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.
- If we have odd number of terms, we have to take common value out from all the terms.
- If we have even number of terms, divide them into smaller groups and take common value out.
- Factoring quadratic polynomial with leading coefficient as 1
- Factoring quadratic polynomial with leading coefficient is not equal to 1
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,
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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