Factoring Formulas in Algebra
Factoring Formulas in Algebra
Factoring formulas in algebra are special algebraic identities used to break down complex polynomials into the product of two or more simpler expressions. Factoring is the exact opposite of expanding. Mastering these 10 special factoring formulas is vital for CBSE Class 8 to 10 board exams and competitive exams like JEE.
Topic: Algebra & Polynomials
Exams: CBSE · ICSE · JEE Foundation
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What are Special Factoring Formulas?
We use standard algebraic identities as special factoring formulas. Instead of expanding two brackets, factoring requires us to look at an expanded polynomial and shrink it back into its multiplied factors. These formulas can be verified by solving the Left-Hand Side (LHS) and Right-Hand Side (RHS) of the expression.
1. Basic Quadratic Factoring Formulas
These are the foundational identities taught in CBSE Class 8. They deal with powers of 2 (squares).
2. Cubic Factoring Formulas
Introduced in Class 9, these formulas handle polynomials with a power of 3. Students often confuse the perfect cube formulas with the sum/difference of cubes.
a3 + b3 = (a + b)(a2 − ab + b2)
a3 − b3 = (a − b)(a2 + ab + b2)
a3 + b3 + 3ab(a + b) = (a + b)3
a3 − b3 − 3ab(a − b) = (a − b)3
3. 3-Variable Factoring Formulas (Class 9 CBSE)
For complex polynomials containing variables x, y, and z (or a, b, c), these standard identities are essential for advanced factorization.
a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2
x3 + y3 + z3 − 3xyz = (x + y + z)(x2 + y2 + z2 − xy − yz − xz)Note: If x + y + z = 0, then x3 + y3 + z3 = 3xyz.
List of All 10 Special Factoring Formulas
| Formula No. | Algebraic Identity (Expanded to Factored) |
|---|---|
| Formula 1 | a2 + 2ab + b2 = (a + b)2 |
| Formula 2 | a2 − 2ab + b2 = (a − b)2 |
| Formula 3 | a2 − b2 = (a + b)(a − b) |
| Formula 4 | x2 + (a + b)x + ab = (x + a)(x + b) |
| Formula 5 | a3 + b3 + 3ab(a + b) = (a + b)3 |
| Formula 6 | a3 − b3 − 3ab(a − b) = (a − b)3 |
| Formula 7 | a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2 |
| Formula 8 | x3 + y3 + z3 − 3xyz = (x + y + z)(x2 + y2 + z2 − xy − yz − xz) |
| Formula 9 | x3 + y3 = (x + y)(x2 − xy + y2) |
| Formula 10 | x3 − y3 = (x − y)(x2 + xy + y2) |
Examples Using Factoring Formulas
Example 1: Factoring Sum of Cubes
Factorize the expression: 8x3 + 27
We can rewrite the given expression as perfect cubes: (2x)3 + (3)3.
Now substitute a = 2x and b = 3 into the formula a3 + b3 = (a + b)(a2 − ab + b2).= (2x + 3) [ (2x)2 − (2x)(3) + (3)2 ]
= (2x + 3)(4x2 − 6x + 9)
Answer: 8x3 + 27 = (2x + 3)(4x2 − 6x + 9)
Example 2: Perfect Square Trinomial
Factorize the algebraic expression: x2 + 4xy + 4y2
We can write the given expression as: (x)2 + 2(x)(2y) + (2y)2.
Using the special formula a2 + 2ab + b2 = (a + b)2.
Substitute a = x and b = 2y into this formula:(x)2 + 2(x)(2y) + (2y)2 = (x + 2y)2
Answer: x2 + 4xy + 4y2 = (x + 2y)2
Example 3: Difference Perfect Square
Factorize: x2 − 6x + 9
We have, x2 − 6x + 9 = x2 − 2(3)(x) + 32.
Using the factoring formula a2 − 2ab + b2 = (a − b)2, where a = x and b = 3:x2 − 2(3)(x) + 32 = (x − 3)2
Answer: x2 − 6x + 9 = (x − 3)2