Factorise: (i) `x^3-2x^2-x+2` (ii)`x^3-3x^2-9x-5` (iii)`x^3+13 x^2+32 x+20` (iv) `2y^3+y^2-2y-1`
`x^3+2x^2-x-2`
`= x(x^2 – 1) – 2(x^2 – 1)`
`x^3 – 2x^2 – x + 2 = (x^2 – 1) (x – 2)`
Using the identity:` a^2 − b^2 = (a + b) (a- b)`
`∴ (x^2 – 1) = (x + 1)(x – 1)`
`= (x – 1) (x + 1) (x- 2)`
ii)`x^3-3x^2-9x-5`
`x(x^2+2x+1)-5((x^2+2x+1)`
`(x-5)(x^2+2x+1)`
The value of `x^3+13x^2+32x+20 is (x^3+x^2+12x^2+12x+20x+20)`
`=>x^2(x+1)+12x(x+1)+20(x+1)`
`=>(x−1)(x^2+12x+20)`
`=>(x−1)(x^2+10x+2x+20)`
`=>(x+1)(x(x+10)+2(x+10))`
`=>(x+1)(x+2)(x+10)`
iv)Now Factorizing,`2y³ + y² – 2y – 1`
`= y² (2y + 1) -1(2y+1)`
`= (2y + 1)(y² – 1)`
`= (2y + 1)(y² – 1²)`
`(y² – 1²) = (y + 1)(y – 1)`
`(2y + 1)(y² – 1²)`
` (2y + 1) (y + 1) (y – 1)`

Explanation

`x^3+2x^2-x-2`
`= x(x^2 – 1) – 2(x^2 – 1)`
`x^3 – 2x^2 – x + 2 = (x^2 – 1) (x – 2)`
Using the identity:` a^2 − b^2 = (a + b) (a- b)`
`(x^2 – 1) = (x + 1)(x – 1)`
`= (x – 1) (x + 1) (x- 2)`

ii)`x^3-3x^2-9x-5`
`x(x^2+2x+1)-5((x^2+2x+1)`
`(x-5)(x^2+2x+1)`
`(x-5)(x+1)(x+1)`

iii)The value of `x^3+13x^2+32x+20 is (x^3+x^2+12x^2+12x+20x+20)`
`=>x^2(x+1)+12x(x+1)+20(x+1)`
`=>(x−1)(x^2+12x+20)`
`=>(x−1)(x^2+10x+2x+20)`
`=>(x+1)(x(x+10)+2(x+10))`
`=>(x+1)(x+2)(x+10)`

iv)Now Factorizing,`2y³ + y² – 2y – 1`
`= y² (2y + 1) -1(2y+1)`
`= (2y + 1)(y² – 1)`
`= (2y + 1)(y² – 1²)`
`(y² – 1²) = (y + 1)(y – 1)`
`(2y + 1)(y² – 1²)`
` (2y + 1) (y + 1) (y – 1)`