Sec 3 · Additional Mathematics

Polynomials, cubic equations and partial fractions

Practise polynomials, cubic equations and partial fractions with step-by-step help. Free account.

1. Dividing polynomials

Name	Formula	Example
division	$f(x) = (\text{divisor})(\text{quotient}) + \text{remainder}$	$x^{3} - x^{2} - 9x + 9 = (x - 3)(x^{2} + 2x - 3)$
degree of the remainder	less than the degree of the divisor	divide by $x - 3$ : the remainder is a number

📖 Worked example

Find the quotient when $x^3 - x^2 - 9x + 9$ is divided by $x - 3$ .

Step 1 of 2

Long division: divide the leading terms, multiply the divisor by that term, subtract, bring down the next term; repeat

\def\arraystretch{1.6}\begin{array}{rrrrr} & & \;x^2\; & \;+\,2x\; & \;-\,3\; \\ x - 3\,\Big) & \overline{\;x^3\;} & \overline{\;-\,x^2\;} & \overline{\;-\,9x\;} & \overline{\;+\,9\;} \\ & \underline{\;x^3\;} & \underline{\;-\,3x^2\;} & & \\ & & \;2x^2\; & \;-\,9x\; & \\ & & \underline{\;2x^2\;} & \underline{\;-\,6x\;} & \\ & & & \;-3x\; & \;+\,9\; \\ & & & \underline{\;-3x\;} & \underline{\;+\,9\;} \\ & & & & \;0\; \end{array}

Identities: compare coefficients, or put in a value of x

\displaystyle x^{2} - 7x + 2 \equiv A(x - 1)^{2} + B(x - 1) + C

$x^{2}$ terms: $A = 1$
Put $x = 1$ (the brackets are $0$ ): $C = 1 - 7 + 2$
so $C = -4$
$x$ terms: $-7 = -2A + B$ , so $B = -5$

📖 Worked example

Given that $x^{2} - 7x + 2 \equiv A(x - 1)^2 + B(x - 1) + C$ for all values of $x$ , find the values of $A$ , $B$ and $C$ .

Step 1 of 3

Compare the coefficients of x² (only the squared bracket has an x² term)

A = 1

Practise this →

2. Remainder and factor theorems

Theorem	Formula
remainder theorem	$f(x) \div (x - a)$ leaves remainder $f(a)$
	$f(x) \div (bx - a)$ leaves remainder $f\left(\dfrac{a}{b}\right)$
factor theorem	$(x - a)$ is a factor when $f(a) = 0$

Remainder: x³ − 6x + 4 divided by x − 1

\displaystyle f(1)

e.g.

\displaystyle 1^{3} - 6(1) + 4

\displaystyle = -1

Remainder: 4x³ − 2x + 3 divided by 2x − 1

\displaystyle f\left(\dfrac{1}{2}\right)

e.g.

\displaystyle 4\left(\dfrac{1}{2}\right)^{3} - 2\left(\dfrac{1}{2}\right) + 3

\displaystyle = \dfrac{5}{2}

Factor: x − 2 is a factor of x³ − 6x + k

\displaystyle f(2) = 0

e.g.

\displaystyle 2^{3} - 6(2) + k = 0

\displaystyle k = 4

📖 Worked example

Find the remainder when $x^3 - 6x + 4$ is divided by $2x - 1$ .

Step 1 of 2

Remainder theorem: the remainder is f(the value of x that makes the divisor 0)

2x - 1 = 0

when

x = \dfrac{1}{2}

Dividing by a quadratic: the remainder is ax + b

\displaystyle f(x) = (x - 1)(x + 1)Q(x) + ax + b

$f(1) = -4$ : $a + b = -4$
$f(-1) = -6$ : $-a + b = -6$
$a = 1$ , $b = -5$ : the remainder is $x - 5$

📖 Worked example

When the polynomial $f(x)$ is divided by $x - 1$ the remainder is $-4$ , and when it is divided by $x + 1$ the remainder is $-6$ . Find the remainder when $f(x)$ is divided by $x^{2} - 1$ .

Step 1 of 4

Dividing by a quadratic leaves a remainder of degree at most 1: call it ax + b

f(x) = (x - 1)(x + 1)Q(x) + ax + b

Practise this →

3. Solving cubic equations

Find a root by trying factors of the constant: f(1) = 0

\displaystyle x^{3} - 2x^{2} - 5x + 6 = 0

e.g.

\displaystyle x^{3} - 2x^{2} - 5x + 6 = 0

\displaystyle (x - 1)(x^{2} - x - 6) = 0

\displaystyle (x - 1)(x - 3)(x + 2) = 0

\displaystyle x = 1 \text{ or } x = 3 \text{ or } x = -2

$f(1) = 1 - 2 - 5 + 6$ , which is $0$

Find the quadratic factor by comparing coefficients

\displaystyle x^{3} - 2x^{2} - 5x + 6 = (x - 1)(x^{2} + px + q)

$x^{2}$ terms, $p - 1$ and $-2$ : $p = -1$
numbers, $-q$ and $6$ : $q = -6$

Name	Formula	Example
sum of cubes	$a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})$	$x^{3} + 8 = (x + 2)(x^{2} - 2x + 4)$
difference of cubes	$a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})$	$x^{3} - 27 = (x - 3)(x^{2} + 3x + 9)$

📖 Worked example

Solve $x^3 + 2x^2 - 13x + 10 = 0$ .

Step 1 of 4

Try factors of the constant term until f(a) = 0

\text{A root: } x = 1

f(1) = (1)^3 + 2(1)^2 - 13(1) + 10

= 1 + 2 - 13 + 10

= 0

(x - 1)

is a factor.

Sum or difference of two cubes	Example
$a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})$	$x^{3} + 27 = (x + 3)(x^{2} - 3x + 9)$
$a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})$	$8x^{3} - 1 = (2x - 1)(4x^{2} + 2x + 1)$

📖 Worked example

Factorise $8x^{3} - 1$ completely.

Step 1 of 2

Write each term as a cube: a³ and b³

(2x)^3 - 1^3

Practise this →

4. Partial fractions

Denominator	Partial fractions
$(ax + b)(cx + d)$	$\dfrac{A}{ax + b} + \dfrac{B}{cx + d}$
$(ax + b)(cx + d)^{2}$	$\dfrac{A}{ax + b} + \dfrac{B}{cx + d} + \dfrac{C}{(cx + d)^{2}}$
$(ax + b)(x^{2} + c^{2})$	$\dfrac{A}{ax + b} + \dfrac{Bx + C}{x^{2} + c^{2}}$ ( $x^{2} + c^{2}$ does not factorise)

Two linear factors

\displaystyle \dfrac{5x - 11}{(x - 3)(x - 1)} = \dfrac{A}{x - 3} + \dfrac{B}{x - 1}

$5x - 11 = A(x - 1) + B(x - 3)$
put $x = 3$ : $A = 2$
put $x = 1$ : $B = 3$

📖 Worked example

Express $\dfrac{5x - 11}{(x - 3)(x - 1)}$ in partial fractions.

Step 1 of 4

One fraction for each linear factor

\dfrac{5x - 11}{(x - 3)(x - 1)} = \dfrac{A}{x - 3} + \dfrac{B}{x - 1}

📖 Worked example

Express $\dfrac{2x^2 + x + 5}{(x + 1)(x^2 + 1)}$ in partial fractions.

Step 1 of 5

Write it out with the factors of the denominator

\dfrac{2x^2 + x + 5}{(x + 1)(x^2 + 1)} = \dfrac{A}{x + 1} + \dfrac{Bx + C}{x^2 + 1}

Improper (top's degree ≥ bottom's): divide first

\displaystyle \dfrac{x^{2} + 4x + 1}{(x - 1)(x + 2)}

e.g.

\displaystyle \dfrac{x^{2} + 4x + 1}{(x - 1)(x + 2)}

\displaystyle = 1 + \dfrac{3x + 3}{(x - 1)(x + 2)}

\displaystyle = 1 + \dfrac{2}{x - 1} + \dfrac{1}{x + 2}

📖 Worked example

Express $\dfrac{x^{2} + 4x + 1}{(x - 1)(x + 2)}$ in partial fractions.

Step 1 of 4

The top's degree is not less than the bottom's: expand the bottom and divide first

(x - 1)(x + 2) = x^{2} + x - 2

Practise this →

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Sec 3 · Additional Mathematics

Polynomials, cubic equations and partial fractions

Practise polynomials, cubic equations and partial fractions with step-by-step help. Free account.

Start free Worksheet (PDF)

1. Dividing polynomials

Name	Formula	Example
division	$f(x) = (\text{divisor})(\text{quotient}) + \text{remainder}$	$x^{3} - x^{2} - 9x + 9 = (x - 3)(x^{2} + 2x - 3)$
degree of the remainder	less than the degree of the divisor	divide by $x - 3$ : the remainder is a number

📖 Worked example

Find the quotient when $x^3 - x^2 - 9x + 9$ is divided by $x - 3$ .

Step 1 of 2

Long division: divide the leading terms, multiply the divisor by that term, subtract, bring down the next term; repeat

\def\arraystretch{1.6}\begin{array}{rrrrr} & & \;x^2\; & \;+\,2x\; & \;-\,3\; \\ x - 3\,\Big) & \overline{\;x^3\;} & \overline{\;-\,x^2\;} & \overline{\;-\,9x\;} & \overline{\;+\,9\;} \\ & \underline{\;x^3\;} & \underline{\;-\,3x^2\;} & & \\ & & \;2x^2\; & \;-\,9x\; & \\ & & \underline{\;2x^2\;} & \underline{\;-\,6x\;} & \\ & & & \;-3x\; & \;+\,9\; \\ & & & \underline{\;-3x\;} & \underline{\;+\,9\;} \\ & & & & \;0\; \end{array}

Identities: compare coefficients, or put in a value of x

\displaystyle x^{2} - 7x + 2 \equiv A(x - 1)^{2} + B(x - 1) + C

$x^{2}$ terms: $A = 1$
Put $x = 1$ (the brackets are $0$ ): $C = 1 - 7 + 2$
so $C = -4$
$x$ terms: $-7 = -2A + B$ , so $B = -5$

📖 Worked example

Given that $x^{2} - 7x + 2 \equiv A(x - 1)^2 + B(x - 1) + C$ for all values of $x$ , find the values of $A$ , $B$ and $C$ .

Step 1 of 3

Compare the coefficients of x² (only the squared bracket has an x² term)

A = 1

Practise this →

2. Remainder and factor theorems

Theorem	Formula
remainder theorem	$f(x) \div (x - a)$ leaves remainder $f(a)$
	$f(x) \div (bx - a)$ leaves remainder $f\left(\dfrac{a}{b}\right)$
factor theorem	$(x - a)$ is a factor when $f(a) = 0$

Remainder: x³ − 6x + 4 divided by x − 1

\displaystyle f(1)

e.g.

\displaystyle 1^{3} - 6(1) + 4

\displaystyle = -1

Remainder: 4x³ − 2x + 3 divided by 2x − 1

\displaystyle f\left(\dfrac{1}{2}\right)

e.g.

\displaystyle 4\left(\dfrac{1}{2}\right)^{3} - 2\left(\dfrac{1}{2}\right) + 3

\displaystyle = \dfrac{5}{2}

Factor: x − 2 is a factor of x³ − 6x + k

\displaystyle f(2) = 0

e.g.

\displaystyle 2^{3} - 6(2) + k = 0

\displaystyle k = 4

📖 Worked example

Find the remainder when $x^3 - 6x + 4$ is divided by $2x - 1$ .

Step 1 of 2

Remainder theorem: the remainder is f(the value of x that makes the divisor 0)

2x - 1 = 0

when

x = \dfrac{1}{2}

Dividing by a quadratic: the remainder is ax + b

\displaystyle f(x) = (x - 1)(x + 1)Q(x) + ax + b

$f(1) = -4$ : $a + b = -4$
$f(-1) = -6$ : $-a + b = -6$
$a = 1$ , $b = -5$ : the remainder is $x - 5$

📖 Worked example

When the polynomial $f(x)$ is divided by $x - 1$ the remainder is $-4$ , and when it is divided by $x + 1$ the remainder is $-6$ . Find the remainder when $f(x)$ is divided by $x^{2} - 1$ .

Step 1 of 4

Dividing by a quadratic leaves a remainder of degree at most 1: call it ax + b

f(x) = (x - 1)(x + 1)Q(x) + ax + b

Practise this →

3. Solving cubic equations

Find a root by trying factors of the constant: f(1) = 0

\displaystyle x^{3} - 2x^{2} - 5x + 6 = 0

e.g.

\displaystyle x^{3} - 2x^{2} - 5x + 6 = 0

\displaystyle (x - 1)(x^{2} - x - 6) = 0

\displaystyle (x - 1)(x - 3)(x + 2) = 0

\displaystyle x = 1 \text{ or } x = 3 \text{ or } x = -2

$f(1) = 1 - 2 - 5 + 6$ , which is $0$

Find the quadratic factor by comparing coefficients

\displaystyle x^{3} - 2x^{2} - 5x + 6 = (x - 1)(x^{2} + px + q)

$x^{2}$ terms, $p - 1$ and $-2$ : $p = -1$
numbers, $-q$ and $6$ : $q = -6$

Name	Formula	Example
sum of cubes	$a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})$	$x^{3} + 8 = (x + 2)(x^{2} - 2x + 4)$
difference of cubes	$a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})$	$x^{3} - 27 = (x - 3)(x^{2} + 3x + 9)$

📖 Worked example

Solve $x^3 + 2x^2 - 13x + 10 = 0$ .

Step 1 of 4

Try factors of the constant term until f(a) = 0

\text{A root: } x = 1

f(1) = (1)^3 + 2(1)^2 - 13(1) + 10

= 1 + 2 - 13 + 10

= 0

(x - 1)

is a factor.

Sum or difference of two cubes	Example
$a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})$	$x^{3} + 27 = (x + 3)(x^{2} - 3x + 9)$
$a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})$	$8x^{3} - 1 = (2x - 1)(4x^{2} + 2x + 1)$

📖 Worked example

Factorise $8x^{3} - 1$ completely.

Step 1 of 2

Write each term as a cube: a³ and b³

(2x)^3 - 1^3

Practise this →

4. Partial fractions

Denominator	Partial fractions
$(ax + b)(cx + d)$	$\dfrac{A}{ax + b} + \dfrac{B}{cx + d}$
$(ax + b)(cx + d)^{2}$	$\dfrac{A}{ax + b} + \dfrac{B}{cx + d} + \dfrac{C}{(cx + d)^{2}}$
$(ax + b)(x^{2} + c^{2})$	$\dfrac{A}{ax + b} + \dfrac{Bx + C}{x^{2} + c^{2}}$ ( $x^{2} + c^{2}$ does not factorise)

Two linear factors

\displaystyle \dfrac{5x - 11}{(x - 3)(x - 1)} = \dfrac{A}{x - 3} + \dfrac{B}{x - 1}

$5x - 11 = A(x - 1) + B(x - 3)$
put $x = 3$ : $A = 2$
put $x = 1$ : $B = 3$

📖 Worked example

Express $\dfrac{5x - 11}{(x - 3)(x - 1)}$ in partial fractions.

Step 1 of 4

One fraction for each linear factor

\dfrac{5x - 11}{(x - 3)(x - 1)} = \dfrac{A}{x - 3} + \dfrac{B}{x - 1}

📖 Worked example

Express $\dfrac{2x^2 + x + 5}{(x + 1)(x^2 + 1)}$ in partial fractions.

Step 1 of 5

Write it out with the factors of the denominator

\dfrac{2x^2 + x + 5}{(x + 1)(x^2 + 1)} = \dfrac{A}{x + 1} + \dfrac{Bx + C}{x^2 + 1}

Improper (top's degree ≥ bottom's): divide first

\displaystyle \dfrac{x^{2} + 4x + 1}{(x - 1)(x + 2)}

e.g.

\displaystyle \dfrac{x^{2} + 4x + 1}{(x - 1)(x + 2)}

\displaystyle = 1 + \dfrac{3x + 3}{(x - 1)(x + 2)}

\displaystyle = 1 + \dfrac{2}{x - 1} + \dfrac{1}{x + 2}

📖 Worked example

Express $\dfrac{x^{2} + 4x + 1}{(x - 1)(x + 2)}$ in partial fractions.

Step 1 of 4

The top's degree is not less than the bottom's: expand the bottom and divide first

(x - 1)(x + 2) = x^{2} + x - 2

Practise this →

Try a question: Polynomials, cubic equations and partial fractions

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