Theory & Concepts

SSC CGL Polynomials and Factorization Questions & Tricks

Get comprehensive theory, expert shortcuts, and hand-picked practice questions for Polynomials & Factorization specifically designed for the SSC CGL 2025-26 pattern.

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25 min readDifficulty: Intermediate

Polynomials in SSC CGL often look terrifyingly complex, but they are designed to be solved in under 30 seconds. By mastering the Remainder Theorem and the "Value Putting" shortcut, you can instantly bypass pages of algebra and jump straight to the answer.

Learning path

The Value Putting Trick
The Remainder Theorem
The Factor Theorem
10 Exam-Level Examples

1. The Value Putting Shortcut

If an algebraic question has variables in the problem but only numbers in the options, the answer is independent of the variables. You can substitute any value you want!

Substitution Rules

To instantly solve complex expressions:

1. Set Variables to 0 or 1: This immediately eliminates massive chunks of the equation.

2. Avoid Zero Denominators: Never pick a value that makes any denominator equal to 0 (Infinity is invalid).

3. Symmetry: If the equation is perfectly symmetrical (e.g., $x^2+y^2+z^2 = xy+yz+zx$ ), simply assume $x = y = z$ .

2. Remainder & Factor Theorems

Instead of doing long polynomial division, use these exact substitution theorems to instantly find remainders or missing variables.

The Remainder Theorem

If a polynomial $P(x)$ is divided by $(x - a)$ , the remainder is exactly $P(a)$ .

Just set the divisor to 0, find $x$ , and plug it into the main equation!

The Factor Theorem

If $(x - a)$ is a perfect factor of $P(x)$ , it leaves NO remainder.

Therefore, $P(a) = 0$ . Set the equation to 0 to find missing coefficients.

3. The 4th Power Master Identity

SSC Tier 2 exams frequently feature this highly specific factorization. Memorize it to save crucial minutes.

The Split Structure

x^4 + x^2y^2 + y^4 = (x^2 - xy + y^2)(x^2 + xy + y^2)

They will always give you the value of the main expression and ONE of the factors, asking you to find $xy$ .

4. 10 Solved examples

Question 01Exam Pattern

Find the remainder when $ x^3 - 3x^2 + 4x - 5 $ is divided by $ (x - 2) $.

-1

-3

Correct answer: a) -1

Solution

Step 1: Set the divisor to zero: $ x - 2 = 0 \implies x = 2 $.

Step 2: By Remainder Theorem, the remainder is $ P(2) $.

Step 3: Substitute $ x=2 $ into the equation: $ (2)^3 - 3(2)^2 + 4(2) - 5 $.

Step 4: Evaluate: $ 8 - 12 + 8 - 5 $.

Final calculation: $ 16 - 17 = -1 $.

Question 02Exam Pattern

If $ (x+1) $ is a factor of $ 2x^3 + ax^2 + 2x - 1 $, find the value of $ a $.

Correct answer: c) 5

Solution

Step 1: Since it is a factor, $ P(-1) = 0 $.

Step 2: Substitute $ x = -1 $: $ 2(-1)^3 + a(-1)^2 + 2(-1) - 1 = 0 $.

Step 3: Evaluate: $ -2 + a - 2 - 1 = 0 $.

Step 4: Combine constants: $ a - 5 = 0 $.

Final calculation: $ a = 5 $.

Question 03Exam Pattern

What must be added to $ 4x^2 - 12x $ to make it a perfect square?

Correct answer: a) 9

Solution

Step 1: Compare to the identity $ (a - b)^2 = a^2 - 2ab + b^2 $.

Step 2: Here, $ a^2 = 4x^2 \implies a = 2x $.

Step 3: The middle term is $ -2ab = -12x \implies -2(2x)(b) = -12x $.

Step 4: Solve for $ b $: $ -4xb = -12x \implies b = 3 $.

Final calculation: The missing term is $ b^2 = 3^2 = 9 $.

Question 04Exam Pattern

If $ x^4 + x^2y^2 + y^4 = 21 $ and $ x^2 - xy + y^2 = 3 $, find the value of $ xy $.

Correct answer: b) 2

Solution

Step 1: Use the identity $ x^4 + x^2y^2 + y^4 = (x^2 - xy + y^2)(x^2 + xy + y^2) $.

Step 2: Substitute the knowns: $ 21 = (3)(x^2 + xy + y^2) $.

Step 3: Solve for the other factor: $ x^2 + xy + y^2 = 7 $.

Step 4: You now have two equations: $ x^2 + xy + y^2 = 7 $ and $ x^2 - xy + y^2 = 3 $.

Final calculation: Subtract the second from the first: $ 2xy = 4 \implies xy = 2 $.

Question 05Exam Pattern

If $ x^2 + y^2 + z^2 = xy + yz + zx $, find the value of $ \frac{x+y}{z} $.

Correct answer: b) 2

Solution

Step 1: The condition $ x^2 + y^2 + z^2 = xy + yz + zx $ is only possible if $ x = y = z $.

Step 2: This is a perfect symmetry scenario. Substitute 1 for all variables.

Step 3: $ x = 1, y = 1, z = 1 $.

Final calculation: $ \frac{1 + 1}{1} = \frac{2}{1} = 2 $.

Question 06Exam Pattern

Simplify: $ a(b-c) + b(c-a) + c(a-b) $.

abc

a+b+c

Correct answer: d) 0

Solution

Step 1: Use the Value Putting method. Since options are constants, assume $ a=1, b=2, c=3 $.

Step 2: Substitute: $ 1(2-3) + 2(3-1) + 3(1-2) $.

Step 3: Evaluate: $ 1(-1) + 2(2) + 3(-1) $.

Final calculation: $ -1 + 4 - 3 = 0 $. Alternatively, just expanding brackets cancels everything perfectly to 0.

Question 07Exam Pattern

Find the value of $ \frac{(a-b)^2}{(b-c)(c-a)} + \frac{(b-c)^2}{(a-b)(c-a)} + \frac{(c-a)^2}{(a-b)(b-c)} $.

abc

Correct answer: c) 3

Solution

Step 1: Take the LCM of the denominators: $ (a-b)(b-c)(c-a) $.

Step 2: The numerator becomes $ (a-b)^3 + (b-c)^3 + (c-a)^3 $.

Step 3: Let $ X=a-b $, $ Y=b-c $, $ Z=c-a $. Note that $ X+Y+Z = 0 $.

Step 4: Since $ X+Y+Z=0 $, $ X^3+Y^3+Z^3 = 3XYZ $. Numerator is $ 3(a-b)(b-c)(c-a) $.

Final calculation: Divide by denominator: $ \frac{3(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)} = 3 $.

Question 08Exam Pattern

If $ a+b+c = 10 $, find the value of $ \frac{a^3+b^3+c^3-3abc}{(a-b)^2+(b-c)^2+(c-a)^2} $.

Correct answer: b) 5

Solution

Step 1: Use the modified 3-variable identity for the numerator.

Step 2: Num = $ \frac{1}{2}(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2] $.

Step 3: Notice that the entire bracket perfectly matches the denominator.

Step 4: The brackets cancel out, leaving just $ \frac{1}{2}(a+b+c) $.

Final calculation: Since $ a+b+c = 10 $, the value is $ \frac{1}{2} \times 10 = 5 $.

Question 09Exam Pattern

Factorize: $ x^4 + 4 $.

$ (x^2+2x+2)(x^2-2x+2) $

$ (x^2+2)^2 $

$ (x^2+2)(x^2-2) $

$ x^2(x^2+4) $

Correct answer: a) $ (x^2+2x+2)(x^2-2x+2) $

Solution

Step 1: Add and subtract $ 4x^2 $ to create a perfect square. $ x^4 + 4x^2 + 4 - 4x^2 $.

Step 2: The first three terms form $ (x^2 + 2)^2 $.

Step 3: The expression is now $ (x^2 + 2)^2 - (2x)^2 $.

Step 4: Use $ A^2 - B^2 = (A+B)(A-B) $.

Final calculation: $ (x^2 + 2x + 2)(x^2 - 2x + 2) $.

Question 010Exam Pattern

Simplify $ \frac{x^2 - y^2}{x-y} $ evaluated at $ x=5 $ and $ y=3 $.

Correct answer: c) 8

Solution

Step 1: Expand the numerator using difference of squares: $ x^2 - y^2 = (x+y)(x-y) $.

Step 2: The expression is $ \frac{(x+y)(x-y)}{x-y} $.

Step 3: The $ (x-y) $ terms cancel out.

Step 4: You are left with $ x + y $.

Final calculation: Substitute $ x=5, y=3 $. $ 5 + 3 = 8 $.

SSC CGL Polynomials and Factorization Questions & Tricks

Learning path

1. The Value Putting Shortcut

2. Remainder & Factor Theorems

The Remainder Theorem

The Factor Theorem

3. The 4th Power Master Identity

The Split Structure

4. 10 Solved examples

Find the remainder when \( x^3 - 3x^2 + 4x - 5 \) is divided by \( (x - 2) \).

If \( (x+1) \) is a factor of \( 2x^3 + ax^2 + 2x - 1 \), find the value of \( a \).

What must be added to \( 4x^2 - 12x \) to make it a perfect square?

If \( x^4 + x^2y^2 + y^4 = 21 \) and \( x^2 - xy + y^2 = 3 \), find the value of \( xy \).

If \( x^2 + y^2 + z^2 = xy + yz + zx \), find the value of \( \frac{x+y}{z} \).

Simplify: \( a(b-c) + b(c-a) + c(a-b) \).

Find the value of \( \frac{(a-b)^2}{(b-c)(c-a)} + \frac{(b-c)^2}{(a-b)(c-a)} + \frac{(c-a)^2}{(a-b)(b-c)} \).

If \( a+b+c = 10 \), find the value of \( \frac{a^3+b^3+c^3-3abc}{(a-b)^2+(b-c)^2+(c-a)^2} \).

Factorize: \( x^4 + 4 \).

Simplify \( \frac{x^2 - y^2}{x-y} \) evaluated at \( x=5 \) and \( y=3 \).