.
Section
Section 5
Concept
Once one root is known, dividing out its factor produces a polynomial of lower degree. Repeating this unwinds a polynomial into linear factors.
The last point is the practical one. After dividing out one factor you work with the quotient from then on; going back to the original for the second root wastes the reduction you just achieved and usually reintroduces the first root as a spurious repeat.
Figure (svg): A cubic being reduced to a quadratic by dividing out a known linear factor, shown as a chain with the degree dropping at each step until ordinary factoring applies
OpenStax, Precalculus, §3.5 Dividing Polynomials §3.5, pp. 373-374
Picture it
One known root converts an unfamiliar problem into a familiar one.
Figure (svg): A cubic being reduced to a quadratic by dividing out a known linear factor, shown as a chain with the degree dropping at each step until ordinary factoring applies
Finding that first root is the missing piece, and the next section supplies it. This section is the machinery that makes a found root worth having.
Worked example
One root, one division, then a quadratic.
\[ \text{Factor } f(x)=x^3-4x^2+x+6 \text{ given that } x=-1 \text{ is a root.} \]
Divide by the corresponding factor
Why: The root is negative 1, so c is negative 1.
\[ \text{divide by } x + 1 \]
Run synthetic division
Why: Coefficients 1, -4, 1, 6 with c equal to -1.
\[ \text{quotient } x ^{2} - 5 x + 6 \]
Factor the quadratic
Why: Two numbers multiplying to 6, adding to -5.
\[ (x - 2) (x - 3) \]
Assemble
Why: The linear factor and the two from the quadratic.
\[ (x + 1) (x - 2) (x - 3) \]
Figure (svg): A cubic being reduced to a quadratic by dividing out a known linear factor, shown as a chain with the degree dropping at each step until ordinary factoring applies
\[ f(x)=(x+1)(x-2)(x-3) \]
Verify: check the roots and the degree
Why: Substituting 2 gives 8 minus 16 plus 2 plus 6, which is zero, and 3 gives 27 minus 36 plus 3 plus 6, also zero. Three linear factors give degree 3, matching. The one given root did all the work: without it the cubic has no obvious factoring and no standard formula.
OpenStax, Precalculus, §3.5 Dividing Polynomials §3.5, pp. 373-374
Prediction
A degree 5 polynomial has a known root, and its factor is divided out.
Predict first
What degree is the quotient?
Correct: 4.
Why: Dividing by a linear factor reduces the degree by exactly one. Repeating with another known root would give a cubic, and a third would give a quadratic, at which point standard methods finish it. Each division buys exactly one degree.
Worked example
Each division works on the previous quotient.
\[ \text{Factor } f(x)=x^4-5x^3+5x^2+5x-6 \text{ given roots at } 1 \text{ and } 2. \]
Divide by the first factor
Why: Root 1, so divide by x minus 1.
\[ \text{quotient } x ^{3} - 4 x ^{2} + x + 6 \]
Divide the QUOTIENT by the second
Why: Root 2, applied to the cubic just found.
\[ \text{quotient } x ^{2} - 2 x - 3 \]
Factor the remaining quadratic
Why: Numbers multiplying to -3, adding to -2.
\[ (x - 3) (x + 1) \]
Assemble all four factors
Why: Two from the divisions, two from the quadratic.
\[ (x - 1) (x - 2) (x - 3) (x + 1) \]
Figure (svg): The solution to Worked example reduce twice shown as a ladder of expressions, one row per legal move
\[ f(x)=(x-1)(x-2)(x-3)(x+1) \]
Verify: check the second division was on the quotient
Why: The second synthetic division used the cubic's coefficients, not the quartic's. Going back to the quartic would have tested 2 against the original, which works but wastes the first reduction and leaves a cubic to factor rather than a quadratic. Each reduction should build on the last.
OpenStax, Precalculus, §3.5 Dividing Polynomials §3.5, pp. 374-374
Trap
\[ \text{after dividing out } (x-1), \text{ divide the ORIGINAL quartic by } (x-2) \]
Apply the second known root to the polynomial you started with
Why: The root 2 works on the original, so dividing by x minus 2 leaves no remainder.
The result is a cubic, and the process has to continue from there.
Divide the quotient, not the original. The first division already removed one factor, and working on the quotient removes a second, leaving a quadratic.
Going back to the original produces a cubic that still contains the first factor, so nothing has been gained by the first division at all.
Each reduction should build on the last. Two divisions applied in sequence take a quartic to a quadratic; two applied independently to the original take it to two different cubics, which is no progress.
Faded example
A cubic with a root at 3 is divided by the corresponding factor.
Fill in the blanks
\text3 (x - quadratic) \;\Longrightarrow\; \text______
Why: A root at 3 gives the factor x minus 3, and dividing a cubic by a linear factor leaves a quadratic. That quadratic can then be factored or handed to the quadratic formula, which is the point of the whole exercise: the problem has been converted into one with a standard method.
Sorting
Each division reduces the degree by one, and a quadratic needs none.
Sort into buckets
Sort each polynomial by how many divisions get you to a quadratic.
Real world
Division comes immediately before the section on finding roots.
Discussion prompt
Why is this section placed where it is, and what would be missing if the order were reversed?
Hint: What does a found root let you do, and what tool does that require?
Answer:
Finding a root is only useful if you can do something with it, and what you do is divide it out to reduce the degree. Without division, a found root of a quartic tells you one answer and leaves a quartic.
With division, that root converts the quartic into a cubic, which is a strictly easier problem — and repeating gets to a quadratic, where standard methods finish.
So this section is the tool and the next is the search. Reversing the order would leave students able to find roots and unable to exploit them, which is why every textbook puts division first even though it looks like the less interesting half.
Comparison
Fill the blanks from memory. One is general and one is fast.
Comparison matrix
| long division | synthetic division | |
|---|---|---|
| divisor allowed | any polynomial | only x minus c |
| what you write | full polynomials | coefficients only |
| the repeated step | divide, multiply, subtract | multiply by c and add |
| placeholders needed | yes, for every missing degree | yes, for the same reason |
| gives f(c) directly | yes, as the remainder | yes, as the last entry |
The fourth row catches people out in both methods, and for the same reason: both are column-aligned procedures, and a missing column shifts everything after it.
Pattern
Six steps, and the second is the one the next section supplies.
Step 4 is a free check that costs nothing: if the remainder is not zero, either the number tested was not a root or the division went wrong, and either way it is caught immediately rather than three steps later.
OpenStax Algebra and Trigonometry 2e, §5.4 Dividing Polynomials §5.4
Check
The sign of c.
Check your understanding
To divide a polynomial by x plus 6 using synthetic division, what number goes at the left?
Answer: A
Why: The layout uses c from the form x minus c, and x plus 6 is x minus negative 6. Equivalently, dividing by x plus 6 tests whether negative 6 is a root, so negative 6 is the value to use.
Check
The theorem in one step.
Check your understanding
Dividing f by x minus 4 leaves a remainder of 9. What is f(4)?
Answer: A
Why: The Remainder Theorem says the remainder on dividing by x minus c is exactly the value at c. So f(4) is 9, and since it is not zero, 4 is not a root.
Check
Degrees subtract.
Check your understanding
A degree 6 polynomial is divided by a linear factor. What is the quotient's degree?
Answer: A
Why: Dividing by a degree 1 factor reduces the degree by 1, giving a quotient of degree 5. This is the reduction that makes division worth performing: each known root buys one degree.
Real world
Synthetic division is how a computer evaluates a polynomial, under a different name.
Discussion prompt
Evaluating a degree 10 polynomial directly needs many multiplications. How does the synthetic method reduce that, and why does it matter?
Hint: Count the multiplications each way.
Answer:
Direct substitution computes each power separately — x squared, x cubed, and so on up to the tenth — then multiplies each by its coefficient. That is dozens of multiplications, and the powers get large enough to lose precision.
The synthetic method does one multiplication and one addition per coefficient: ten of each for a degree 10 polynomial. No powers are ever formed, so nothing large is computed and nothing overflows.
This is known as Horner's method, and it is what numerical libraries actually use. It is the same three-row computation from this section, which is a good example of a schoolroom shortcut turning out to be the professional algorithm rather than a simplification of it.
Commit first
State your confidence along with your answer.
Predict first
Synthetic division of a polynomial by x minus 2 gives a remainder of 0. What follows?
Correct: 2 is a root and x - 2 is a factor.
Why: A zero remainder means the value at 2 is zero, which is what being a root means, and it also means the divisor divides exactly, which is what being a factor means. The two statements are equivalent — that equivalence is the Factor Theorem, and it is why a single division answers both questions at once.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate why the Remainder Theorem is true, and why it makes division useful for factoring.
Hint: Substitute c into the division identity.
Answer:
Start from the identity: f of x equals x minus c times the quotient, plus the remainder. It holds for every input because both sides are the same polynomial.
Put x equal to c. The factor x minus c becomes zero, so the whole first term vanishes whatever the quotient is, leaving f of c equal to the remainder. That is the entire proof.
It is useful because a zero remainder now means a root, and a root means a factor. So one division both tests a candidate and, when it succeeds, hands you the quotient — which is the reduced polynomial you needed. The test and the reduction are the same computation, which is why the section pays for itself.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The second is where most arithmetic errors in the section occur, and the sign of c accounts for most of those. The fourth is the reason the whole section exists, and it is the piece the next section will assume you can perform without thinking about it.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Write the division identity with all four parts labelled, and note the degree condition on the remainder. Then perform one synthetic division of a cubic by a linear factor, laying out all three rows, and circle the remainder. Beside it, write what that remainder tells you about the polynomial's value and about whether you have found a factor.
If your circled remainder is annotated with both readings — the value at c, and the factor test — you have the theorem that makes this section more than arithmetic.
Recap
Five things, and the last is what the next section will be waiting for.
| if you remember one thing | it should be this |
|---|---|
| about the algorithm | the remainder's degree must be below the divisor's |
| about synthetic division | c is the root being tested, so x plus 3 means negative 3 |
| about the theorem | the remainder on dividing by x minus c is f(c) |
| about the purpose | each known root buys exactly one degree of reduction |
Section 3.6 supplies the missing half: a systematic way to find that first root, so the reduction machinery here has something to work on.
OpenStax, Precalculus, §3.5 Dividing Polynomials §3.5, pp. 362-374 — everything on these slides traces back here
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