2.6.1 Rational Expressions

Rational Functions

Definition:

A rational function is a function that can be written as $\dfrac{p(x)}{q(x)}$ where $p(x)$ and $q(x)$ are polynomials in $x$ .

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Example: $\dfrac{x^2 + 5}{x - 8}$ is a rational function.

Simplifying Rational Functions

Rational functions can be simplified using polynomial factorization.

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Example 1:

\frac{4y^2}{2y^2 + y} \times \frac{y^2 + 2y - 15}{2y^2 + 11y + 5}

Factorize the polynomials:

= \frac{4y^2 \times (y-1)(y+5)}{(2y+1)(y)(y+5)(y-3)}

Cancel common factors:

= \frac{4y \times (y-1)}{(2y+1)(y-3)}

Result:

\frac{4y}{y-3}

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Example 2:

\frac{5t + 12}{2t^2 + 7t + 3} - \frac{4}{2t+1}

Find a common denominator:

= \frac{5t + 12}{(2t+1)(t+3)} - \frac{4(t+3)}{(2t+1)(t+3)}

Combine into a single fraction:

= \frac{5t + 12 - 4t - 12}{(2t+1)(t+3)}

Simplify the expression:

= \frac{t}{(2t+1)(t+3)}

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Example 3: Simplify fully

\frac{3x^2 - 8x - 3}{2x^2 - 6x}

Factorize both numerator and denominator:

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

Cancel common factors:

= \frac{3x+1}{2x}

Final result:

\frac{3}{2} + \frac{1}{2x}

Applications to Partial Fractions

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Express $\dfrac{9x^2 + 20x - 10}{(x+2)(3x-1)}$ in partial fractions. Notice that the denominator has the same order as the numerator. This means the division algorithm will give rise to a constant term.

(x+2)(3x-1) = 3x^2 + 6x - x - 2 = 3x^2 + 5x - 2

Do division by grid method:

\frac{9x^2 + 20x - 10}{3x^2 + 5x - 2}

Divide $9x^2$ by $3x^2$ to get $3$ (Quotient):

3 \times (3x^2 + 5x - 2) = 9x^2 + 15x - 6

Subtract to find the remainder:

(9x^2 + 20x - 10) - (9x^2 + 15x - 6) = 5x - 4

\Rightarrow \frac{5x-4}{(x+2)(3x-1)}

Key Point: If the numerator is top-heavy, i.e., has a degree $\geq$ the denominator, a division must be performed before partial fractions can be found.

Concentrating on just the partial fractions and ignoring the quotient:

\frac{5x-4}{(x+2)(3x-1)} = \frac{A}{x+2} + \frac{B}{3x-1}

Multiply both sides by $(x+2)(3x-1)$ to get:

5x - 4 = A(3x-1) + B(x+2)

Let $x = -2$ :

5(-2) - 4 = -10 - 4 = -14 \Rightarrow -7A = -14 \Rightarrow A = 2

Let $x = \frac{1}{3}$ :

5 \times \frac{1}{3} - 4 = \frac{5}{3} - \frac{12}{3} = -\frac{7}{3}

\Rightarrow -\frac{7}{3} = \frac{1}{3}B \Rightarrow B = -1

5x - 4 = 2(3x-1) - 1(x+2)]

= 6x - 2 - x - 2

= 5x - 4 \text{ (which verifies our values of \(A\) and \(B\))}

\Rightarrow \frac{9x^2 + 20x - 10}{(x+2)(3x-1)} = 3 + \frac{2}{x+2} - \frac{1}{3x-1}

Partial Fractions

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GCSE Recap:

Write as a single fraction: $\dfrac{3}{x+2} + \dfrac{4}{x-3}$

Need to have the same denominator:

\frac{3(x-3) + 4(x+2)}{(x+2)(x-3)} = \frac{3x-9 + 4x+8}{(x+2)(x-3)} = \frac{7x-1}{(x+2)(x-3)}

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New Material:

Write $\dfrac{7x-1}{(x+2)(x-3)}$ as the sum of separate fractions.

Write an identity with the LHS as our combined fraction and the RHS as our target form:

\frac{7x-1}{(x+2)(x-3)} \equiv \frac{A}{x+2} + \frac{B}{x-3}

Multiply both sides by the denominator of the LHS:

7x-1 \equiv A(x-3) + B(x+2)

Substitute in 'tactical' values of $x$ to help find the values of $A$ and $B$ :

Let $x = 3$ :

20 = A(3-3) + 5B \Rightarrow B = 4

Let $x = -2$ :

-15 = -5A \Rightarrow A = 3

Thus:

\frac{7x-1}{(x+2)(x-3)} = \frac{3}{x+2} + \frac{4}{x-3}

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Example 1:

Write $\dfrac{2}{(x+1)(x+3)}$ as partial fractions.

Start by expressing $\dfrac{2}{(x+1)(x+3)}$ as the sum of two fractions:

\frac{2}{(x+1)(x+3)} \equiv \frac{A}{x+1} + \frac{B}{x+3}

Multiply through by the common denominator $(x+1)(x+3)$ to clear the fractions:

2 \equiv A(x+3) + B(x+1)

Substitute values of $x$ to solve for $A$ and $B$ :

Let $x = -3$ :

2 = A(-3+3) + B(-3+1) \Rightarrow 2 = -2B \Rightarrow B = -1

Let $x = -1$ :

2 = A(-1+3) + B(-1+1) \Rightarrow 2 = 2A \Rightarrow A = 1

Therefore, the partial fraction decomposition is:

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

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Example 2:

Express $\dfrac{1}{x^2 - 1}$ as partial fractions.

Start by expressing $\dfrac{1}{x^2 - 1}$ using the difference of squares:

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

Write the partial fractions decomposition:

\frac{1}{(x+1)(x-1)} \equiv \frac{A}{x+1} + \frac{B}{x-1}

Multiply both sides by the common denominator $(x+1)(x-1)$ :

1 \equiv A(x-1) + B(x+1)

Substitute values of $x$ to solve for $A$ and $B$ :

Let $x = 1$ :

1 = A(1-1) + B(1+1) \Rightarrow 1 = 2B \Rightarrow B = \frac{1}{2}

Let $x = -1$ :

1 = A(-1-1) + B(-1+1) \Rightarrow 1 = -2A \Rightarrow A = -\frac{1}{2}

Therefore, the partial fraction decomposition is:

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

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Example 3:

Express $\dfrac{9x-14}{(x+4)(x-1)^2}$ as partial fractions.

Setup Partial Fractions:

Write the fraction as a sum of simpler fractions:

\frac{9x-14}{(x+4)(x-1)^2} \equiv \frac{A}{x+4} + \frac{B}{x-1} + \frac{C}{(x-1)^2}

Here, $(x-1)^2$ in the denominator leads to factors of $(x-1)$ and $(x-1)^2$ .

Combine into a Single Expression:

9x - 14 \equiv A(x-1)^2 + B(x+4)(x-1) + C(x+4)

Find Coefficients $A$ , $B$ , and $C$ by Substitution:

Let $x = 1$ :

9(1) - 14 = A(1-1)^2 + B(1+4)(1-1) + C(1+4)

-5 = 5C \Rightarrow C = -1

Let $x = -4$ :

9(-4) - 14 = A(-4-1)^2 + B(-4+4)(-4-1) + C(-4+4)

-50 = 25A \Rightarrow A = -2

For another value, let $x = 0$ (a non-clever substitution):

9(0) - 14 = A(0-1)^2 + B(0+4)(0-1) + C(0+4)

-14 = A - 4B + 4C

Substitute $A = -2$ and $C = -1$ into the equation:

-14 = -2 - 4B - 4

-14 = -6 - 4B \Rightarrow 4B = 8 \Rightarrow B = 2

Final Partial Fraction Decomposition:

\frac{9x-14}{(x+4)(x-1)^2} = \frac{-2}{x+4} + \frac{2}{x-1} - \frac{1}{(x-1)^2}

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Q5 (Jun 2010, Q3)

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

Setup Partial Fractions:

Write the fraction as a sum of simpler fractions:

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

Combine into a Single Expression:

x^2 \equiv A(x-1)^2 + B(x-1)(x-2) + C(x-2)

Find Coefficients A, B, and C by Substitution:

Let $x = 2$ :

2^2 = A(2-1)^2 \Rightarrow 4 = A(1)^2 \Rightarrow A = 4

Let $x = 1$ :

1^2 = C(1-2) \Rightarrow 1 = -C \Rightarrow C = -1

For another value, let $x = 0$ (to find $B$ ):

0^2 = A(0-1)^2 + B(0-1)(0-2) + C(0-2)

0 = 4 + 2B + 2(-1)

-6 = 2B \Rightarrow B = -3

Final Partial Fraction Decomposition:

\frac{x^2}{(x-1)^2(x-2)} = \frac{4}{x-2} + \frac{-3}{x-1} + \frac{-1}{(x-1)^2}

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Q9 (Jun 2015, Q1) (i) Express $\dfrac{2}{3-x} + \frac{3}{1+x}$ as a single fraction in its simplest form.

Solution:

\frac{2(1+x) + 3(3-x)}{(3-x)(1+x)} = \frac{2 + 2x + 9 - 3x}{(3-x)(1+x)} = \frac{11-x}{(3-x)(1+x)}

(ii) Hence express $\left(\dfrac{2}{3-x} + \dfrac{3}{1+x}\right) \times \dfrac{x^2 + 8x - 33}{121 - x^2}$ as a single fraction in its lowest terms.

Solution:

\frac{11-x}{(3-x)(1+x)} \times \frac{(x+11)(x-3)}{(11+x)(11-x)} = \frac{x-3}{(3-x)(1+x)}

= \frac{-1}{1+x}

Rational Expressions Simplified Revision Notes for A-Level AQA Maths Pure

2.6.1 Rational Expressions

Rational Functions

Simplifying Rational Functions

Applications to Partial Fractions

Partial Fractions

GCSE Recap:

New Material:

Example 1:

Example 2:

Example 3:

Q5 (Jun 2010, Q3)

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