Polynomial Division: Remainders, Factors and Zeros

From division to remainders, factors and zeros

When a cubic expression does not factor at first sight, a known zero can give you a starting factor. Polynomial division then finds what is left. If you only need to test a proposed factor or find a remainder, substitution can save you the full division.

Start here once you can expand brackets, factor a quadratic and substitute a number into an expression. Work over the real numbers; the variables in these algebra examples have no physical units. This lesson supports selected polynomial topics in Ontario MHF4U. For another high-school, CEGEP or university course, compare the topics with your current course outline. In Québec secondary school, also check your year and mathematics option.

1. Know what the division is asking for

Here, $x$ is the real input that can vary. A coefficient is the number multiplying a power of $x$. Symbols such as $c$ stand for fixed real numbers when they are introduced.

A polynomial in $x$ is a finite sum of terms of the form $ax^n$, where $a$ is a real coefficient and $n$ is a nonnegative integer. For example, $2x^3-3x+5$ is a polynomial; $1/x$ and $\sqrt{x}$ are not polynomials in $x$. The degree of a nonzero polynomial is the largest exponent with a nonzero coefficient. The example has degree 3; a nonzero constant has degree 0. The leading term is the nonzero term with the highest exponent, here $2x^3$.

In polynomial division, $P(x)$ is the dividend, the polynomial being divided. $D(x)$ is the divisor, which cannot be the zero polynomial. The quotient $Q(x)$ and remainder $R(x)$ satisfy

$$P(x)=D(x)Q(x)+R(x).$$

Stop when $R(x)=0$, or when its degree is lower than the divisor's degree. Dividing by a linear polynomial, which has degree 1, therefore leaves a constant remainder or zero. A polynomial remainder is limited by its degree, not by how large its numerical value is.

This is a polynomial identity: it holds for every real $x$. If you write the result as a fraction instead, preserve the restriction:

$$\frac{P(x)}{D(x)}=Q(x)+\frac{R(x)}{D(x)},\qquad D(x)\ne0.$$

Dividing polynomials does not make the fraction defined where its denominator is zero.

2. Long division: remove the highest power first

Divide $2x^3-3x+5$ by $x-2$. We need both the quotient and the remainder, so use long division. Write the missing term explicitly:

$$P(x)=2x^3+0x^2-3x+5.$$

The zero keeps the powers aligned. At each step, divide the leading term of what remains by the leading term of the divisor, multiply the whole divisor by that result, then subtract the whole product.

1. First quotient term: $2x^3\div x=2x^2$. Multiply: $2x^2(x-2)=2x^3-4x^2$. Subtract:

   $$\begin{aligned}(2x^3+0x^2-3x+5)&-(2x^3-4x^2)\\&=4x^2-3x+5.\end{aligned}$$

   The coefficient of $x^2$ is $0-(-4)=4$.

2. Next quotient term: $4x^2\div x=4x$. Multiply: $4x(x-2)=4x^2-8x$. Subtract:

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

3. Last quotient term: $5x\div x=5$. Multiply: $5(x-2)=5x-10$. Subtract:

   $$(5x+5)-(5x-10)=15.$$

   The remainder is constant, so stop. You cannot remove another term using a polynomial quotient.

Collect the quotient terms: $Q(x)=2x^2+4x+5$, with remainder $R=15$. Write the answer in a form you can check:

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

Equivalently,

$$\frac{2x^3-3x+5}{x-2}=2x^2+4x+5+\frac{15}{x-2},\qquad x\ne2.$$

Check by expanding: the product is $2x^3-3x-10$; adding 15 recovers $2x^3-3x+5$. This verifies the identity for all $x$.

Independent number check: at $x=-1$, the original polynomial is $-2+3+5=6$. The divided form gives $(-3)(2-4+5)+15=-9+15=6$. At $x=0$, it gives $(-2)(5)+15=5$, matching the constant term. A few spot checks are useful for catching arithmetic errors; use expansion for the complete identity check.

3. Why substitution finds the remainder

For division by $x-c$, the remainder is a constant $r$:

$$P(x)=(x-c)Q(x)+r.$$

Substitute $x=c$ into this identity. The product disappears because $c-c=0$:

$$P(c)=0\cdot Q(c)+r=r.$$

This is the remainder theorem: the remainder when $P(x)$ is divided by $x-c$ is $P(c)$. In the worked division, $P(2)=2(2)^3-3(2)+5=15$, agreeing with the long division. We evaluated the polynomial identity at 2; we did not substitute 2 into the undefined fraction $P(x)/(x-2)$.

Read the divisor carefully. For $x+2=x-(-2)$, use $P(-2)$. Substitution gives the remainder, but does not by itself give the quotient.

4. A zero remainder gives a factor

The factor theorem says that $x-c$ is a factor of $P(x)$ exactly when $P(c)=0$. A factor divides the polynomial with no remainder. A nonzero remainder still answers the division question, but the tested divisor is not a factor.

A zero of $P$, or a root of $P(x)=0$, is an input $c$ for which $P(c)=0$. On the graph $y=P(x)$, it gives the point $(c,0)$ on the horizontal axis. The factor is the expression $x-c$; the zero is the number $c$. A zero does not always mean the graph crosses the axis: for example, $y=(x-1)^2$ touches it at $(1,0)$.

Worked example: factor $F(x)=x^3-2x^2-5x+6$, then solve $F(x)=0$. Testing $x=1$ gives $1-2-5+6=0$, so $x-1$ is a factor. Divide to find the other factor:

• The first quotient term is $x^2$. Subtract $x^2(x-1)$ to leave $-x^2-5x+6$.
• The next term is $-x$. Subtract $-x(x-1)$ to leave $-6x+6$.
• The last term is $-6$. Subtract $-6(x-1)$ to leave 0.

Thus

$$\begin{aligned}F(x)&=(x-1)(x^2-x-6)\\&=(x-1)(x-3)(x+2).\end{aligned}$$

The two numbers $-3$ and 2 multiply to $-6$ and add to $-1$, which factors the quadratic. The zero-product rule says that a product is zero exactly when at least one factor is zero. It now gives $x=1$, $x=3$ or $x=-2$. These are all the solutions because the full polynomial has been written as a product of these three linear factors.

Checks in the original polynomial: $F(1)=1-2-5+6=0$, $F(3)=27-18-15+6=0$, and $F(-2)=-8-8+10+6=0$. The corresponding intercepts are $(1,0)$, $(3,0)$ and $(-2,0)$. Finding the first zero alone would not have found all three.

When a question does not supply a zero, first check for a common factor, grouping or a familiar quadratic pattern. Trying a few simple inputs can find a useful zero, but unsuccessful guesses do not prove that no zero exists.

5. When the divisor starts with a coefficient other than 1

For the linear divisor $ax-b$, where $a$ and $b$ are fixed real numbers and $a\ne0$, solve $ax-b=0$ first. Its zero is $b/a$, so the remainder is $P(b/a)$. For $2x-1$, use $x=1/2$, not $x=1$.

Take $G(x)=2x^2+3x-1$. Division by $2x-1$ starts with $2x^2\div2x=x$. Subtract $x(2x-1)$ to leave $4x-1$. The next quotient term is 2; subtract $2(2x-1)$ to leave 1. Therefore

$$G(x)=(2x-1)(x+2)+1.$$

Check the remainder: $G(1/2)=1/2+3/2-1=1$. Thus $2x-1$ is not a factor. Expanding the right side also recovers $2x^2+3x-1$.

Keep the divisor's coefficient in the quotient: dividing the same polynomial by $x-1/2$ instead gives quotient $2x+4$, with the same remainder 1:

$$G(x)=(x-1/2)(2x+4)+1.$$

Since $2x-1=2(x-1/2)$, divide the quotient for $x-1/2$ by 2 to obtain the quotient for $2x-1$. If you use the usual synthetic-division method for $x-c$, it returns the quotient for $x-c$. To obtain the quotient for $a(x-c)$, divide that quotient by $a$; the remainder stays the same. Long division using the original divisor keeps this coefficient visible.

Before practising, choose the shortest complete method

• Only a remainder? For a linear divisor, substitute its zero into the polynomial.
• Is it a factor? Find that remainder and check whether it is exactly zero.
• Need the quotient? Use division. To factor from a known zero, divide by its corresponding factor; when the remainder is zero, factor the quotient further if possible.
• All solutions of an equation? Set the complete factorization equal to zero and check every candidate in the original polynomial.

Four steps of polynomial practice

Try each question before opening its solution. Write why you chose substitution or long division, and include a check. These are original practice questions; use your teacher's topic list to decide what belongs in your revision.

1. Find the remainder without dividing

Let $H(x)=x^3-4x+1$. Find the remainder on division by (a) $x+2$ and (b) $x-1$. Is either divisor a factor?

2. Divide when one power is missing

Divide $x^3+2x^2-5$ by $x+1$. Give the quotient, remainder and an identity that checks your result.

3. Turn one zero into a full solution

Let $J(x)=x^3-4x^2-x+4$. Check that 1 is a zero, use division to factor $J$ completely, and solve $J(x)=0$. Give the corresponding horizontal-axis intercepts.

4. Find a coefficient from a stated remainder

For $P_k(x)=2x^3+kx^2-3x+2$, where $k$ is an unknown real constant, division by $2x-1$ leaves remainder 4. Find $k$, then divide the resulting polynomial by $2x-1$ to verify the quotient and remainder.

Catch the mistakes that change the answer

• Using the wrong sign: $x+2$ calls for $P(-2)$. Solve “divisor = 0” before substituting.
• Dropping a missing power: keep $0x^2$ or $0x$ in the division layout so columns still represent the same power.
• Subtracting only one term: write parentheses around each product you subtract. For example, $-3x-(-8x)=5x$.
• Stopping too soon: a nonzero remainder must have degree lower than the divisor. With a linear divisor, a leftover $5x+5$ still needs another step.
• Losing a denominator restriction: simplifying $P(x)/D(x)$ does not restore a zero of $D(x)$. State excluded inputs before cancelling factors.
• Confusing a remainder with a factor: $P(c)=4$ means remainder 4; only $P(c)=0$ gives factor $x-c$.
• Changing the divisor but keeping the quotient: $2x-1$ and $x-1/2$ have the same zero, but their quotients differ by a factor of 2.

Before moving on: explain why $P(c)$ gives the remainder, complete question 2 without looking back, and distinguish the factor $x-c$ from the zero $c$. If signs caused difficulty, return to expanding brackets. If the quadratic quotient caused difficulty, return to factorization. For fractions built from these factors, continue with rational equations or rational inequalities and excluded inputs.

If you would like individual help, bring a topic or a question, even if you are starting from scratch. A tutor can explain the first step, work through division with you and help you check a factor. Include your course code when you have one so the practice can match your class. The algebra and precalculus tutoring page helps you choose the relevant starting point.

Original practice and course scope

The examples, four main questions and four changed-data retries are original teaching practice. This resource supports selected polynomial-division, remainder/factor-theorem and polynomial-equation topics in Ontario MHF4U. It does not provide an official exam or a complete course review; technology investigations and wider applications require further work. For another course, compare these topics with your current course outline. In Québec secondary school, also check your year and mathematics option.

Topic reference: Ontario Ministry of Education, Grades 11 and 12 Mathematics (2007), MHF4U, printed page 93, expectations C3.1–C3.4 and C3.7; checked 19 October 2026.