Lesson Objective

Dear learner,

by the end of this lesson you will be able to:

• Define rational expressions
• Identify the universe of a given rational expression
• Show the simplified form of rational expressions
• Perform the four basic operations on rational expressions

Key Terms

• Rational expressions
• Domain of rational Expressions
• Operations on rational expressions
• Decomposition of rational expressionms

Brainstorming Activity

Dear learner, think about what you have learnt in the previous grades and try to give your answer on the following two questions.

1). What are rational numbers?

2).How do you describe polynomial expressions?

Solution:

I think your description about the two questions includes the following main points.

1). A rational number is a number writtten in the form $\frac{a}{b}$ where a and b are integers $b \ne 0$.

2). polynomial expressions are expressions of the form $a_{n} x^{n} + a_{n-1}x^{n-1} + …. + a_{1}x + a_{0}$, where $a_{0},….,a_{n}$ are real numbers and $n$ is a non negative intger.

1.1 Rational Expressions

Definition:

An expression written in the form of $\frac{P(x)}{Q(x)}$ but $Q(x) \ne 0$ and $P(x), Q(x)$ are polynomials is called a rational expression. $P(X)$ is called a numerator, while $Q(X)$ is called the denominator.

Example 1: Which of the following are rational expressions?

a). $\frac{2x+3}{x^{2}-5}$

b). $\frac{5}{\sqrt{ X^{2} +1}}$

c). $\frac{\sqrt {x^{2} +1}^{2}}{x-2}$

Solution:

a). $\frac{2x+3}{x^{2}-5}$ is a rational expression since $2x+3$ and $x^{2} -5$ are polynomials

b). $\frac{5}{\sqrt {X^{2} +1}}$ is not a rational expression because $\sqrt(x^{2} +1)$ is not a polynomial expression

c). $\frac{\sqrt {x^{2} +1}^{2}}{x-2}$ is a rational expression, since x-2 and $\sqrt{x^{2}+1}^{2} = |x^{2} +1| = x^{2} +1 $are polynomials.

1.2 Domain of rational expressions

The domain of arational expression$\frac{P(x)}{Q(X)}$ is the set $\mathbb{R}$ in which ${Q(x) \ne 0}$.

Example 2:

determine domain of the following expressions

a). $\frac{2x^{2} + 1}{3x -5}$

b). $\frac{x^{2} +1}{x^{2} +5x+6}$

c). $\frac{5x+6}{x^{2} +5}$

Solution:

a. $\frac{2x^{2} + 1}{3x -5}$ to determine the domain solve $Q(x) = 0$:

$3x – 5 =0$ $\implies$ $x = \frac{5}{3}$

then $domain = \mathbb{R}$/ {${\frac{5}{3}}$}

b). $\frac{x^{2} +1}{x^{2} +5x+6}$

$Q(x) = x^{2} +5x+6$ and solve $x^{2} +5x +6 =0$ $iff$ ${x+2)(x+3) = 0$

$\implies$ $x+2 =0$ or $x+3 = 0$ $\implies $ $ x = -2 or x = -3$

therefore $Domain = \mathbb{R}$/ {${{-3,-2}}$}

c). $\frac{5x+6}{x^{2} +5}$

$Q(x) = x^{2} +5 $ $\implies$ $x^{2} +2 $ is always positive, it cannot be zero for any real number.

therefore $domain = \mathbb{R}$

1.3 Simplifying rational expressions

Definition: A rational expression $\frac{P(x)}{Q(x)}$ is in its lowest form when the common factor of $P|(x)$ and $Q(x)$ is 1.

Steps to simplify:

     Step 1: Restrict the domain

     Step2: Factorize the numerator and denominator if possible

     Step 3: Cancel out common factors

     Step 4: Re- write the expression in its simplest form

Example 3:

simplify the following rational expressions

a). $\frac{x+2}{x^{2}-4}$

b). $\frac{x^{2} +5x+6}{x^{2}-9}$

c). $\frac{x^{2} +1}{x^{4}-1}$

Solution:

a. $\frac{x+2}{x^{2}-4}$, lets determine the domain first $Q(x) = x^{2} -4 = 0$

$\implies$ $(x+2)(x-) = 0 $ $\implies$ $x=-2 or x = 2$

$domain = \mathbb{R}/_{-2,2}$

Then factorize both the numerator and denominators and cancel out common factors

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

Therefore $\frac{x+2}{x^{2}-4} = \frac{1}{x-2}$, $x \ne -2,2$

b. $\frac{x^{2} +5x+6}{x^{2}-9}$

Domain: $ x^{2}-9 = 0$ $\implies$ $x= \pm 3$, so $domain = \mathbb{R}/_{-3,3}$

factorize and cancelout: $\frac{x^{2} +5x+6}{x^{2}-9} = \frac{(x+2)(x+3)}{x-3)(x+3)} = \frac{x+2}{x-3}$ where $x+3$ is the common factor and is cancelled out.

Therefore $\frac{x^{2} +5x+6}{x^{2}-9} = \frac{x+2}{x-3}$, $x \ne -3,3$

c. $\frac{x^{2} +1}{x^{4}-1}$:

domain: $x^{4}-1 = (x^{2} -1)(x^{2} +1) = 0$ $\implies$ $ x^{2} -1 = 0$ since $x^{2}+1$ cannot be zero, then $x = \pm 1$ $\implies$ $domain = \mathbb{R}/_{-1,1}$,

Factorize and simplify:

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

therefore $\frac{x^{2} +1}{x^{4}-1}= \frac{1}{x^{2}-1}$, $x \ne {-1,1}$

1.4. Operations on Rational Expressions

A) Addition and Subtraction of  Rational Expressions

Let $ \frac{P(x)}{Q(x)} $ and $\frac{R(x)}{S(x)}$ are two rational expressions and $ Q(x), S(x) \neq 0 $ then:

$\frac{P(x)}{Q(x)} + \frac{R(x)}{S(x)} = \frac{P(x)S(x) + R(x)Q(x)}{Q(x)S(x)}$

$\frac{P(x)}{Q(x)} – \frac{R(x)}{S(x)} = \frac{P(x)S(x) – R(x)Q(x)}{Q(x)S(x)}$.

Example 4:

1. Find the sum for the following expressions.

a. $ \frac{1}{x} + \frac{x(x+3)}{x(x-1)} $.

b. $\frac{x}{x+1} + \frac{2x^3 + 3}{x+1} $

c. $ \frac{2}{x-3} + \frac{x+5}{x^2 – 9} $

d. $\frac{2}{a^2} + \frac{3}{ab} + \frac{4}{b^2} $

e. $u + 1 + \frac{1}{u+1} $

2. Find the difference

a. $ \frac{x^2}{x^2 + x – 2} – \frac{x + 3}{x + 2} $
b. $\frac{2}{x – 1} – \frac{5}{x – 1} $
c. $ \frac{1}{x – y} – \frac{x^2 + xy + y^2}{x^2 – 2xy + y^2} $.

Solutions:

1. a. $ \frac{1}{x} + \frac{x(x+3)}{x(x-1)} = \frac{x(x-1) + x(x+3)}{x(x-1)} = \frac{x – 1 + 2 + 3x}{x^2 – x}, \quad x \neq 0,1 $

b. $\frac{x}{x+1} + \frac{2x^3 + 3}{x+1} = \frac{2x^3 + x + 3}{x + 1}, \quad x \neq -1 $.

c. $ \frac{2}{x-3} + \frac{x+5}{x^2 – 9} = \frac{2(x-3) + x + 5}{x^2 – 9} = \frac{2x – 6 + x + 5}{x^2 – 9} = \frac{3x – 1}{x^2 – 9}, \quad x \neq 3, -3 $

d. $ \frac{2}{a^2} + \frac{3}{ab} + \frac{4}{b^2} = \frac{2b^2 + 3ab + 4a^2}{a^2 b^2}, \quad a, b \neq 0 $

e. $u + 1 + \frac{1}{u+1} = \frac{(u+1)^2 + 1}{u+1}, \quad u \neq -1 $.

2.a. $\frac{x^2}{x^2 + x – 2} – \frac{x + 3}{x + 2} $

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

$= \frac{((x+2)(x^2 + x – 2)(x – 1))}{(x+2)(x^2 + x – 2)}, \quad x \neq 2,1 $.

$=\frac{-2x + 3}{(x + 2)(x – 1)} \quad x \neq 2,1 $.

b. $ \frac{2}{x – 1} – \frac{5}{x – 1} = \frac{2-5}{(x – 1)} $

$= \frac{-3}{x-1}, \quad x \neq 1, -1 $.

c. $\frac{1}{x – y} – \frac{x^2 + xy + y^2}{x^2 – 2xy + y^2} $

$= \frac{(x-y) – (x^2 + xy + y^2)}{(x^2 – 2xy + y^2)}, \quad x \neq y $

$=\frac{-x^2 – xy + x – y – y^2}{(x – y)^2}\quad x \neq y$

b) Multiplication and Division of Rational Expressions

Let $\frac{P(x)}{Q(x)}$ and $ \frac{R(x)}{S(x)} $ be two rational expressions and $ Q(x), S(x) \neq 0 $ then:

$ \frac{P(x)}{Q(x)} \times \frac{R(x)}{S(x)} = \frac{P(x) \times R(x)}{Q(x) \times S(x)} = \frac{P(x)R(x)}{Q(x)S(x)}$

$ \frac{P(x)}{Q(x)} \div \frac{R(x)}{S(x)} = \frac{P(x)}{Q(x)} \times \frac{S(x)}{R(x)} = \frac{P(x)S(x)}{Q(x)R(x)}, \ R(x) \neq 0$.

Example 5:

Simplify the following expressions

a. $\frac{x^2 – x – 12}{x^2 – 9} \times \frac{3 + x}{4 – x} $

b. $\frac{2x^2 – 3x – 2}{x^2 – 1} \times \frac{x^2 +x- 2}{2x^2 + 5x + 2} $

c. $\frac{xy – xz}{xy + xz} \times \frac{y}{yz} \times \frac{y+z}{y} $

d. $\frac{x^2 – 3x – 4}{x^2 – 1} \div \frac{x^2 – 16}{x + 4} $

e. $\frac{x^2 – 3}{x-\sqrt{6}} \div \frac{x^2 – \sqrt{3}x}{x^2 – 6} $.

Solutions:

a. $\frac{x^2 – x – 12}{x^2 – 9} \times \frac{3 + x}{4 – x} = -\frac{(x + 3)(x – 4)(3 + x)}{(x – 4)(x + 3)(x – 3)} $

$= \frac{x + 3}{3-x} , \quad x \neq -3, 3, 4$.

b. $\frac{2x^2 – 3x – 2}{x^2 – 1} \times \frac{x^2 + 2}{2x^2 + 5x + 2} = \frac{(2x + 1)(x – 2)(x – 1)}{(x + 1)(x – 1)(2x + 1)} = \frac{x – 2}{x + 1}, \quad x \neq -2, -1, \frac{1}{2}$.

c. $\frac{xy – xz}{xy + xz} \times \frac{y}{yz} \times \frac{y+z}{y} = \frac{x(y – z)}{x(y + z)} \times \frac{y+z}{yz}= \frac{y – z}{yz}, \quad y, z \neq 0 $.

d. $\frac{x^2 – 3x – 4}{x^2 – 1} \div \frac{x^2 – 16}{x + 4}=\frac{(x-4)(x+1)}{(x-1)(x+1)}\times \frac{(x+4)}{(x + 4)(x-4)}= \frac{1}{x-1}, \quad x \neq -4, -1, 1 , 4$.

e. $\frac{x^2 – 3}{x-\sqrt{6}} \div \frac{x^2 – \sqrt{3}x}{x^2 – 6}= \frac{(x-\sqrt{3})(x+\sqrt{3})}{x-\sqrt{6}} \times \frac{(x- \sqrt{6})(x+ \sqrt{6})}{x\times (x-\sqrt{3})}=\frac{(x+\sqrt{3})(x+\sqrt{6})}{x}$.

1.5.Decomposition of  Rational Expressions into Partial Fractions

Definition:

Let $\frac{P(x)}{Q(x)}$ be a proper rational expression, the method of splitting $\frac{P(x)}{Q(x)}$ in to simplest fractions(partial fractions) is said to be decomposition. That is,

$\frac{P(x)}{Q(x)}$ be proper when degree of $P(x) < Q(x).$

Example 6:

Which of the following expressions are proper?

a. $\frac {7x+3}{x^2+5x-4}$

b. $\frac{6x+5}{x+8}$

c. $\frac{5}{x}$

d. $\frac{3x^2+9}{5x}$

Solution:

a. $\frac {7x+3}{x^2+5x-4}$ is a proper rational expression, since degree of $7x+3$ is less than degree of $x^2+5x-4$.

b. $\frac{6x+5}{x+8}$ is an improper rational expression, since $6x+5$ and $x+8$ have the same dgree.

c. $\frac{5}{x}$ is proper, since degree of constants is zero which is less than degree of $x$ that is one.

d. $\frac{3x^2+9}{5x}$ is an improper rational expression, since degree of $3x^2+9$ is greater than degree of $5x$.

Methods of Decomposing Rational Expressions into Partial Fractions

Expressing a simple rational expression into the sum of two or more simple rational fractions is called Partial fraction resolution.  

Method 1: Linear Factor

When the factor describing denominator are linear (That is, of the form $ax+b, a \neq 0$.

Assume the partial fraction as $\frac{constant}{factor}$.

i. Different factors of the form $\frac{P(x)}{(ax+b)(cx+d)}$, assume $\frac{P(x)}{(ax+b)(cx+d)}=\frac{A}{ax+b}+\frac{B}{cx+d}+…$

ii. Factors with multiplicity, $\frac{P(x)}{(ax+b)^n}, n \in \mathbb{N}$, assume $\frac{P(x)}{(ax+b)^n}=\frac{A}{ax+b}+\frac{B}{(ax+b)^2}+…+\frac{z}{(ax+b)^n}$.

Example 7:

Decompose rational expressions into partial fractions

a. $\frac{5x+6}{x^{2}+4x+3}=\frac{5x+6}{(x+1)(x+3)}=\frac{A}{x+1}+\frac{B}{x+3}=\frac{A(x+3)+B(x+1)}{(x+1)(x+3)}$

$\Rightarrow\frac{5x+6}{x^{3}+4x+3}=\frac{A(x+3)+B(x+1)}{(x+1)(x+3)}$

$\Rightarrow5x+6=A(x+3)+B(x+1)=Ax+3A+Bx+B$

$\Rightarrow5x=Ax+Bx~and~6=3A+B.\text{ Take the corresponding terms from the equality}$

$\Rightarrow5=A+B~and~6=3A+B\Rightarrow2A=1\Rightarrow A=\frac{1}{2}\text{ and}~B=\frac{9}{2}$.

$\text{OR use the shortest trick:}$

$\text{To find A, make}~x=-1$ and $\text{to find B, make}~x=-3$.

$\Rightarrow5(-1)+6=A(-1+3)+B(-1+1)$

$\Rightarrow-5+6=2A+0$

$\Rightarrow2A=1\Rightarrow A=\frac{1}{2}$

$\Rightarrow5(-3)+6=A((-3)+3)+B(-3+1)$

$\Rightarrow-15+6=-2B$

$\Rightarrow B=\frac{9}{2}$

Therefore,

$\frac{(5x+6)}{x^{2}+4x+6}=\frac{\frac{1}{2}}{x+1}+\frac{\frac{9}{2}}{x+3}=\frac{1}{2(x+1)}+\frac{9}{2(x+1)}$

b. $\frac{3x+2}{x^{2}+6x+9}=\frac{3x+2}{(x+3)^{2}}=\frac{A}{x+3}+\frac{B}{(x+3)^{2}} ……\text{LCM}=(x+3)^{2}$

$\Rightarrow\frac{3x+2}{(x+3)^{2}}=\frac{A(x+3)+B}{(x+3)^{2}}$

$\Rightarrow A(x+3)+B=3x+2$

$\Rightarrow Ax+3A+B=3x+2$

$\Rightarrow Ax=3x~and~3A+B=2\Rightarrow A=3~and~B=2-3A\Rightarrow B=-7$.

$\text{Therefore,}\frac{3x+2}{x^{2}+6x+9}=\frac{A}{x+3}+\frac{B}{(x+3)^{2}}=\frac{3}{x+3}+\frac{-7}{(x+3)^{2}}$.

Method 2: Quadratic Factor

$\text{When the factors describing denominator are quadratic (i.e. of the form}~ax^{2}+bx+c,~a\neq0\text{)}$

$\bullet\text{ Assume the partial fraction as} \frac{\text{linear }}{factor}=\frac{Ax+B}{ax^{2}+bx+c}$

$\text{i. Different factors of the form}~\frac{Q(x)}{ax^{2}+bx+c}$

$\Rightarrow\text{ assume:}~\frac{Q(x)}{(ax^{2}+bx+c)(px^{2}+rx+s)}=\frac{Ax+B}{ax^{2}+bx+c}+\frac{Cx+D}{px^{2}+rx+s}+\cdots$

$\text{ii. Factors with multiplicity,}~\frac{Q(x)}{(ax^{2}+bx+c)^{n}},~n\in\mathbb{N}$

$\Rightarrow\text{ assume:}~\frac{Q(x)}{(ax^{2}+bx+c)^{n}}=\frac{Ax+B}{ax^{2}+bx+c}+\frac{Cx+D}{(ax^{2}+bx+c)^{2}}+\cdots+\frac{Ex+F}{(ax^{2}+bx+c)^{n}}$

Example 8:

Decompose the following

$\text{a.} \frac{x^{2}+2x+1}{(x^2+1)(x^2+2)}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+2}=\frac{(Ax+B)(x+2)+(Cx+D)(x+1)}{(x^2+1)(x^2+2)}$

$\Rightarrow\frac{x^{2}+2x+1}{(x^2+1)(x^2+2)}=\frac{Ax^{3}+2Ax+Bx^{2}+2B+Cx^{3}+Cx+Dx^{2}+D}{(x^2+1)(x^2+2)}$

$\Rightarrow x^{2}+2x+1=Ax^{3}+2Ax+Bx^{2}+2B+Cx^{3}+Cx+Dx^{2}+D$

$\Rightarrow x^{2}=Ax^{2}+Bx^{2}+Dx^{2},~2x=2Ax+Cx,~1=2B+D~\text{and}~0=Ax^{3}+Cx^{3}$

$\Rightarrow x^{2}=x^{2}(A+B+D),~2x=x(2A+C),~1=2B+D~\text{and}~0=A+C$

$\Rightarrow1=A+B+D,~2=2A+C,~1=2B+D~\text{and}~A=-C$

$\Rightarrow2A-A=2\Rightarrow A=2~\text{and}~C=-2$

$\Rightarrow1=2+B+D~\text{and}~1=2B+D\Rightarrow B=2~\text{and}~D=-3$

$\text{Therefore}$

$\frac{x^{2}+2x+1}{(x^2+1)(x^2+2)}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+2}=\frac{2x+2}{x^2+1}-\frac{2x+3}{x^2+2}$.

$\text{b.} \frac{5x^{3}+3x^{2}+1}{(x^{2}+2)^{2}}=\frac{Ax+B}{x^{2}+2}+\frac{Cx+D}{(x^{2}+2)^{2}}=\frac{(Ax+B)(x^{2}+2)+Cx+D}{(x^{2}+2)^{2}}$

$\Rightarrow5x^{3}+3x^{2}+1=(Ax+B)(x^{2}+2)+Cx+D$

$\Rightarrow5x^{3}+3x^{2}+1=Ax^{3}+Bx^{2}+2Ax+Cx+2B+D$

$\Rightarrow5x^{3}=Ax^{3},~3x^{2}=Bx^{2},~0=2Ax+Cx~\text{and}~1=2B+D$

$\Rightarrow A=5,~B=3,~C=-10~\text{and}~D=-5$

$\text{Therefore},$

$\frac{5x^{3}+3x^{2}+1}{(x^{2}+2)^{2}}=\frac{Ax+B}{x^{2}+2}+\frac{Cx+D}{(x^{2}+2)^{2}}=\frac{5x+3}{x^{2}+2}-\frac{10x+5}{(x^{2}+2)^{2}}$

$\text{c.} \frac{x^{2}}{(1-x)(x^{2}+1)}=\frac{A}{1-x}+\frac{Bx+C}{x^{2}+1}=\frac{A(x^{2}+1)+(Bx+C)(1-x)}{(1-x)(x^{2}+1)}$

$\Rightarrow x^{2}=A(x^{2}+1)+(Bx+C)(1-x)$

$\Rightarrow x^{2}=Ax^{2}-Bx^{2}+Bx-Cx+A+C$

$\Rightarrow x^{2}=Ax^{2}-Bx^{2},~0=Bx-Cx~\text{and}~0=A+C$

$\Rightarrow1=A-B,~B=C~\text{and}~A=-C$

$\Rightarrow A=\frac{1}{2},~B=\frac{1}{2}~\text{and}~C=-\frac{1}{2}$

$\text{Therefore},$

$\frac{x^{2}}{(1-x)(x^{2}+1)}=\frac{A}{1-x}+\frac{Bx+C}{x^{2}+1}=\frac{\frac{1}{2}}{1-x}+\frac{\frac{x}{2}-\frac{1}{2}}{x^{2}+1}=\frac{1}{2(1-x)}+\frac{x-1}{2(x^{2}+1)}$