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SSC CGL Maths · Algebraic Identities and Simplification

Rational Algebraic Expressions

Factorising the top and bottom of a fraction and cancelling, joining fractions over one denominator, and using a given condition to replace a whole piece.

All Algebraic Identities and Simplification notes

Why this matters

19 questions in the bank, 13 of them EASY. The identities from the earlier pages do the factorising here: a difference of squares, a perfect square, a cube. Since the options are expressions, putting in one or two small values checks the answer in seconds.

Concept 1 of 2: Factorise, then cancel

A fraction of polynomials simplifies the way 1218\frac{12}{18} does: write the top and the bottom as products, then strike out the factors they share.

Definition

  • Look for a common factor first: 6x2−9x=3x(2x−3)6x^2 - 9x = 3x(2x - 3), and (5p)3=125p3(5p)^3 = 125p^3.
  • Then the patterns: a2−b2a^2 - b^2, a2±2ab+b2a^2 \pm 2ab + b^2, a3±b3a^3 \pm b^3, the expanded cube, and x2+(p+q)x+pq=(x+p)(x+q)x^2 + (p + q)x + pq = (x + p)(x + q).
  • A leading coefficient: split the middle term. 6a2−ab−12b2=(3a+4b)(2a−3b)6a^2 - ab - 12b^2 = (3a + 4b)(2a - 3b), since 3×(−3)+4×2=−13 \times (-3) + 4 \times 2 = -1.
  • A factorisation that is easy to miss: a4+a2b2+b4=(a2+ab+b2)(a2−ab+b2)a^4 + a^2b^2 + b^4 = (a^2 + ab + b^2)(a^2 - ab + b^2).
  • Cancel only factors, never terms joined by + or −.
  • A factor written backwards is its negative: 3y−2x=−(2x−3y)3y - 2x = -(2x - 3y), while (3y−2x)2=(2x−3y)2(3y - 2x)^2 = (2x - 3y)^2.
  • Fast route: put x=2x = 2 (or a=2,b=1a = 2, b = 1) into the question and the options. Avoid a value that makes a bottom zero.

Factorisations for cancelling

x2+(p+q)x+pq=(x+p)(x+q)a4+a2b2+b4=(a2+ab+b2)(a2−ab+b2)x^2 + (p + q)x + pq = (x + p)(x + q) \qquad a^4 + a^2b^2 + b^4 = (a^2 + ab + b^2)(a^2 - ab + b^2)

Worked example

Simplify x2−x−20x2−25\frac{x^2 - x - 20}{x^2 - 25}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 19 Sep 2025, Shift 2 · Q75Easy

Example 1 · Algebraic Identities and Simplification · Rational Algebraic Expressions

If p=0.8p = 0.8 and q=0.02q = 0.02, find the value of (4p)3+(4q)3p3+q3\frac{(4p)^3 + (4q)^3}{p^3 + q^3}.

Cancelling terms, not factors

x2+5x+5\frac{x^2 + 5}{x + 5} does not simplify to x. Only a factor of the whole top and the whole bottom can cancel, and x2+5x^2 + 5 has no factor x+5x + 5.

A backwards factor changes the sign

2x−77−2x=−1\frac{2x - 7}{7 - 2x} = -1, not 1, because 7−2x=−(2x−7)7 - 2x = -(2x - 7).
Drill 11 more on factorise, then cancel

Concept 2 of 2: Combining fractions and using a given condition

Fractions with different bottoms join over one common denominator, just as with numbers. When the question gives a condition, use it to replace a whole piece of the expression instead of solving for the letters.

Definition

  • Common denominator: ap+bq=aq+bppq\frac{a}{p} + \frac{b}{q} = \frac{aq + bp}{pq}. Multiply each top by what its bottom is missing.
  • Split a fraction: 3a−4a=3−4a\frac{3a - 4}{a} = 3 - \frac{4}{a}. This turns an equation in such fractions into one in 1a\frac{1}{a}, 1b\frac{1}{b}, 1c\frac{1}{c}.
  • Replace using the condition: if x2+9=7xx^2 + 9 = 7x, then x2−3x+9=4xx^2 - 3x + 9 = 4x. A quadratic in the question is often the given one plus a simple extra term.
  • If a+b=ka + b = k, then 1a+1b=kab\frac{1}{a} + \frac{1}{b} = \frac{k}{ab}.
  • If a+b=2ca + b = 2c, then a−c=c−ba - c = c - b: a, c, b are equally spaced.
  • Fast route for a constant answer: put in simple values that satisfy the condition (equal values, or a triple such as 1, 2, −3 when the sum must be 0) and compute.
  • Limit: the values must satisfy the condition. Values that break it prove nothing.

Adding two fractions

ap+bq=aq+bppq\frac{a}{p} + \frac{b}{q} = \frac{aq + bp}{pq}

Worked example

If x2+6=5xx^2 + 6 = 5x, find x2+x+6x2+6\frac{x^2 + x + 6}{x^2 + 6}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 25 Sep 2024, Shift 1 · Q68Easy

Example 2 · Algebraic Identities and Simplification · Rational Algebraic Expressions

If x2+7x+8=0x^2 + 7x + 8 = 0, then find the value of 4xx2−5x+8\frac{4x}{x^2 - 5x + 8}.

The minus sign before a fraction

In a2−b−34\frac{a}{2} - \frac{b - 3}{4}, the minus applies to the whole top of the second fraction: the result is 2a−b+34\frac{2a - b + 3}{4}, not 2a−b−34\frac{2a - b - 3}{4}.

Test values that break the condition

If the question says a+b+c=0a + b + c = 0, testing with a=b=c=1a = b = c = 1 is not allowed, because those values add to 3. Use values that fit, such as 1, 2 and −3.
Drill 6 more on combining fractions and using a given condition

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 (2)

  • Factorise, then cancel

    Factorisations for cancelling

    x2+(p+q)x+pq=(x+p)(x+q)a4+a2b2+b4=(a2+ab+b2)(a2−ab+b2)x^2 + (p + q)x + pq = (x + p)(x + q) \qquad a^4 + a^2b^2 + b^4 = (a^2 + ab + b^2)(a^2 - ab + b^2)
  • Combining fractions and using a given condition

    Adding two fractions

    ap+bq=aq+bppq\frac{a}{p} + \frac{b}{q} = \frac{aq + bp}{pq}

Watch out for (4)

Test yourself on Algebraic Identities and Simplification

20 past SSC CGL questions from this chapter, timed at 12 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Algebraic Identities and Simplification chapter testDrill all 19 questions