FACTOR POLYNOMIALS BY GROUPING

Problem 2 :

x⁵ - x⁴ - 2x³ + 2x²

Solution :

= x⁵ - x⁴ - 2x³ + 2x²

Factoring x²

= x²(x³ - x² - 2x + 2)

= x²[x²(x - 1) - 2(x - 1)]

= x²(x - 1) (x² - 2)

So, factors are x²(x - 1) (x² - 2).

Problem 3 :

x⁴ - 3x³ - 5x² + 15x

Solution :

= x⁴ - 3x³ - 5x² + 15x

= x(x³ - 3x² - 5x + 15)

= x[x²(x - 3) - 5(x - 3)]

= x(x² - 5) (x - 3) 

So, factors are x(x² - 5) (x - 3).

Problem 4 :

x² (x - 1) - 9(x - 1)

Solution :

 x² (x - 1) - 9(x - 1)

 (x - 1) (x² - 9)

= (x - 1) (x² - 3²)

Using algebraic identity,

a² - b² = (a + b) (a - b)

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

So, factors are (x - 1) (x + 3) (x - 3).

Problem 5 :

x³ - 2x² - 16x + 32

Solution :

= x³ - 2x² - 16x + 32

= (x³ - 2x²) + (-16x + 32)

= x²(x - 2) - 16 (x - 2)

= (x - 2) (x² - 16)

= (x - 2) (x² - 4²)

Using algebraic identity,

a² - b² = (a + b) (a - b)

= (x - 2) (x + 4) (x - 4)

So, factors are (x - 2) (x + 4) (x - 4).

Problem 6 :

4x²(x - 3) - 25(x - 3)

Solution :

= 4x²(x - 3) - 25(x - 3)

= (x - 3) (4x² - 25)

= (x - 3) [(2x)² - 5²]

Using algebraic identity,

a² - b² = (a + b) (a - b)

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

So, factors are (x - 3) (2x + 5) (2x - 5).

Problem 7 :

x³ + 4x² - 36x - 144

Solution :

= x³ + 4x² - 36x - 144

= (x³ + 4x²) + (- 36x - 144)

= x²(x + 4) - 36(x + 4)

= (x + 4) (x² - 36)

= (x + 4) (x² - 6²)

Using algebraic identity,

a² - b² = (a + b) (a - b)

= (x + 4) (x + 6) (x - 6)

So, factors are (x + 4) (x + 6) (x - 6).