What is handling a non-monic linear factor?

A factor like 2x - 1 is linear but not of the form x - a. The fix is to find the value of x that makes it zero. Setting 2x - 1 = 0 gives x = 12, so 2x - 1 is a factor of P(x) exactly when P\!(12) = 0. More generally bx - c is a factor when P\!(cb) = 0, because cb is the zero of that linear expression.

§-Syllabus dot point

NSWMaths Extension 1Syllabus dot point

How can the remainder and factor theorems tell us the result of a division, and even fully factor a polynomial, without carrying out the long division?

Use the remainder theorem (the remainder on division by x - a is P(a)) and the factor theorem ((x - a) is a factor if and only if P(a) = 0), test divisors of the constant term to locate integer zeroes, and combine these with division to find unknown coefficients and to fully factor a polynomial

A first-contact answer to the HSC Maths Extension 1 dot point on the remainder and factor theorems. The remainder on division by (x - a) is P(a); (x - a) is a factor exactly when P(a) = 0; an integer zero must divide the constant term. Find remainders without dividing, find unknown coefficients, and fully factor a cubic, with worked examples.

Generated by Claude Opus 4.817 min answerUpdated 2026-06-21

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Quick answer

The remainder theorem says the remainder when a polynomial $P(x)$ is divided by $x - a$ is just $P(a)$ , so you can find a remainder by substitution with no long division (for $x + a$ , use $P(-a)$ ). The factor theorem follows: $x - a$ is a factor of $P(x)$ exactly when $P(a) = 0$ , linking each zero $x = a$ to a factor $x - a$ . For integer coefficients, every integer zero divides the constant term, so to factor a cubic or quartic you list the divisors of the constant, test them until one gives zero, divide out that factor, and factor the quotient, leaving any negative-discriminant quadratic irreducible. A non-monic factor like $2x - 1$ is handled by substituting its zero $x = \tfrac{1}{2}$ . Always check by expanding the factors back to $P(x)$ .

What this dot point is asking

On the previous page you divided one polynomial by another with the full long-division tableau, and you saw that the remainder appears at the foot of the working. That is reliable but slow, and for the most common case, a linear divisor $x - a$ , there is a shortcut that gets the remainder in a single line. The remainder theorem says that the remainder on dividing $P(x)$ by $x - a$ is simply $P(a)$ , the value of the polynomial at $x = a$ . From that one idea everything else on this page follows: the factor theorem (a divisor leaves no remainder exactly when that value is zero), a quick test for which whole numbers could possibly be zeroes, and a clean method to factor a cubic or quartic by hand.

This is the Year 11 first-contact version, reasoned from the division algorithm $P(x) = D(x)Q(x) + R(x)$ you have already met. The applied, exam-heavy treatment that leans on these theorems lives on the Year 12 page on polynomial division and the factor theorem, and the link from factors to the sum and product of roots is developed at roots and coefficients. Here the goal is to make these two theorems automatic and to use them to factor a polynomial without grinding through repeated long divisions.

The answer

The remainder theorem

Start from the division algorithm with a linear divisor. Dividing $P(x)$ by $x - a$ gives a quotient $Q(x)$ and a remainder whose degree is less than $1$ , that is, a constant $r$ :

P(x) = (x - a)\,Q(x) + r.

This identity is true for every value of $x$ , so substitute the one value that kills the quotient term. Putting $x = a$ makes the bracket $(a - a) = 0$ , and the whole $(x - a)Q(x)$ piece vanishes, leaving $P(a) = r$ . The remainder is exactly $P(a)$ .

The power of this is that a four-line tableau collapses into one substitution. On the previous page the long division of $x^3 - 2x^2 - 5x + 9$ by $x - 3$ ended with remainder $3$ . The remainder theorem gets the same number immediately:

P(3) = 3^3 - 2(3)^2 - 5(3) + 9 = 27 - 18 - 15 + 9 = 3.

The two methods agree, as they must, because $P(a)$ is the remainder. Long division still earns its keep when you also need the quotient (for example, to finish a factorisation), but for the remainder alone, substitution wins every time.

The factor theorem

A remainder of zero is special: it means the divisor goes in exactly, so $x - a$ is a factor of $P(x)$ . Combine that with the remainder theorem, which says the remainder is $P(a)$ , and you get a one-step test for factors.

This is the bridge between the two ways of describing a polynomial, its zeroes (where the graph meets the $x$ -axis) and its factors (how it splits into a product). Every zero $x = a$ corresponds to a factor $x - a$ , and vice versa, which is the engine behind factoring by hand and behind reading a factored form straight off a graph (developed at graphs of polynomials and multiplicity).

The integer-zero test: which numbers to try

The factor theorem gives a test for a given number, but it does not say which numbers to try. For a polynomial with integer coefficients there is a sharp restriction that cuts the search to a short list.

The reason is short. Suppose $x = k$ is an integer zero of $P(x) = a_n x^n + \dots + a_1 x + a_0$ . Then

a_n k^n + a_{n-1}k^{n-1} + \dots + a_1 k + a_0 = 0,

so $a_0 = -k\left(a_n k^{n-1} + \dots + a_1\right)$ . Every term on the right has a factor of $k$ , so $k$ divides $a_0$ . An integer zero therefore cannot be anything except a divisor of the constant term, which is why you only ever test those divisors. (This is the integer case of the wider rational-root idea: a rational zero $\tfrac{p}{q}$ in lowest terms has $p$ dividing the constant term and $q$ dividing the leading coefficient. For a monic polynomial $q = 1$ , so the only rational zeroes are integers dividing $a_0$ .)

So to start factoring $x^3 - 4x^2 + x + 6$ you do not guess wildly: the constant term is $6$ , so the only possible integer zeroes are $\pm 1, \pm 2, \pm 3, \pm 6$ . You test those eight numbers with the remainder theorem until one gives zero.

Fully factoring a polynomial

Putting the three ideas together gives a dependable hand method for factoring a cubic or quartic that does not factor by inspection:

List the candidates. Write down the divisors of the constant term (both signs). These are the only possible integer zeroes.
Find one zero. Test the candidates with the remainder theorem until some $P(a) = 0$ . By the factor theorem, $x - a$ is then a factor.
Divide it out. Long-divide $P(x)$ by $x - a$ to get $P(x) = (x - a)Q(x)$ , dropping the degree by one.
Repeat or finish. Factor the quotient $Q(x)$ : if it is a quadratic, use the usual methods (or the quadratic formula); if it is still a cubic, repeat the search on $Q(x)$ . Stop when you have a product of linear factors and, if any quadratic factor has a negative discriminant, leave that quadratic as an irreducible factor over the reals.

Each pass turns one zero into one linear factor and shrinks the problem, so a cubic takes one division and a quartic at most two. The division step is exactly the long division from the previous page; the new content here is using the theorems to find that first factor without trial division.

Handling a non-monic linear factor

A factor like $2x - 1$ is linear but not of the form $x - a$ . The fix is to find the value of $x$ that makes it zero. Setting $2x - 1 = 0$ gives $x = \tfrac{1}{2}$ , so $2x - 1$ is a factor of $P(x)$ exactly when $P\!\left(\tfrac{1}{2}\right) = 0$ . More generally $bx - c$ is a factor when $P\!\left(\tfrac{c}{b}\right) = 0$ , because $\tfrac{c}{b}$ is the zero of that linear expression. You test the fraction, not a whole number, but the logic is identical to the factor theorem.

Exam technique

The theorems are fast, but the marks come from saying why each line is true. Lock them in like this:

Quote the theorem by name before you substitute. Write "by the remainder theorem, remainder $= P(2)$ " or "by the factor theorem, $x - 3$ is a factor since $P(3) = 0$ ." Markers reward the justification, not just the arithmetic.
Mind the sign in $x + a$ . A divisor $x + 2$ means $a = -2$ , so evaluate $P(-2)$ , not $P(2)$ . Reversing this sign is the single most common error in the topic.
For a non-monic factor, substitute its zero. For $2x - 1$ , evaluate $P\!\left(\tfrac{1}{2}\right)$ ; for $3x + 2$ , evaluate $P\!\left(-\tfrac{2}{3}\right)$ . State the zero you are using.
List the divisors of the constant before testing. Show the candidate set, then test in order. It demonstrates the integer-zero test and stops you guessing past the answer.
Always expand the factors back. Multiply your final factored form out and confirm it equals $P(x)$ . It is quick and catches a slip in any division.

How exam questions ask about these theorems

The wording is predictable once you map it to the right theorem.

"Find the remainder when $P(x)$ is divided by $x - a$ ." Remainder theorem: evaluate $P(a)$ and write it down. No division.
"Show that $x - a$ is a factor of $P(x)$ " or "Show that $x = a$ is a zero." Factor theorem: compute $P(a)$ and state it equals $0$ , so $x - a$ is a factor.
"When $P(x)$ is divided by $x - a$ the remainder is $r$ ; find the unknown." Set $P(a) = r$ and solve for the unknown coefficient. Two such conditions give two equations to solve simultaneously.
"Given $x - a$ is a factor, find $k$ ." Set $P(a) = 0$ and solve for $k$ ; "divisible by" and "is a factor" both mean remainder zero.
" $2x - a$ is a factor; use this to factor $P(x)$ ." Confirm with $P\!\left(\tfrac{a}{2}\right) = 0$ , divide it out, then factor the quotient.
"Factor $P(x)$ completely." Run the full method: divisors of the constant, find a zero, divide out, factor the quotient, and check by expanding.

Worked example

Find a remainder without dividing

Find the remainder when $P(x) = 2x^3 - 7x^2 + 4x + 1$ is divided by $x - 3$ .

Name the theorem. By the remainder theorem, the remainder on division by $x - a$ is $P(a)$ . The divisor is $x - 3$ , so $a = 3$ and the remainder is $P(3)$ .

Substitute $x = 3$ .

P(3) = 2(3)^3 - 7(3)^2 + 4(3) + 1 = 54 - 63 + 12 + 1 = 4.

State the answer. The remainder is $4$ . A full long division would end with the same constant at the foot of the tableau, but the substitution gets there in one line.

Test whether a linear expression is a factor

For $P(x) = x^3 - 4x^2 + x + 6$ , decide whether $x - 2$ is a factor and whether $x - 1$ is a factor.

Test $x - 2$ with the factor theorem. The factor theorem says $x - 2$ is a factor exactly when $P(2) = 0$ .

P(2) = 2^3 - 4(2)^2 + 2 + 6 = 8 - 16 + 2 + 6 = 0,

so $x - 2$ is a factor.

Test $x - 1$ .

P(1) = 1^3 - 4(1)^2 + 1 + 6 = 1 - 4 + 1 + 6 = 4 \ne 0,

so $x - 1$ is not a factor. The non-zero value $4$ is the remainder on division by $x - 1$ , which is consistent: a factor leaves remainder zero, a non-factor leaves the remainder $P(1) = 4$ .

Find unknown coefficients from two remainder conditions

The polynomial $P(x) = x^3 + ax^2 + bx - 4$ leaves a remainder of $6$ when divided by $x - 1$ , and is exactly divisible by $x + 2$ . Find $a$ and $b$ .

Translate each condition. By the remainder theorem, dividing by $x - 1$ leaves $P(1)$ , so $P(1) = 6$ . "Exactly divisible by $x + 2$ " means the remainder there is zero, and $x + 2 = x - (-2)$ , so by the factor theorem $P(-2) = 0$ .

First equation from $P(1) = 6$ .

P(1) = 1 + a + b - 4 = a + b - 3 = 6 \implies a + b = 9. \quad (1)

Second equation from $P(-2) = 0$ .

P(-2) = -8 + 4a - 2b - 4 = 4a - 2b - 12 = 0 \implies 2a - b = 6. \quad (2)

Solve simultaneously. Adding $(1)$ and $(2)$ gives $3a = 15$ , so $a = 5$ ; then $b = 9 - 5 = 4$ from $(1)$ .

Check both conditions. With $P(x) = x^3 + 5x^2 + 4x - 4$ : $P(1) = 1 + 5 + 4 - 4 = 6$ and $P(-2) = -8 + 20 - 8 - 4 = 0$ . Both hold, so $a = 5$ and $b = 4$ .

Fully factor a cubic by testing divisors of the constant term

Factor $P(x) = x^3 - 4x^2 + x + 6$ completely.

List the candidate zeroes: The coefficients are integers, so by the integer-zero test any integer zero divides the constant term $6$ . The candidates are $\pm 1, \pm 2, \pm 3, \pm 6$ .
Find one zero: Test in order: $P(1) = 1 - 4 + 1 + 6 = 4 \ne 0$ , then $P(-1) = -1 - 4 - 1 + 6 = 0$ . So $x = -1$ is a zero and, by the factor theorem, $x + 1$ is a factor.
Divide out the factor: Long division of $P(x)$ by $x + 1$ gives the quotient $x^2 - 5x + 6$ , so $P(x) = (x + 1)(x^2 - 5x + 6)$ .
Factor the quotient: $x^2 - 5x + 6 = (x - 2)(x - 3)$ , hence

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

Check by expanding back. $(x + 1)(x^2 - 5x + 6) = x^3 - 5x^2 + 6x + x^2 - 5x + 6 = x^3 - 4x^2 + x + 6 = P(x)$ . The three zeroes $-1, 2, 3$ are all divisors of $6$ , exactly as the integer-zero test predicted.

Use a non-monic factor (2x - 1) by substituting its zero

Show that $2x - 1$ is a factor of $P(x) = 2x^3 + x^2 - 13x + 6$ , then factor $P(x)$ completely.

Find the zero of the linear factor. $2x - 1 = 0$ at $x = \tfrac{1}{2}$ , so $2x - 1$ is a factor exactly when $P\!\left(\tfrac{1}{2}\right) = 0$ .

Evaluate at the fraction.

P\!\left(\tfrac{1}{2}\right) = 2 \cdot \tfrac{1}{8} + \tfrac{1}{4} - 13 \cdot \tfrac{1}{2} + 6 = \tfrac{1}{4} + \tfrac{1}{4} - \tfrac{13}{2} + 6 = \tfrac{1}{2} - \tfrac{13}{2} + 6 = -6 + 6 = 0,

so $2x - 1$ is a factor.

Divide it out. Dividing $P(x)$ by $2x - 1$ gives the quotient $x^2 + x - 6$ , so $P(x) = (2x - 1)(x^2 + x - 6)$ .

Factor the quotient. $x^2 + x - 6 = (x + 3)(x - 2)$ , hence

2x^3 + x^2 - 13x + 6 = (2x - 1)(x + 3)(x - 2).

Check by expanding back. $(2x - 1)(x^2 + x - 6) = 2x^3 + 2x^2 - 12x - x^2 - x + 6 = 2x^3 + x^2 - 13x + 6 = P(x)$ . The factorisation is verified, and the zeroes are $\tfrac{1}{2}, -3, 2$ .

Common traps

Getting the sign wrong for $x + a$: A divisor $x + 3$ is $x - (-3)$ , so you evaluate $P(-3)$ , not $P(3)$ . This sign slip is the most frequent error with both theorems; read off $a$ as the value that makes the divisor zero.
Treating a non-monic factor like $x - a$: For $2x - 1$ you must substitute its zero $x = \tfrac{1}{2}$ , not $x = 1$ or $x = 2$ . Always solve "divisor $= 0$ " first and substitute that value.
Testing numbers that cannot be zeroes: Only divisors of the constant term can be integer zeroes (for integer coefficients). Trying $x = 5$ when the constant is $6$ wastes time, because $5$ does not divide $6$ .

Confusing "remainder is zero" with "remainder is the constant term". The remainder on division by $x - a$ is $P(a)$ , a single number you compute; it is not the constant term of $P(x)$ unless $a = 0$ .

Stopping before fully factoring. After dividing out one factor, the quotient usually factors further. "Factor completely" means continue until you have a product of linear factors (and any irreducible quadratics over the reals), then check by expanding.

Exam technique

A few habits turn these theorems into guaranteed marks:

Write the theorem, then the substitution, then the conclusion. "By the factor theorem, $P(-1) = -1 - 4 - 1 + 6 = 0$ , so $x + 1$ is a factor." Three lines, full marks.
Set up simultaneous equations cleanly for unknown coefficients. Each remainder or factor condition gives one equation in the unknowns; label them $(1)$ and $(2)$ and solve by elimination.
Show the divisor list before factoring. Listing $\pm 1, \pm 2, \pm 3, \pm 6$ proves you are using the integer-zero test and keeps the search short.
Reach for the remainder theorem, not long division, when only the remainder is asked. Substitution is faster and harder to get wrong.
Finish with the expand-back check. Multiplying the factors out and recovering $P(x)$ is the cheapest insurance against an arithmetic slip in the division.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksUse the remainder theorem to find the remainder when

P(x) = x^3 + 2x^2 - 5x + 7

is divided by

x - 2

Show worked solution →

State the theorem. The remainder on division by $x - a$ is $P(a)$ . Here the divisor is $x - 2$ , so $a = 2$ and the remainder is $P(2)$ .

Substitute $x = 2$ .

P(2) = 2^3 + 2(2^2) - 5(2) + 7 = 8 + 8 - 10 + 7 = 13.

State the remainder. The remainder is $13$ . No long division is needed; one substitution does it.

foundation4 marksThe polynomial is

P(x) = x^3 + 2x^2 - 9x - 18

. Show that

x + 3

is a factor, then factor

P(x)

completely.

Show worked solution →

Apply the factor theorem. The divisor $x + 3$ equals $x - (-3)$ , so test $x = -3$ . If $P(-3) = 0$ then $x + 3$ is a factor.

P(-3) = (-3)^3 + 2(-3)^2 - 9(-3) - 18 = -27 + 18 + 27 - 18 = 0,

so $x + 3$ is a factor.

Divide out the known factor. Dividing $P(x)$ by $x + 3$ gives the quotient $x^2 - x - 6$ , so $P(x) = (x + 3)(x^2 - x - 6)$ .

Factor the quadratic. $x^2 - x - 6 = (x - 3)(x + 2)$ , hence

x^3 + 2x^2 - 9x - 18 = (x + 3)(x - 3)(x + 2).

Check by expanding back. $(x + 3)(x^2 - x - 6) = x^3 - x^2 - 6x + 3x^2 - 3x - 18 = x^3 + 2x^2 - 9x - 18 = P(x)$ . The factorisation is correct.

core4 marksFind the value of

k

for which

x - 2

is a factor of

P(x) = x^3 - kx^2 + 2x + 8

. Using that value of

k

, factor

P(x)

completely.

Show worked solution →

Use the factor theorem as an equation. If $x - 2$ is a factor then $P(2) = 0$ . Substitute $x = 2$ and treat $k$ as the unknown:

P(2) = 8 - 4k + 4 + 8 = 20 - 4k = 0,

so $4k = 20$ and $k = 5$ .

Write the polynomial with $k = 5$: $P(x) = x^3 - 5x^2 + 2x + 8$ .
Divide out $x - 2$: Dividing gives the quotient $x^2 - 3x - 4$ , so $P(x) = (x - 2)(x^2 - 3x - 4)$ .
Factor the quadratic: $x^2 - 3x - 4 = (x - 4)(x + 1)$ , hence

x^3 - 5x^2 + 2x + 8 = (x - 2)(x - 4)(x + 1).

Check by expanding back. $(x - 2)(x^2 - 3x - 4) = x^3 - 3x^2 - 4x - 2x^2 + 6x + 8 = x^3 - 5x^2 + 2x + 8 = P(x)$ . Correct, with $k = 5$ .

core4 marksThe polynomial

P(x) = x^3 + ax^2 + bx + 6

leaves a remainder of

24

when divided by

x - 2

, and is divisible by

x + 1

. Find

a

and

b

Show worked solution →

Turn each condition into an equation. By the remainder theorem, dividing by $x - 2$ gives remainder $P(2)$ , so $P(2) = 24$ . Divisible by $x + 1$ means the remainder there is zero, so by the factor theorem $P(-1) = 0$ .

Form the first equation from $P(2) = 24$ .

P(2) = 8 + 4a + 2b + 6 = 4a + 2b + 14 = 24 \implies 2a + b = 5. \quad (1)

Form the second equation from $P(-1) = 0$ .

P(-1) = -1 + a - b + 6 = a - b + 5 = 0 \implies a - b = -5. \quad (2)

Solve the pair. Adding $(1)$ and $(2)$ gives $3a = 0$ , so $a = 0$ ; then $b = 5$ from $(2)$ .

Check both conditions. With $P(x) = x^3 + 5x + 6$ : $P(2) = 8 + 10 + 6 = 24$ and $P(-1) = -1 - 5 + 6 = 0$ . Both hold, so $a = 0$ and $b = 5$ .

exam5 marksFactor

P(x) = x^4 - 2x^3 - 7x^2 + 20x - 12

completely over the real numbers.

Show worked solution →

List the candidate zeroes: Every coefficient is an integer, so any integer zero must divide the constant term $-12$ . The candidates are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$ .
Test small candidates with the remainder theorem: $P(1) = 1 - 2 - 7 + 20 - 12 = 0$ , so $x - 1$ is a factor. $P(2) = 16 - 16 - 28 + 40 - 12 = 0$ , so $x - 2$ is a factor. $P(-3) = 81 + 54 - 63 - 60 - 12 = 0$ , so $x + 3$ is a factor.
Account for the degree: Three distinct factors $x - 1$ , $x - 2$ , $x + 3$ multiply to a cubic, but $P$ has degree $4$ , so one zero is repeated. Their product $(x - 1)(x - 2)(x + 3)$ has constant term $6$ , while $P$ has constant term $-12$ ; the missing factor must have constant term $-2$ , identifying the repeat as a second $x - 2$ .
Write the full factorisation

x^4 - 2x^3 - 7x^2 + 20x - 12 = (x - 1)(x - 2)^2(x + 3).

Check by expanding back. $(x - 1)(x + 3) = x^2 + 2x - 3$ and $(x - 2)^2 = x^2 - 4x + 4$ ; multiplying, $(x^2 + 2x - 3)(x^2 - 4x + 4) = x^4 - 2x^3 - 7x^2 + 20x - 12 = P(x)$ . The factorisation is verified.

exam5 marksThe polynomial

P(x) = 2x^3 + ax^2 + bx + 6

has

2x - 1

as a factor, and leaves a remainder of

-6

when divided by

x - 1

. Find

a

and

b

, then factor

P(x)

completely.

Show worked solution →

Convert the non-monic factor into a substitution. $2x - 1 = 0$ at $x = \tfrac{1}{2}$ , so $2x - 1$ being a factor means $P\!\left(\tfrac{1}{2}\right) = 0$ .

P\!\left(\tfrac{1}{2}\right) = 2 \cdot \tfrac{1}{8} + a \cdot \tfrac{1}{4} + b \cdot \tfrac{1}{2} + 6 = \tfrac{1}{4} + \tfrac{a}{4} + \tfrac{b}{2} + 6 = 0.

Multiplying through by $4$ gives $1 + a + 2b + 24 = 0$ , so $a + 2b = -25. \quad (1)$

Use the remainder condition. The remainder on division by $x - 1$ is $P(1) = -6$ :

P(1) = 2 + a + b + 6 = a + b + 8 = -6 \implies a + b = -14. \quad (2)

Solve the pair. Subtracting $(2)$ from $(1)$ gives $b = -11$ ; then $a = -3$ from $(2)$ .

Factor with $a = -3$ , $b = -11$ . So $P(x) = 2x^3 - 3x^2 - 11x + 6$ . Dividing out $2x - 1$ gives the quotient $x^2 - x - 6 = (x - 3)(x + 2)$ , hence

2x^3 - 3x^2 - 11x + 6 = (2x - 1)(x - 3)(x + 2).

Check by expanding back. $(x - 3)(x + 2) = x^2 - x - 6$ , and $(2x - 1)(x^2 - x - 6) = 2x^3 - 2x^2 - 12x - x^2 + x + 6 = 2x^3 - 3x^2 - 11x + 6 = P(x)$ . The factorisation is verified, with $a = -3$ , $b = -11$ and zeroes $\tfrac{1}{2}, 3, -2$ .