CDS Mathematics · Algebraic Identities and Simplification

Rational Algebraic Expressions

Factor every numerator and denominator, cancel common factors, combine over a common denominator, and clear denominators to solve.

All Algebraic Identities and Simplification notes CDS Mathematics strategy

Why this matters

Nineteen PYQs, the most of any page in the chapter. Nothing here is a trick: the work is factorising cleanly and keeping signs straight. Most answers collapse to a single term or a constant, so if yours does not, look for a factor you missed.

Concept 1 of 3: Factorise and cancel

A fraction of polynomials simplifies only through common factors, so factor everything first. The factors that appear are the ones from the identity pages: differences of squares, sums and differences of cubes, and split-middle-term quadratics.

Definition

Factor each quadratic: $x^2 - 9x + 14 = (x - 2)(x - 7)$ .
Use $a^2 - b^2$ , $a^3 \pm b^3$ , and perfect squares like $x^2 + 4\sqrt3\,x + 12 = (x + 2\sqrt3)^2$ .
Cancel only whole factors, never single terms.
With a numeric value like $x = 9999$ , simplify the algebra first and substitute last.

Cancel a common factor

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

Worked example

Simplify

\dfrac{(x^2 - 9)(x^2 + 2x - 8)}{(x^2 + x - 6)(x^2 + 7x + 12)}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2021 · CDS (I) 2021 — Elementary Mathematics · Q11Moderate

Example 1 · Algebraic Identities and Simplification · Rational Algebraic Expressions

The expression

\frac{(x^3 - 1)(x^2 - 9x + 14)}{(x^2 + x + 1)(x^2 - 8x + 7)}

simplifies to

Cancel factors, not terms

\dfrac{x^2 + 3x}{x + 3}

you may cancel

(x + 3)

after writing the top as

x(x + 3)

. You may not cross out the

3x

against the

3

Drill 5 more on factorise and cancel

Concept 2 of 3: Combining fractions

Adding fractions needs a common denominator, and the smallest one comes from the factorised denominators. Work in pairs: two of the fractions often combine into something that cancels the third.

Definition

Factor denominators, then use their least common multiple.
'What must be added to $P$ to get $Q$ ?' is $Q - P$ .
Given $A + B$ and $A - B$ : $B = \dfrac{(A + B) - (A - B)}{2}$ .
$\dfrac{x + y}{x - y} = \dfrac{2x}{x - y} - 1$ : rewriting a term this way can expose a copy of another expression.
$b - a = -(a - b)$ : flip a factor and flip the sign.

Two fractions

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

Worked example

Simplify

\dfrac{1}{x - 1} - \dfrac{1}{x + 1} - \dfrac{2}{x^2 + 1}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2021 · CDS (I) 2021 — Elementary Mathematics · Q12Moderate

Example 2 · Algebraic Identities and Simplification · Rational Algebraic Expressions

What should be added to

\frac{1}{(x - 2)(x - 4)}

to get

\frac{2x - 5}{(x^2 - 5x + 6)(x - 4)}

Order of subtraction

'What should be added to

P

to get

Q

' is

Q - P

, not

P - Q

. The reversed answer differs only in sign and is usually printed.

Drill 7 more on combining fractions

Concept 3 of 3: Equations with fractions

Clear the denominators, and the equation becomes a polynomial. Before multiplying out everything, move terms so that matching pieces sit on the same side; big terms often cancel.

Definition

Multiply through by the common denominator, then simplify.
Group first: put the fractions with $x$ in the denominator together.
A parameter that appears everywhere can be scaled out: put $x = tk$ .
Find an easy root by trial and factor it out.
Reject any root that makes an original denominator zero.

Excluded values

\dfrac{P(x)}{Q(x)} = 0 \;\Rightarrow\; P(x) = 0,\ Q(x) \ne 0

Worked example

Solve

\dfrac{1}{x - 1} + \dfrac{1}{x - 2} = \dfrac{2}{x - 3}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2026 · CDS (II) 2026 — Elementary Mathematics · Q11Moderate

Example 3 · Algebraic Identities and Simplification · Rational Algebraic Expressions

\frac{1}{x + k} + \frac{1}{x + 2k} + \frac{1}{x + 5k} = \frac{1}{k}

, then what is the solution of the equation ?

Check the denominators

A root that makes an original denominator zero is not a solution, even though it solves the cleared polynomial. And

x = 0

often satisfies a symmetric equation trivially; the question usually wants the other roots.

Drill 4 more on equations with fractions

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Factorise and cancel
Cancel a common factor
$\dfrac{(x - 1)(x^2 + x + 1)}{x^2 + x + 1} = x - 1$
Combining fractions
Two fractions
$\dfrac{1}{x - 1} - \dfrac{1}{x + 1} = \dfrac{2}{x^2 - 1}$
Equations with fractions
Excluded values
$\dfrac{P(x)}{Q(x)} = 0 \;\Rightarrow\; P(x) = 0,\ Q(x) \ne 0$

Watch out for (3)

Cancel factors, not terms
→ Factorise and cancel
Order of subtraction
→ Combining fractions
Check the denominators
→ Equations with fractions

Test yourself on Algebraic Identities and Simplification

20 past CDS questions from this chapter, timed at 24 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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