Algebraic Methods

What you'll learn

How to split algebraic fractions into partial fractions.
How to handle repeated factors and “top-heavy” algebraic fractions.
How to use substitution and equating coefficients cleanly.
How to write proof by contradiction arguments for integers, irrational numbers and primes.

The algebra you need first

A polynomial is an expression made from powers of $x$ , such as $3x^2-5x+1$ . The degree of a polynomial is the highest power of $x$ with a non-zero coefficient.

In this topic, you often create an equation that must be true for every allowed value of $x$ . That is stronger than just solving for one value.

Definition

Identity

An identity is an equation that is true for all allowed values of the variable. We often write $≡\equiv$ instead of $=$ to emphasise that it is always true.

Equating coefficients means comparing the coefficient of each power of $x$ on both sides of an identity.

Example

Solving an identity

Find $A$ and $B$ if $\equiv A(x+2)+B(x-1)$ .

Choose values of $x$ that make terms disappear. Put $x = 1$ :
$\Rightarrow A=2$
Put $x = - 2$ :
$\Rightarrow B=3$
Check by expanding the right-hand side:
$2 (x + 2) + 3 (x - 1) = 5 x + 1$

Partial fractions

Definition

Partial fractions

A rational function is a fraction of polynomials. A partial fraction decomposition rewrites it as a sum of simpler algebraic fractions.

Partial fractions work backwards from putting fractions over a common denominator. They are useful later for integration, binomial expansions and simplifying algebraic expressions.

Distinct linear factors

A linear factor is a factor of degree one, such as $x + 4$ or $3 x - 1$ .

For a proper fraction with two different linear factors in the denominator, use one constant numerator above each factor.

$A template showing how each distinct linear denominator factor gets its own constant numerator in a partial fraction decomposition.$

Key Idea

Distinct linear factors

For a fraction with denominator $(x - a) (x - b)$ , try $Ax−a+Bx−b\frac{A}{x-a}+\frac{B}{x-b}$ . Multiply through by the full denominator, then substitute values that make factors zero.

Example

Two distinct linear factors

Express $3x+11(x+2)(x+5)\frac{3x+11}{(x+2)(x+5)}$ in partial fractions.

The substitution values x = -2 and x = -5 each make one denominator factor zero, isolating one unknown coefficient.

Write the correct form:
$3x+11(x+2)(x+5)≡Ax+2+Bx+5\frac{3x+11}{(x+2)(x+5)} \equiv \frac{A}{x+2}+\frac{B}{x+5}$
Multiply by $(x + 2) (x + 5)$ :
$\equiv A(x+5)+B(x+2)$
Put $x = - 2$ to remove the $B$ term:
$\Rightarrow A=\frac{5}{3}$
Put $x = - 5$ to remove the $A$ term:
$\Rightarrow B=\frac{4}{3}$
Write the final answer:
$3x+11(x+2)(x+5)≡53(x+2)+43(x+5)\frac{3x+11}{(x+2)(x+5)} \equiv \frac{5}{3(x+2)}+\frac{4}{3(x+5)}$

Tip

Factors like $3 x - 1$

Use the value that makes the factor zero. For $3 x - 1$ , use $x=13x=\frac{1}{3}$ ; for $2 - x$ , use $x = 2$ .

Repeated linear factors

If a denominator contains a repeated factor, you must include every power of that factor.

So a denominator containing $x-a)^2$ needs terms with both $x - a$ and $x-a)^2$ .

$Repeated linear factors require a separate partial fraction term for each power up to the repeated power.$

Common Mistake

Missing a repeated term

If the denominator contains $x-a)^2$ , do not write only $C(x−a)2\frac{C}{(x-a)^2}$ . You also need $Bx−a\frac{B}{x-a}$ .

Example

A repeated linear factor

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

Since $x+2)^2$ is repeated, include both powers:
$4x+1(x−1)(x+2)2≡Ax−1+Bx+2+C(x+2)2\frac{4x+1}{(x-1)(x+2)^2} \equiv \frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}$
Multiply by $x-1)(x+2)^2$ :
$\equiv A(x+2)^2+B(x-1)(x+2)+C(x-1)$
Put $x = 1$ :
$\Rightarrow A=\frac{5}{9}$
Put $x = - 2$ :
$\Rightarrow C=\frac{7}{3}$
Put $x = 0$ to find $B$ :
$1=4A−2B−C1=209−2B−73B=−59\begin{aligned} 1 &= 4A-2B-C \\ 1 &= \frac{20}{9}-2B-\frac{7}{3} \\ B &= -\frac{5}{9} \end{aligned}$
Write the final answer:
$4x+1(x−1)(x+2)2≡59(x−1)−59(x+2)+73(x+2)2\frac{4x+1}{(x-1)(x+2)^2} \equiv \frac{5}{9(x-1)}-\frac{5}{9(x+2)}+\frac{7}{3(x+2)^2}$

Improper rational functions

A rational function is proper if the degree of the numerator is less than the degree of the denominator.

If the numerator has degree greater than or equal to the denominator, the fraction is improper. You need a polynomial part first: a constant, a linear expression, or sometimes higher.

$A degree comparison shows when a rational function is improper and why a polynomial part is needed before partial fractions.$

Key Idea

Top-heavy fractions

If the numerator is not smaller in degree, include a polynomial part before the partial fractions. If the numerator degree is one more than the denominator degree, use $A x + B$ .

Example

An improper fraction with a linear part

Express $2x3+7x2+3x+4(x+1)(x+3)\frac{2x^3+7x^2+3x+4}{(x+1)(x+3)}$ in the form $Ax+B+Cx+1+Dx+3Ax+B+\frac{C}{x+1}+\frac{D}{x+3}$ .

$The improper fraction is decomposed into a linear polynomial part plus two simple partial fractions.$

Start with the required form:
$2x3+7x2+3x+4(x+1)(x+3)≡Ax+B+Cx+1+Dx+3\frac{2x^3+7x^2+3x+4}{(x+1)(x+3)} \equiv Ax+B+\frac{C}{x+1}+\frac{D}{x+3}$
Multiply by $(x + 1) (x + 3)$ :
$2x3+7x2+3x+4≡(Ax+B)(x+1)(x+3)+C(x+3)+D(x+1)2x^3+7x^2+3x+4 \equiv (Ax+B)(x+1)(x+3)+C(x+3)+D(x+1)$
Since $x+1)(x+3)=x^2+4x+3$ , compare the highest powers:
$A=2,4A+B=7⇒B=−1A=2,\qquad 4A+B=7 \Rightarrow B=-1$
Put $x = - 1$ :
$\Rightarrow C=3$
Put $x = - 3$ :
$\Rightarrow D=-2$
Write the final answer:
$2x3+7x2+3x+4(x+1)(x+3)≡2x−1+3x+1−2x+3\frac{2x^3+7x^2+3x+4}{(x+1)(x+3)} \equiv 2x-1+\frac{3}{x+1}-\frac{2}{x+3}$

Proof by contradiction

Definition

Proof by contradiction

A proof by contradiction starts by assuming the statement you want to prove is false. You then show that this assumption leads to an impossibility, so the original statement must be true.

The structure is usually:

Proof by contradiction follows a repeated four-step logical sequence from assuming the opposite to stating the conclusion.

Assume the opposite.
Use algebra or number facts.
Reach a contradiction.
Clearly state the conclusion.

Divisibility contradictions

An integer is a whole number: positive, negative or zero.

Many integer contradiction proofs rely on divisibility. If every term on one side is a multiple of 7, then the whole side is a multiple of 7.

Example

No integer solutions

Use contradiction to show that there are no integers $m$ and $n$ such that $14 m + 21 n = 5$ .

Assume, for contradiction, that integers $m$ and $n$ do satisfy $14 m + 21 n = 5$ .
Factor the left-hand side:
$14 m + 21 n = 7 (2 m + 3 n)$
Since $2 m + 3 n$ is an integer, the left-hand side is a multiple of 7.
The equation would therefore imply that 5 is a multiple of 7, which is impossible.
This contradiction shows that no such integers $m$ and $n$ exist.

Irrationality proofs

Definition

Rational and irrational numbers

A rational number can be written as $ab\frac{a}{b}$ , where $a$ and $b$ are integers and $b≠0b\neq0$ . An irrational number cannot be written in this form.

When proving a square root is irrational, assume it equals a fraction in lowest terms. “Lowest terms” means the numerator and denominator are coprime, so they share no common factor greater than one.

Example

Proving $5$ is irrational

Prove by contradiction that $5\sqrt{5}$ is irrational.

The contradiction comes from showing both a and b would share a factor of 5 despite being chosen coprime.

Assume the opposite: $5\sqrt{5}$ is rational, so $5=ab\sqrt{5}=\frac{a}{b}$ for coprime integers $a$ and $b$ , with $b≠0b\neq0$ .
Square both sides:
$a^2=5b^2$
So 5 divides $a^2$ . Since 5 is prime, 5 divides $a$ , so write $a = 5 k$ for some integer $k$ .
Substitute $a = 5 k$ :
$25k2=5b2⇒b2=5k225k^2=5b^2 \Rightarrow b^2=5k^2$
So 5 divides $b$ as well. This contradicts $a$ and $b$ being coprime.
Therefore $5\sqrt{5}$ is irrational.

Common Mistake

Lowest terms matters

The contradiction is that the numerator and denominator share a factor. That only works if you first choose the fraction in lowest terms.

Other classic contradiction patterns

Example

Rational plus irrational

Show that the sum of a rational number and an irrational number is irrational.

Let $r$ be rational and $s$ be irrational.
Assume, for contradiction, that $r + s$ is rational.
Since the difference of two rational numbers is rational, $(r + s) - r$ would be rational.
But $(r + s) - r = s$ , so $s$ would be rational, contradicting the fact that $s$ is irrational.
Therefore $r + s$ is irrational.

Definition

Prime number

A prime number is an integer greater than one with exactly two positive factors: one and itself.

Example

Infinitely many primes

Prove by contradiction that there are infinitely many prime numbers.

Euclid’s contradiction constructs a number one more than the product of all listed primes, so no listed prime divides it.

Assume there are only finitely many primes, called $p1,p2,…,pnp_1,p_2,\ldots,p_n$ .
Form a new integer:
$N=p1p2⋯pn+1N=p_1p_2\cdots p_n+1$
Dividing $N$ by any listed prime $p_i$ leaves remainder one, so none of the listed primes divides $N$ .
Since $N > 1$ , it must have at least one prime factor.
That prime factor is not in the list, which is a contradiction.
Therefore there are infinitely many prime numbers.

Exam technique

In the exam

For partial fractions, write the required form first; repeated factors need every power.
Multiply through by the full denominator before substituting values of $x$ .
In contradiction proofs, clearly state the assumption, the contradiction, and the final conclusion.

Self review

Check yourself

If a denominator contains $x+4)^3$ , which partial fraction terms are needed?
Why must an improper rational function have a polynomial part?
In a proof that $7\sqrt{7}$ is irrational, why do you begin with a fraction in lowest terms?

Algebraic Methods

What you'll learn

How to split algebraic fractions into partial fractions.
How to handle repeated factors and “top-heavy” algebraic fractions.
How to use substitution and equating coefficients cleanly.
How to write proof by contradiction arguments for integers, irrational numbers and primes.

The algebra you need first

A polynomial is an expression made from powers of $x$ , such as $3x^2-5x+1$ . The degree of a polynomial is the highest power of $x$ with a non-zero coefficient.

In this topic, you often create an equation that must be true for every allowed value of $x$ . That is stronger than just solving for one value.

Definition

Identity

An identity is an equation that is true for all allowed values of the variable. We often write $≡\equiv$ instead of $=$ to emphasise that it is always true.

Equating coefficients means comparing the coefficient of each power of $x$ on both sides of an identity.

Example

Solving an identity

Find $A$ and $B$ if $\equiv A(x+2)+B(x-1)$ .

Choose values of $x$ that make terms disappear. Put $x = 1$ :
$\Rightarrow A=2$
Put $x = - 2$ :
$\Rightarrow B=3$
Check by expanding the right-hand side:
$2 (x + 2) + 3 (x - 1) = 5 x + 1$

Partial fractions

Definition

Partial fractions

A rational function is a fraction of polynomials. A partial fraction decomposition rewrites it as a sum of simpler algebraic fractions.

Partial fractions work backwards from putting fractions over a common denominator. They are useful later for integration, binomial expansions and simplifying algebraic expressions.

Distinct linear factors

A linear factor is a factor of degree one, such as $x + 4$ or $3 x - 1$ .

For a proper fraction with two different linear factors in the denominator, use one constant numerator above each factor.

$A template showing how each distinct linear denominator factor gets its own constant numerator in a partial fraction decomposition.$

Key Idea

Distinct linear factors

For a fraction with denominator $(x - a) (x - b)$ , try $Ax−a+Bx−b\frac{A}{x-a}+\frac{B}{x-b}$ . Multiply through by the full denominator, then substitute values that make factors zero.

Example

Two distinct linear factors

Express $3x+11(x+2)(x+5)\frac{3x+11}{(x+2)(x+5)}$ in partial fractions.

The substitution values x = -2 and x = -5 each make one denominator factor zero, isolating one unknown coefficient.

Write the correct form:
$3x+11(x+2)(x+5)≡Ax+2+Bx+5\frac{3x+11}{(x+2)(x+5)} \equiv \frac{A}{x+2}+\frac{B}{x+5}$
Multiply by $(x + 2) (x + 5)$ :
$\equiv A(x+5)+B(x+2)$
Put $x = - 2$ to remove the $B$ term:
$\Rightarrow A=\frac{5}{3}$
Put $x = - 5$ to remove the $A$ term:
$\Rightarrow B=\frac{4}{3}$
Write the final answer:
$3x+11(x+2)(x+5)≡53(x+2)+43(x+5)\frac{3x+11}{(x+2)(x+5)} \equiv \frac{5}{3(x+2)}+\frac{4}{3(x+5)}$

Tip

Factors like $3 x - 1$

Use the value that makes the factor zero. For $3 x - 1$ , use $x=13x=\frac{1}{3}$ ; for $2 - x$ , use $x = 2$ .

Repeated linear factors

If a denominator contains a repeated factor, you must include every power of that factor.

So a denominator containing $x-a)^2$ needs terms with both $x - a$ and $x-a)^2$ .

$Repeated linear factors require a separate partial fraction term for each power up to the repeated power.$

Common Mistake

Missing a repeated term

If the denominator contains $x-a)^2$ , do not write only $C(x−a)2\frac{C}{(x-a)^2}$ . You also need $Bx−a\frac{B}{x-a}$ .

Example

A repeated linear factor

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

Since $x+2)^2$ is repeated, include both powers:
$4x+1(x−1)(x+2)2≡Ax−1+Bx+2+C(x+2)2\frac{4x+1}{(x-1)(x+2)^2} \equiv \frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}$
Multiply by $x-1)(x+2)^2$ :
$\equiv A(x+2)^2+B(x-1)(x+2)+C(x-1)$
Put $x = 1$ :
$\Rightarrow A=\frac{5}{9}$
Put $x = - 2$ :
$\Rightarrow C=\frac{7}{3}$
Put $x = 0$ to find $B$ :
$1=4A−2B−C1=209−2B−73B=−59\begin{aligned} 1 &= 4A-2B-C \\ 1 &= \frac{20}{9}-2B-\frac{7}{3} \\ B &= -\frac{5}{9} \end{aligned}$
Write the final answer:
$4x+1(x−1)(x+2)2≡59(x−1)−59(x+2)+73(x+2)2\frac{4x+1}{(x-1)(x+2)^2} \equiv \frac{5}{9(x-1)}-\frac{5}{9(x+2)}+\frac{7}{3(x+2)^2}$

Improper rational functions

A rational function is proper if the degree of the numerator is less than the degree of the denominator.

If the numerator has degree greater than or equal to the denominator, the fraction is improper. You need a polynomial part first: a constant, a linear expression, or sometimes higher.

$A degree comparison shows when a rational function is improper and why a polynomial part is needed before partial fractions.$

Key Idea

Top-heavy fractions

If the numerator is not smaller in degree, include a polynomial part before the partial fractions. If the numerator degree is one more than the denominator degree, use $A x + B$ .

Example

An improper fraction with a linear part

Express $2x3+7x2+3x+4(x+1)(x+3)\frac{2x^3+7x^2+3x+4}{(x+1)(x+3)}$ in the form $Ax+B+Cx+1+Dx+3Ax+B+\frac{C}{x+1}+\frac{D}{x+3}$ .

$The improper fraction is decomposed into a linear polynomial part plus two simple partial fractions.$

Start with the required form:
$2x3+7x2+3x+4(x+1)(x+3)≡Ax+B+Cx+1+Dx+3\frac{2x^3+7x^2+3x+4}{(x+1)(x+3)} \equiv Ax+B+\frac{C}{x+1}+\frac{D}{x+3}$
Multiply by $(x + 1) (x + 3)$ :
$2x3+7x2+3x+4≡(Ax+B)(x+1)(x+3)+C(x+3)+D(x+1)2x^3+7x^2+3x+4 \equiv (Ax+B)(x+1)(x+3)+C(x+3)+D(x+1)$
Since $x+1)(x+3)=x^2+4x+3$ , compare the highest powers:
$A=2,4A+B=7⇒B=−1A=2,\qquad 4A+B=7 \Rightarrow B=-1$
Put $x = - 1$ :
$\Rightarrow C=3$
Put $x = - 3$ :
$\Rightarrow D=-2$
Write the final answer:
$2x3+7x2+3x+4(x+1)(x+3)≡2x−1+3x+1−2x+3\frac{2x^3+7x^2+3x+4}{(x+1)(x+3)} \equiv 2x-1+\frac{3}{x+1}-\frac{2}{x+3}$

Proof by contradiction

Definition

Proof by contradiction

A proof by contradiction starts by assuming the statement you want to prove is false. You then show that this assumption leads to an impossibility, so the original statement must be true.

The structure is usually:

Proof by contradiction follows a repeated four-step logical sequence from assuming the opposite to stating the conclusion.

Assume the opposite.
Use algebra or number facts.
Reach a contradiction.
Clearly state the conclusion.

Divisibility contradictions

An integer is a whole number: positive, negative or zero.

Many integer contradiction proofs rely on divisibility. If every term on one side is a multiple of 7, then the whole side is a multiple of 7.

Example

No integer solutions

Use contradiction to show that there are no integers $m$ and $n$ such that $14 m + 21 n = 5$ .

Assume, for contradiction, that integers $m$ and $n$ do satisfy $14 m + 21 n = 5$ .
Factor the left-hand side:
$14 m + 21 n = 7 (2 m + 3 n)$
Since $2 m + 3 n$ is an integer, the left-hand side is a multiple of 7.
The equation would therefore imply that 5 is a multiple of 7, which is impossible.
This contradiction shows that no such integers $m$ and $n$ exist.

Irrationality proofs

Definition

Rational and irrational numbers

A rational number can be written as $ab\frac{a}{b}$ , where $a$ and $b$ are integers and $b≠0b\neq0$ . An irrational number cannot be written in this form.

Example

Proving $5$ is irrational

Prove by contradiction that $5\sqrt{5}$ is irrational.

The contradiction comes from showing both a and b would share a factor of 5 despite being chosen coprime.

Assume the opposite: $5\sqrt{5}$ is rational, so $5=ab\sqrt{5}=\frac{a}{b}$ for coprime integers $a$ and $b$ , with $b≠0b\neq0$ .
Square both sides:
$a^2=5b^2$
So 5 divides $a^2$ . Since 5 is prime, 5 divides $a$ , so write $a = 5 k$ for some integer $k$ .
Substitute $a = 5 k$ :
$25k2=5b2⇒b2=5k225k^2=5b^2 \Rightarrow b^2=5k^2$
So 5 divides $b$ as well. This contradicts $a$ and $b$ being coprime.
Therefore $5\sqrt{5}$ is irrational.

Common Mistake

Lowest terms matters

The contradiction is that the numerator and denominator share a factor. That only works if you first choose the fraction in lowest terms.

Other classic contradiction patterns

Example

Rational plus irrational

Show that the sum of a rational number and an irrational number is irrational.

Let $r$ be rational and $s$ be irrational.
Assume, for contradiction, that $r + s$ is rational.
Since the difference of two rational numbers is rational, $(r + s) - r$ would be rational.
But $(r + s) - r = s$ , so $s$ would be rational, contradicting the fact that $s$ is irrational.
Therefore $r + s$ is irrational.

Definition

Prime number

A prime number is an integer greater than one with exactly two positive factors: one and itself.

Example

Infinitely many primes

Prove by contradiction that there are infinitely many prime numbers.

Euclid’s contradiction constructs a number one more than the product of all listed primes, so no listed prime divides it.

Assume there are only finitely many primes, called $p1,p2,…,pnp_1,p_2,\ldots,p_n$ .
Form a new integer:
$N=p1p2⋯pn+1N=p_1p_2\cdots p_n+1$
Dividing $N$ by any listed prime $p_i$ leaves remainder one, so none of the listed primes divides $N$ .
Since $N > 1$ , it must have at least one prime factor.
That prime factor is not in the list, which is a contradiction.
Therefore there are infinitely many prime numbers.

Exam technique

In the exam

For partial fractions, write the required form first; repeated factors need every power.
Multiply through by the full denominator before substituting values of $x$ .
In contradiction proofs, clearly state the assumption, the contradiction, and the final conclusion.

Self review

Check yourself

If a denominator contains $x+4)^3$ , which partial fraction terms are needed?
Why must an improper rational function have a polynomial part?
In a proof that $7\sqrt{7}$ is irrational, why do you begin with a fraction in lowest terms?

Algebraic Methods

What you'll learn

The algebra you need first

Partial fractions

Distinct linear factors

Repeated linear factors

Improper rational functions

Proof by contradiction

Divisibility contradictions

Irrationality proofs

Other classic contradiction patterns

Recap questions

You've reached the end

Algebraic Methods

What you'll learn

The algebra you need first

Partial fractions

Distinct linear factors

Repeated linear factors

Improper rational functions

Proof by contradiction

Divisibility contradictions

Irrationality proofs

Other classic contradiction patterns

Recap questions

You've reached the end

Algebraic Methods

What you'll learn

The algebra you need first

Partial fractions

Distinct linear factors

Repeated linear factors

Improper rational functions

Proof by contradiction

Divisibility contradictions

Irrationality proofs

Other classic contradiction patterns

Recap questions

You've reached the end

Algebraic Methods Revision Guide

Algebraic Methods

What you'll learn

The algebra you need first

Partial fractions

Distinct linear factors

Repeated linear factors

Improper rational functions

Proof by contradiction

Divisibility contradictions

Irrationality proofs

Other classic contradiction patterns

Recap questions

You've reached the end

Algebraic Methods Revision Guide