Factor Theorem and Remainder Theorem: Polynomial Division Step by Step, with a Box-Volume Case

In 30 seconds

The remainder theorem: when a polynomial $p(x)$ is divided by $(x – a)$, the remainder is $p(a)$. No long division needed.
The factor theorem: $(x – a)$ is a factor of $p(x)$ exactly when $p(a) = 0$.
To factorise a cubic, test small values such as $\pm 1, \pm 2$ to find one root, divide by that factor, then factorise the quadratic that is left.
In our case, an open box folded from a 20 cm by 12 cm sheet holds 256 cm³ when the corner cut is 2 cm or about 2.88 cm; the third root, about 11.12 cm, is impossible.
Exam point: for a divisor like $(2x + 1)$, substitute $x = -\tfrac{1}{2}$, the value that makes the divisor zero.

The case: how deep should the box be?

A small packaging workshop makes open-topped trays from rectangular card sheets 20 cm long and 12 cm wide. A square of side $x$ cm is cut from each corner, and the four flaps are folded up to make the sides. A customer wants trays that hold exactly 256 cm³ of loose tea. (The figures are illustrative.)

After cutting, the base measures $(20 – 2x)$ cm by $(12 – 2x)$ cm and the tray is $x$ cm deep. So the volume is

\[ V(x) = x(20 – 2x)(12 – 2x) = 4x^3 – 64x^2 + 240x. \]

Setting $V(x) = 256$ gives a cubic equation. Unlike a quadratic, there is no short formula that students are expected to use for cubics. The workshop needs a method that finds every valid cut size, so that it can choose the one that suits its folding machine. This is exactly what the factor theorem and the remainder theorem are for.

The concept: what a remainder tells you about a polynomial

A polynomial is an expression made of whole-number powers of $x$ with constant coefficients, such as $2x^3 – 3x^2 + 4x – 5$. Its degree is the highest power. When you divide a polynomial $p(x)$ by a linear divisor $(x – a)$, you get a quotient $q(x)$ and a remainder $r$, just as dividing 17 by 5 gives quotient 3 and remainder 2. Because the divisor has degree 1, the remainder is a constant.^[1]

We can write the result as an identity, true for every value of $x$:

\[ p(x) \equiv (x – a)\,q(x) + r \]

Now substitute $x = a$. The bracket $(x – a)$ becomes zero, so the whole first term vanishes and we are left with $p(a) = r$. That one line is the proof of the remainder theorem. If the remainder is zero, $(x – a)$ divides $p(x)$ exactly, and so it is a factor. That is the factor theorem.^[1]

Key termFactor theorem: for a polynomial $p(x)$, $(x – a)$ is a factor of $p(x)$ if and only if $p(a) = 0$. More generally, $(bx – c)$ is a factor if and only if $p\!\left(\tfrac{c}{b}\right) = 0$. The related remainder theorem says the remainder on dividing $p(x)$ by $(x – a)$ is $p(a)$, and on dividing by $(bx – c)$ it is $p\!\left(\tfrac{c}{b}\right)$.

\[ p(x) = (x – a)\,q(x) + p(a), \qquad (x – a) \text{ is a factor} \iff p(a) = 0 \]

Here $p(x)$ is the polynomial being divided, $a$ is the number that makes the divisor zero, $q(x)$ is the quotient (one degree lower than $p$), and $p(a)$ is the remainder. The double arrow $\iff$ means “if and only if”: the statement works in both directions. A zero remainder proves a factor, and a factor guarantees a zero remainder.

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A quick way to divide by $(x – a)$ is synthetic division. Write the coefficients in a row. Bring down the first one. Multiply it by $a$, add the result to the next coefficient, and repeat. The last number is the remainder, and the others are the coefficients of the quotient.^[3] It is the same arithmetic as long division, laid out more compactly. You can also find the quotient by comparing coefficients, which many examiners accept and some students find safer.

Why this method fits the case

The box equation, once tidied, is a cubic with whole-number coefficients. For such cubics, any whole-number root must divide the constant term. That gives a short list of candidates to test. One substitution per candidate tells you instantly, through the factor theorem, whether you have found a root. Once one factor is known, division leaves a quadratic, and quadratics are easy.

The method also suits design work because it finds every root, not just one. A trial-and-error search on a spreadsheet might find the cut of 2 cm and stop there, missing the second valid answer near 2.88 cm. The workshop might prefer the deeper tray, so knowing both matters.

The remainder theorem adds something else: it lets you evaluate how far you are from the target. If $p(x) = V(x) – 256$, then the remainder on dividing by $(x – a)$ tells you how many cubic centimetres the tray with cut $a$ is over or under the target.

When is it not the right tool? If a cubic has no rational roots, testing candidates will never give zero. For example, $x^3 – 2x – 5 = 0$ has one real root near 2.09, and no simple value works. Then you need a numerical method such as iteration or the Newton-Raphson method, or a graphing tool. The factor theorem also only helps with linear factors; it will not split an irreducible quadratic such as $x^2 + 1$ into real factors.

Setting it up

Here are the quantities in the case.

Quantity	Symbol	Value or expression
Sheet length	–	20 cm
Sheet width	–	12 cm
Side of square cut from each corner	$x$	unknown, cm
Base length	–	$20 – 2x$ cm
Base width	–	$12 – 2x$ cm
Volume	$V(x)$	$4x^3 – 64x^2 + 240x$ cm³
Target volume	–	256 cm³
Valid cuts	–	$0 < x < 6$ (the width must stay positive)

Setting $V(x) = 256$ and dividing every term by 4 gives the equation we must solve:

\[ x^3 – 16x^2 + 60x – 64 = 0. \]

Solving it

We first practise the remainder theorem, then factorise a cubic fully, then solve the box case.

Worked example

Find the remainder when $p(x) = 2x^3 – 3x^2 + 4x – 5$ is divided by (a) $(x – 2)$ and (b) $(2x + 1)$.

Step 1For $(x – 2)$, the divisor is zero at $x = 2$. So the remainder is $p(2) = 2(8) – 3(4) + 4(2) – 5 = 16 – 12 + 8 – 5 = 7$.

Step 2For $(2x + 1)$, the divisor is zero at $x = -\tfrac{1}{2}$.

Step 3$p\!\left(-\tfrac{1}{2}\right) = 2\left(-\tfrac{1}{8}\right) – 3\left(\tfrac{1}{4}\right) + 4\left(-\tfrac{1}{2}\right) – 5 = -0.25 – 0.75 – 2 – 5 = -8$.

The remainders are 7 and −8. Neither is zero, so neither divisor is a factor.

Worked example

Factorise $p(x) = x^3 – 2x^2 – 5x + 6$ completely, and solve $p(x) = 0$.

Step 1Candidates for whole-number roots divide 6: $\pm 1, \pm 2, \pm 3, \pm 6$. Test $x = 1$: $1 – 2 – 5 + 6 = 0$. So $(x – 1)$ is a factor.

Step 2Divide using synthetic division with $a = 1$ on the coefficients $1, -2, -5, 6$: bring down 1; $1 \times 1 = 1$, $-2 + 1 = -1$; $-1 \times 1 = -1$, $-5 – 1 = -6$; $-6 \times 1 = -6$, $6 – 6 = 0$.

Step 3The quotient is $x^2 – x – 6$, remainder 0. Factorise: $x^2 – x – 6 = (x – 3)(x + 2)$.

Step 4So $p(x) = (x – 1)(x – 3)(x + 2)$, and the roots are $x = 1, 3, -2$.

Check by expanding: $(x – 1)(x^2 – x – 6) = x^3 – 2x^2 – 5x + 6$. Correct.

Worked example

Solve $x^3 – 16x^2 + 60x – 64 = 0$ to find every corner cut that gives a 256 cm³ tray.

Step 1Whole-number roots must divide 64. Test $x = 1$: $1 – 16 + 60 – 64 = -19$. Not zero.

Step 2Test $x = 2$: $8 – 64 + 120 – 64 = 0$. So $(x – 2)$ is a factor.

Step 3Synthetic division with $a = 2$ on $1, -16, 60, -64$: bring down 1; $2$, giving $-14$; $-28$, giving $32$; $64$, giving $0$. Quotient: $x^2 – 14x + 32$.

Step 4Solve $x^2 – 14x + 32 = 0$ with the quadratic formula: $x = \dfrac{14 \pm \sqrt{196 – 128}}{2} = 7 \pm \sqrt{17}$.

Step 5So $x = 2$, $x = 7 – \sqrt{17} \approx 2.877$ or $x = 7 + \sqrt{17} \approx 11.123$.

Step 6Only $0 < x < 6$ is possible, so reject 11.123.

The workshop can cut 2 cm squares (base 16 cm by 8 cm) or squares of about 2.88 cm (base about 14.25 cm by 6.25 cm).

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What the answer tells us

Two different trays hold the same 256 cm³. The 2 cm cut gives a shallow, wide tray: 16 cm by 8 cm by 2 cm, and $16 \times 8 \times 2 = 256$. The 2.88 cm cut gives a deeper, narrower tray: about 14.25 cm by 6.25 cm by 2.88 cm, which also multiplies to 256 cm³. A shallow tray is easier to fill and stack; a deeper one has a smaller footprint on a shelf. The algebra does not choose for the workshop, but it lays out both options.

The third root, about 11.12 cm, is a real solution of the cubic but makes no sense here. A cut that size would make the width $12 – 2x$ negative. This is a common pattern: a model equation can have roots that the real situation rules out, so always check them against the context.

Reasonableness check. Try nearby whole-number cuts: $V(1) = 1 \times 18 \times 10 = 180$, $V(2) = 256$, $V(3) = 3 \times 14 \times 6 = 252$ and $V(4) = 4 \times 12 \times 4 = 192$. The volume rises from 180 to 256 between $x = 1$ and $x = 2$, keeps rising a little, then falls back through 256 just before $x = 3$. That fits a second root between 2 and 3, as we found.

Sensitivity. The largest possible volume is only a little above 256. Calculus (setting $V'(x) = 12x^2 – 128x + 240 = 0$) puts the maximum at $x = \frac{32 – \sqrt{304}}{6} \approx 2.43$ cm, with a volume of about 262.7 cm³.^[4] So the target of 256 cm³ sits close to the top of the curve. That is why the two valid cuts are so close together. If the customer asked for 270 cm³, the cubic would have no root between 0 and 6, and no tray from this sheet could do it. A small change in the target near a maximum makes a big change in the answers, which is worth knowing before promising an order.

The graph shows the volume for every cut from 0 to 6 cm. The dashed line is the 256 cm³ target. It meets the curve twice, at the two valid cuts.

Tray volume against corner cut cm³

V(x) = x(20 − 2x)(12 − 2x)Target 256 cm³Maximum

Figure 1. The cubic $V(x)$ meets the 256 cm³ line at $x = 2$ and $x = 7 – \sqrt{17} \approx 2.88$. The third root, $7 + \sqrt{17}$, lies outside the valid range $0 < x < 6$.

Common mistakeUsing the wrong sign. To test whether $(x + 3)$ is a factor, substitute $x = -3$, not $x = 3$. The value to use is always the one that makes the bracket equal zero. For $(2x – 5)$ it is $x = \tfrac{5}{2}$.

Common mistakeLeaving out zero coefficients in division. To divide $x^3 – 7x + 6$ by $(x – 1)$, write the coefficients as $1, 0, -7, 6$. If you skip the 0 for the missing $x^2$ term, every later number shifts and the quotient is wrong.

Exam tipIn Cambridge 9709 Pure Mathematics 2 and 3, a common question type gives a cubic with one or two unknown coefficients and two conditions, such as “$(x – 2)$ is a factor” and “the remainder on dividing by $(x + 1)$ is 6”. Turn each condition into a linear equation using the factor or remainder theorem, then solve the two equations simultaneously.^[2] Show the substitution clearly: the method mark is often for writing $p(2) = 0$ with the numbers in place.

For a full chapter-style walk-through of polynomials, including division, the factor theorem and sketching cubics, see the Mathematics Made Simple textbook. To drill the method until it is quick and reliable, the Mathematics Made Simple Practice Questions book has many graded factor-theorem questions with every step worked.

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Practice questions

Try each one on paper first. Then open the worked answer and compare every step.

EasyQ1. Find the remainder when $x^3 + 2x^2 – 5x + 1$ is divided by $(x – 1)$. Show worked answer

By the remainder theorem, the remainder is $p(1)$.
$p(1) = 1 + 2 – 5 + 1 = -1$.
The remainder is −1.

EasyQ2. Show that $(x + 3)$ is a factor of $x^3 + x^2 – 9x – 9$, and factorise it completely. Show worked answer

$p(-3) = -27 + 9 + 27 – 9 = 0$, so $(x + 3)$ is a factor.
Group the terms: $x^2(x + 1) – 9(x + 1) = (x + 1)(x^2 – 9)$.
So $p(x) = (x + 3)(x – 3)(x + 1)$.

MediumQ3. Factorise $2x^3 + x^2 – 13x + 6$ completely. Show worked answer

Test $x = 2$: $16 + 4 – 26 + 6 = 0$. So $(x – 2)$ is a factor.
Synthetic division on $2, 1, -13, 6$ with $a = 2$: $2$; $1 + 4 = 5$; $-13 + 10 = -3$; $6 – 6 = 0$.
Quotient $2x^2 + 5x – 3 = (2x – 1)(x + 3)$.
So $p(x) = (x – 2)(2x – 1)(x + 3)$.

MediumQ4. The polynomial $p(x) = x^3 + ax^2 + bx – 10$ has $(x – 2)$ as a factor. When $p(x)$ is divided by $(x – 1)$ the remainder is −12. Find $a$ and $b$, then factorise $p(x)$. Show worked answer

$p(2) = 8 + 4a + 2b – 10 = 0$, so $2a + b = 1$.
$p(1) = 1 + a + b – 10 = -12$, so $a + b = -3$.
Subtract: $a = 4$, then $b = -7$.
$p(x) = x^3 + 4x^2 – 7x – 10 = (x – 2)(x^2 + 6x + 5) = (x – 2)(x + 1)(x + 5)$.

ExamQ5. The polynomial $p(x)$ is defined by $p(x) = 6x^3 + 7x^2 – x – 2$. (a) Show that $(x + 1)$ is a factor of $p(x)$. [1] (b) Hence factorise $p(x)$ completely. [3] (c) Find the remainder when $p(x)$ is divided by $(2x + 1)$. [2] Show worked answer

(a) $p(-1) = -6 + 7 + 1 – 2 = 0$, so $(x + 1)$ is a factor [B1].
(b) Divide: $p(x) = (x + 1)(6x^2 + x – 2)$ [M1 A1]. Then $6x^2 + x – 2 = (2x – 1)(3x + 2)$, so $p(x) = (x + 1)(2x – 1)(3x + 2)$ [A1].
(c) Substitute $x = -\tfrac{1}{2}$ [M1]: $6\left(-\tfrac{1}{8}\right) + 7\left(\tfrac{1}{4}\right) + \tfrac{1}{2} – 2 = -0.75 + 1.75 + 0.5 – 2 = -0.5$. Remainder $= -\tfrac{1}{2}$ [A1].

Frequently asked questions

What is the difference between the factor theorem and the remainder theorem?

The remainder theorem tells you the remainder, $p(a)$, when you divide by $(x – a)$. The factor theorem is the special case where that remainder is zero, which means $(x – a)$ is a factor. In practice you use the same substitution for both.

Which values should I try first when looking for a factor?

For a cubic with leading coefficient 1, whole-number roots must divide the constant term, so try its factors: start with $\pm 1$, then $\pm 2$, and so on. If the leading coefficient is not 1, also try fractions such as $\pm \tfrac{1}{2}$ or $\pm \tfrac{2}{3}$, built from factors of the constant over factors of the leading coefficient.

Do I have to use synthetic division?

No. Long division, synthetic division and comparing coefficients all give the same quotient. Use whichever you make fewest errors with, and show enough working for the examiner to follow.

Does every cubic have at least one real root?

Yes. A cubic goes to large positive values at one end and large negative values at the other, so its graph must cross the $x$-axis at least once. That root may not be a neat number, though, in which case you would use a numerical method instead.

References

Abramson, J. et al. (2021) Algebra and Trigonometry 2e. Houston: OpenStax. https://openstax.org/details/books/algebra-and-trigonometry-2e
Cambridge International (2024) Cambridge International AS & A Level Mathematics 9709: Syllabus for examination in 2026 and 2027. https://www.cambridgeinternational.org/programmes-and-qualifications/cambridge-international-as-and-a-level-mathematics-9709/
International Baccalaureate Organization (2019) Mathematics: analysis and approaches guide. Geneva: IBO.
Strang, G. and Herman, E. (2016) Calculus Volume 1. Houston: OpenStax. https://openstax.org/details/books/calculus-volume-1

Factor Theorem and Remainder Theorem: Polynomial Division Step by Step, with a Box-Volume Case

The case: how deep should the box be?

The concept: what a remainder tells you about a polynomial

Why this method fits the case

Setting it up

Solving it

What the answer tells us

Practice questions

Frequently asked questions

What is the difference between the factor theorem and the remainder theorem?

Which values should I try first when looking for a factor?

Do I have to use synthetic division?

Does every cubic have at least one real root?

References

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