FACTORIZATION

CONTENT

1. Factorizing algebraic expressions
2. Factorizing expressions with a common factor bracket and by grouping
3. Special cases of factorization
4. Using factorization to simplify expressions and coefficient of terms
5. Word Problems Involving Factorization

REMOVING BRACKETS (REVISION)

Example 1

Remove brackets from

(a) \(3(2u – v)\)

(b) \((3a+8b)5a\)

(c) \(-2n(7y – 4z)\)

Solutions

(a) \(3(2u – v) = 3 × 2u – 3 × v \\ = 6u – 3v\)

(b) \((3a + 8b)5a =3a × 5a + 8b × 5a \\ =15a^2 + 40ab\)

(c) \(-2n(7y -4z) =(-2n) × 7y -(-2n) × 4z \\ = -14ny + 8ny\)

FACTORIZATION BY TAKING COMMON FACTORS

To factorize an expression is to write it as a product of its factors

Example

Factorize the following:

(a) \(9a – 3z\)

(b) \(5x^2 + 15x\)

(c) Factorize \(2x(5a + 2) -3y(5a + 2)\)

Solution:

(a) The HCF of 9a and 3z is 3

\(9a -3z = 3\Big(\frac{9a}{3} -\frac{3z}{3} \Big) \\ = 3(3a -z)\)

(b) The HCF of 5x² and 15x is 5x

\(5x^2 + 15x = 5x\Big(\frac{5x^2}{5x} + \frac{15x}{5x} \Big) \\ = 5x(x + 3)\)

(c) Factorize \(2x(5a + 2) -3y(5a + 2)\)

In the given expression,

\(2x(5a + 2) = 2x \text{ }times\text{ } (5a + 2) \\ 3y(5a + 2) = 3y \text{ }times\text{ } (5a + 2)\)

Hence the products \(2x(5a + 2) \text{ and }3y(5a + 2)\)

have the factor \((5a + 2)\) in common.