CHAPTER 12 – FACTORIZATION

Introduction

What is Factorization?

Imagine you have a number, say 12. You can break it down into smaller numbers that multiply to give 12:

\(12 = 2 \times 6\) or \(12 = 3 \times 4\) or \(12 = 2 \times 2 \times 3\).

This process of breaking a number into its multiplicative building blocks is called factorization.

Factorization in algebra is the exact same idea, but with algebraic expressions instead of numbers. It is the reverse process of multiplication.

The Key Idea: Factors

When we multiply expressions we get a product. The expressions we multiply are called the factors of that product.

Example:

\((x + 2)\) and \((x + 3)\) are factors because

\((x + 2)(x + 3) = x^2 + 5x + 6\).

So factorization is doing this backwards. If you are given \(x^2 + 5x + 6\), the job is to find its factors: \((x + 2)(x + 3)\).

\(x^2 + 5x + 6 = (x + 2)(x + 3)\)

Why is Factorization Useful?
  • Simplifying calculations. Recognizing patterns can make large computations trivial. For example, \((999)^2 + 2(999)(1) + 1^2\) is \((999+1)^2 = 1000^2.\)
  • Solving equations. To solve \(x^2 + 5x + 6 = 0\) we factorize it as \((x+2)(x+3)=0\) and read off solutions \(x = -2, -3\).
  • Simplifying algebraic fractions. For example, \(\dfrac{x^2-4}{x-2}\) becomes \(\dfrac{(x-2)(x+2)}{x-2} = x+2\) after canceling the common factor.

Main Methods of Factorization

a) Factorization by Common Factor

This is the easiest method. Look for a term (number, variable, or both) common to every term of the expression.

Example: Factorize \(2y + 4\)

Both terms are divisible by 2, so \(2y + 4 = 2(y + 2)\).

b) Factorization by Regrouping Terms

When there is no single common factor, try grouping terms to create common factors.

Example: Factorize \(xy + y + 2x + 2\).

Group terms: \((xy + y) + (2x + 2)\).

From the first group factor out \(y\): \(y(x + 1)\).

From the second group factor out \(2\): \(2(x + 1)\).

Now both groups contain \((x + 1)\). Factor it out:

\((x + 1)(y + 2)\).

c) Factorization using Identities

Use algebraic identities from the previous chapter. The most useful are:

  • \((a + b)^2 = a^2 + 2ab + b^2\)
  • \((a - b)^2 = a^2 - 2ab + b^2\)
  • \((a + b)(a - b) = a^2 - b^2\)

Example 1: Factorize \(x^2 - 9\).

This matches \(a^2 - b^2\) with \(a = x\) and \(b = 3\). So

\(x^2 - 9 = (x + 3)(x - 3)\).

Example 2: Factorize \(4p^2 - 12p + 9\).

This matches \(a^2 - 2ab + b^2\) with \(a = 2p\) and \(b = 3\). Check the middle term: \(2ab = 2 \cdot 2p \cdot 3 = 12p\). So

\(4p^2 - 12p + 9 = (2p - 3)^2\).

Chapter Roadmap

What You Will Practice
  • Finding common factors and taking them out
  • Regrouping and factoring by grouping
  • Applying standard identities for quick factorization
  • Factorizing simple and quadratic polynomials
  • Solving equations using factorization and simplifying algebraic fractions

Tip: Always check for a common factor first. It saves time and prevents mistakes.

Quick check: After factorizing, multiply the factors to verify you get the original expression.

Designed & Developed by Zahid Qayoom