HOW TO FACTOR POLYNOMIALS WITH 4 TERMS WITHOUT GROUPING

Factor the following polynomials without grouping :

Example 1 :

x³ - 2x² - x + 2

Solution :

Let p(x) = x³ - 2x² - x + 2.

Substitute x = -1.

p(-1) = (-1)³ - 2(-1)² - (-1) + 2

= -1 - 2(1) + 1 + 2

= -1 - 2 + 1 + 2

= 0

Since P(-1) = 0, (x + 1) is a factor of P(x).

Since P(x) is a cubic polynomial, the other factor can be assumed as (x² + ax + b).

Then,

(x + 1)((x² + ax + b) = x³ - 2x² - x + 2

Comparing the coefficients of x and constants,

Substitute b = 2 in (1).

2 + a = -1

a = -3

x² + ax + b = x² - 3x + 2

Factors of (x² - 3x + 2) are (x - 1) and (x - 2).

Therefore,

x³ - 2x² - x + 2 = (x + 1)(x - 1)(x - 2)

Example 2 :

x³ + 4x² + x - 6

Solution :

Let p(x) = x³ + 4x² + x - 6.

Substitute x = -1.

p(1) = (-1)³ + 4(-1)² + (-1) - 6

= -1 + 4(1) - 1 - 6

= -1 + 4 - 1 - 6

= -4 ≠ 0

Substitute x = 1.

p(1) = 1³ + 4(1)² + 1 - 6

= 1 + 4 + 1 - 6

= 0

Since P(1) = 0, (x - 1) is a factor of P(x).

Since P(x) is a cubic polynomial, the other factor can be assumed as (x² + ax + b).

Then,

(x - 1)((x² + ax + b) = x³ + 4x² + x - 6

Comparing the coefficients of x and constants,

Substitute b = 6 in (1).

6 - a = 1

-a = -5

a = 5

x² + ax + b = x² + 5x + 6

Factors of (x² + 5x + 6) are (x + 2) and (x + 3).

Therefore,

x³ + 4x² + x - 6 = (x - 1)(x + 2)(x + 3)

Example 3 :

x³ + 3x² - 4x - 12

Solution :

Let p(x) = x³ + 3x² - 4x - 12.

Substitute x = -1.

p(-1) = (-1)³ + 3(-1)² - 4(-1) - 12

= -1 + 3(1) + 4 - 12

= -1 + 3 + 4 - 12

= -6 ≠ 0

Since P(-1) ≠ 0, (x + 1) is not a factor of P(x).

Substitute x = 1.

p(1) = 1³ + 3(1)² - 4(1) - 12

= 1 + 3 - 4 - 12

= -12 ≠ 0

Substitute x = -2.

p(-2) = (-2)³ + 3(-2)² - 4(-2) - 12

= -8 + 3(4) + 8 - 12

= -8 + 12 + 8 - 12

= 0

Since P(-2) = 0, (x + 2) is a factor of P(x).

Since P(x) is a cubic polynomial, the other factor can be assumed as (x² + ax + b).

Then,

(x + 2)((x² + ax + b) = x³ + 3x² - 4x - 12

Comparing the coefficients of x and constants,

b + 2a = -4 ----(1)2b = -12b = -6

Substitute b = -6 in (1).

-6 + 2a = -4

2a = 2

a = 1

x² + ax + b = x² + x - 6

Factors of (x² + x - 6) are (x - 2) and (x + 3).

Therefore,

x³ + 3x² - 4x - 12 = (x + 2)(x - 2)(x + 3)

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