Polynomial division | Year 12

01 / LEARN

Understand the idea.

The division identity

Polynomial division produces $P = D Q + R$ , where $D$ is nonzero and $R$ is zero or has smaller degree than $D$ . Stop only when the remaining polynomial has that required smaller degree.

Arrange powers in descending order, keeping places for zero coefficients. Divide the leading terms to choose the next quotient term, multiply the whole divisor by it, then subtract that whole product. Repeat with the new leading term.

Choosing a method and checking it

For a divisor $x - a$ , synthetic division records the same process using coefficients. Its input is the zero $a$ : division by $x + 2$ uses $- 2$ . Ordinary long division also works for non-monic or higher-degree divisors.

Check the completed division by reconstructing $D Q + R$ . If converting to $P / D = Q + R / D$ , retain the original restriction $D (x) \neq 0$ . A quadratic divisor can leave a linear remainder; it need not leave a constant.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A linear divisor with a remainder

Divide $2 x^{3} + 3 x^{2} - 5 x + 6$ by $x + 2$ .

Worked solution

1
Choose and subtract the first quotient term.
$\begin{aligned} \frac{2 x^{3}}{x} & = 2 x^{2}, \\ (2 x^{3} + 3 x^{2} - 5 x + 6) & - (x + 2) (2 x^{2}) \\ = - x^{2} - 5 x + 6. \end{aligned}$
The choice $2 x^{2}$ cancels the current leading term.
2
Cancel the next leading term.
$\begin{aligned} \frac{- x^{2}}{x} & = - x, \\ (- x^{2} - 5 x + 6) & - (x + 2) (- x) = - 3 x + 6. \end{aligned}$
Subtracting the full product also changes its linear-term sign.
3
Finish the division.
$(- 3 x + 6) - (x + 2) (- 3) = 12.$
The constant remainder has degree less than the linear divisor.
4
State and verify the quotient and remainder.
$2 x^{3} + 3 x^{2} - 5 x + 6 = (x + 2) (2 x^{2} - x - 3) + 12.$
Reconstruction confirms quotient $2 x^{2} - x - 3$ and remainder $12$ .

Answer

Quotient $2 x^{2} - x - 3$ ; remainder $12$ .

EXAMPLE 02

A quadratic divisor

Divide $x^{4} + 2 x^{3} + 3 x^{2} + 4 x + 5$ by $x^{2} + 1$ .

Worked solution

1
Cancel the fourth-degree term.
$\begin{aligned} \frac{x^{4}}{x^{2}} & = x^{2}, \\ (x^{4} + 2 x^{3} + 3 x^{2} + 4 x + 5) & - (x^{2} + 1) x^{2} \\ = 2 x^{3} + 2 x^{2} + 4 x + 5. \end{aligned}$
The first quotient term cancels the dividend leading term.
2
Cancel the cubic term.
$\begin{aligned} \frac{2 x^{3}}{x^{2}} & = 2 x, \\ (2 x^{3} + 2 x^{2} + 4 x + 5) & - (x^{2} + 1) (2 x) \\ = 2 x^{2} + 2 x + 5. \end{aligned}$
Each new quotient term is found from the current leading terms.
3
Cancel the remaining quadratic term.
$(2 x^{2} + 2 x + 5) - 2 (x^{2} + 1) = 2 x + 3.$
The remaining degree one is less than the divisor degree two, so division stops.
4
Record the division identity.
$\begin{aligned} x^{4} + 2 x^{3} + 3 x^{2} + 4 x + 5 & = (x^{2} + 1) (x^{2} + 2 x + 2) \\ + (2 x + 3) . \end{aligned}$
The quotient is $x^{2} + 2 x + 2$ and the remainder is $2 x + 3$ .

Answer

Quotient $x^{2} + 2 x + 2$ ; remainder $2 x + 3$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

A degree- $5$ polynomial is divided by a degree- $2$ polynomial. What are the quotient degree and largest possible remainder degree?

Show answer & reasoning

Quotient degree $3$ ; remainder degree at most $1$ .

The leading quotient term has degree $5 - 2 = 3$ . The remainder degree must be strictly less than two, so its largest possible degree is one; a zero remainder is also possible.

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

Divide $x^{3} - 6 x^{2} + 11 x - 6$ by $x - 1$ .
2 marks
Divide $2 x^{3} - 5 x^{2} + 4 x - 3$ by $x - 2$ .
2 marks
Divide $x^{4} - 16$ by $x^{2} + 4$ .
2 marks
Divide $x^{3} - 2 x + 1$ by $x + 1$ .
2 marks
Divide $3 x^{2} + x - 2$ by $3 x - 2$ .
2 marks
A student writes $x^{3} + 1 = (x + 1) (x^{2} - x) + 1$ . Check the claim and correct the division.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Divide $x^{3} - 6 x^{2} + 11 x - 6$ by $x - 1$ .
2 marks
Synthetic coefficients at $1$ are $1, - 5, 6, 0$ .
$Q = x^{2} - 5 x + 6$ , $R = 0$ .
Divide $2 x^{3} - 5 x^{2} + 4 x - 3$ by $x - 2$ .
2 marks
Synthetic coefficients at $2$ are $2, - 1, 2, 1$ .
$Q = 2 x^{2} - x + 2$ , $R = 1$ .
Divide $x^{4} - 16$ by $x^{2} + 4$ .
2 marks
Use the difference of squares: $(x^{2} - 4) (x^{2} + 4) = x^{4} - 16$ .
$Q = x^{2} - 4$ , $R = 0$ .
Divide $x^{3} - 2 x + 1$ by $x + 1$ .
2 marks
Include the missing $x^{2}$ coefficient: divide coefficients $1, 0, - 2, 1$ using $- 1$ .
$Q = x^{2} - x - 1$ , $R = 2$ .
Divide $3 x^{2} + x - 2$ by $3 x - 2$ .
2 marks
The leading term is $x$ ; the remaining term after subtraction is $3 x - 2$ .
$Q = x + 1$ , $R = 0$ .
A student writes $x^{3} + 1 = (x + 1) (x^{2} - x) + 1$ . Check the claim and correct the division.
3 marks
The claimed product plus remainder is $x^{3} - x + 1$ , not $x^{3} + 1$ . Adding $1$ to the quotient repairs the identity.
The correct quotient is $x^{2} - x + 1$ , remainder $0$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

A recognisable real-polynomial chapter supporting Stage 2 Specialist polynomial factorisation and graph analysis; original lessons use current SACE scope and the school chapter structure requested by the user.

Official curriculum source ↗

In this lesson

Understand the idea.

The division identity

Choosing a method and checking it

See the method in action.

A linear divisor with a remainder

Choose and subtract the first quotient term.

Cancel the next leading term.

Finish the division.

State and verify the quotient and remainder.

A quadratic divisor

Cancel the fourth-degree term.

Cancel the cubic term.

Cancel the remaining quadratic term.

Record the division identity.

Try it yourself.

Check your understanding

Keep practising