HOW TO FACTOR POLYNOMIALS WITH 4 TERMS

Factor the following polynomials by grouping :

Example 1 :

x³ - 2x² - x + 2

Solution :

= x³ - 2x² - x + 2

= (x³ - 2x²) + (-x + 2)

 = x²(x - 2) - 1(x - 2)

 =  (x² - 1)(x - 2)

 =  (x² - 1²)(x - 2)

Using algebraic identity a² - b² = (a + b)(a -b),

= (x + 1)(x - 1)(x - 2)

Example 2 :

x³ + 3x² - x - 3

Solution :

= x³ + 3x² - x - 3

= (x³ + 3x²) + (-x - 3)

= x²(x + 3) - 1(x + 3)

= (x² - 1)(x + 3)

= (x² - 1²)(x + 3)

Using algebraic identity a² - b² = (a + b)(a -b),

= (x + 1)(x - 1)(x + 3)

Example 3 :

x³ + x² - 4x - 4

Solution :

= x³ + x² - 4x - 4

= (x³ + x²) + (-4x - 4)

= x²(x + 1) - 4(x - 1)

= (x² - 4)(x + 1)

= (x² - 2²)(x + 1)

Using algebraic identity a² - b² = (a + b)(a -b),

= (x + 2)(x - 2)(x + 1)

Example 4 :

x³ - 3x² + 2x - 6

Solution :

= x³ - 3x² + 2x - 6

= (x³ - 3x²) + (2x - 6)

= x²(x - 3) + 2(x - 3)

= (x² + 2)(x - 3)

Example 5 :

x⁴ - x³ - x + x²

Solution :

= x⁴ - x³ - x + x²

Arrange the terms with powers in descending order. 

= x⁴ - x³ + x² - x

= (x⁴ - x³) + (x² - x)

= x³(x - 1) + x(x - 1)

= (x³ + x)(x - 1)

= x(x² + 1)(x - 1)

Example 6 :

5a - 5b - xa + xb

Solution :

= 5a - 5b - xa + xb

= (5a - 5b) + (-xa + xb)

= 5(a - b) - x(a - b)

= (a - b)(5 - x)

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 + 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,

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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