FACTORING INTO LINEAR FACTORS

To find factors of polynomial, we use the following ways.

(i) Using grouping method

(ii) Using algebraic identities

(iii) Using synthetic division

Some times we may use more than one of the methods given above.

Using Grouping Method

Taking the common values out, this method is known as grouping method. 

Using Algebraic Identities

The given question will be in the form of 

  • a2 + 2ab + b2
  • a2 - 2ab + b2
  • a2 - b2

The expansion will be in any one of the form, for example

= x2 + 10x + 25

= x2 + 2 (x) (5) + 52

= (x + 5)2

So, the factors are (x + 5)(x + 5).

Using Synthetic Division

The polynomial which is having the highest exponent of 3 or more than 3, we can use synthetic division and find factors.

To see more example problems,

Factoring polynomial using synthetic division

Factor the polynomial into linear factors.

Problem 1 :

x³ - 4x

Solution :

= x³ - 4x

Using grouping method, factoring x we get

= x(x² - 4)

x² - 4 can be written as

= x(x + 2) (x - 2)

Problem 2:

6x² - 54

Solution :

= 6x² - 54

= 6(x² - 9)

x² - 9 can be written as

= 6(x + 3) (x - 3)

Problem 3 :

4x² + 8x - 60

Solution :

= 4x² + 8x - 60

= 4(x² + 2x - 15)

= 4(x - 3) (x + 5)

Problem 4 :

15x³ - 22x² + 8x

Solution :

= 15x³ - 22x² + 8x

= x (15x² - 22x + 8)

= x (15x² - 12x - 10x + 8)

= x [3x(5x - 4) - 2(5x - 4)]

= x(3x - 2) (5x - 4)

Problem 5:

x³ + 2x² - x - 2

Solution :

= x³ + 2x² - x - 2

By grouping,

= (x³ + 2x²) + (-x - 2)

= x²(x + 2) - 1(x + 2)

= (x + 2) (x² - 1)

x² - 1 can be written as

= (x - 1) (x + 1) (x + 2)

Problem 6  :

x4 + x³ - 9x² - 9x

Solution :

= x4 + x³ - 9x² - 9x

= x(x³ + x² - 9x - 9)

By grouping,

= x(x³ - 9x) + (x² - 9)

= x [x(x² - 9) + 1(x² - 9)]

= x [(x + 1) (x² - 9)]

x² - 9 can be written as

= x(x + 1) (x - 3) (x + 3)

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