factorise
(i) `2x^(3)-3x^(2)-17x+30` (ii) `x^(3) -6x^(2)+11x-6`
(iii) `x^(3)+x^(2)-4x-4` (iv) `3x^(2)-x^(2)-3x+1`

Explanation

(i) Let `p(x) =2x^(3)-3x^(3)-17x+30`
Constant term of p(x)=30
`therefore "Factors of 30 are " +-1,+-2,+-3,+_5,+-6,+_10,+_15,+_30`
By trial , we we find that p(2) =0,So (x-2) is a factor of p(x).
`[:' 2(2)^(3)-3(2)^(2)-17(2)+30=16-12-34+30=0]`
Now, we see that `2x^(3)-3x^(2)-17x+30`
`2x^(3)-4x^(2)+x^(2)-2x-15x+30`
`=2x^(3)(x-2)+x(x-2)-15(x-2)`
`=(x-2)(2x^(2)+x-15)` [taking (x-2) common factor]
Now`(2x^(2)+x-15)` can be factorised either by splitting the middle term or by using the factor theorem.
Now, `(2x^(2)+x-15)=2x^(2)+6x-5x-15` [by splitting the middle term]
`=2x(x+3)-5(x+3) `
`=(x+3)(2x-5)`
(ii) let p(x) `=x^(3)-6x^(2)+11-6`
constant term of `p(x) =-6`
Factor of `-6` are `+-1,+_2,+-3,+-6,`
by trial , we find that p(1) =0 ,So (x-1) is a factor of p(x).
`[:' (1)^(3)-6(1)^(2)+11(1)-6+11-6=0]`
Now , we see that `x^(3) -6x^(2)+11x-6`
`=x^(3)-x^(2)-5x^(2)+5x+6x-6`
`=x^(2)(x-1)-5xx-1)+6x-)`
`=(x-1)(x^(2)-5x+6)`[taking (x-1) common factor ]
now , `(x^(2)-5x+6)=x^(2)-3x-2x+6` [by splitting the middle trem]
`=x(x-3)-2(x-3)`
`=(x-3)(x-2)`
` x(x-3)(x-2)`
`therefore x^(3) -6x^(2) +11x-6(x-1)(x-2)(x-3)`
` (iii) Let p(x) x^(3)+x^(2)-4x-4`
Constant term of P(x) =-4
Factors of -4 are `+-1,+_2,,+4,`
By trial ,We find that f P(-1) =0 ,so (x+1) is a factor of p(x).
Now , we see that `x^(3)+x^(2)-4x-4`
`=x^(2)(x+1) -4(x+1)`
`=(x+1)(x^(2)-4)` [taking (x+1) comm factor ]
now `x^(2)-4=x^(2)-2^(2)`
`-(x+2)(x-2) ` [using identilty , `a^(2)-b^(2)=(a-b)(a+b)]`
` therefore x^(3)+x^(2)-4x-4=(x+1)(x-2)(x+2)`
(iii) let `p(x) =3x^(3)-x^(2)-3x+1`
constant term of p(x)=1
Factor of 1 are `+_` 1.
By trial , we find that `p(1) =0, SO (x-1) is a factor of p(x).
`Now , we see that `3x^(3)-x^(2)-3x+1`
`=3x^(3)-3x^(2)+2x^(2)-2x-x+1`
`=3x^(2)(x-1)+2x(x-1)-1(x-1)`
`=(x-1)(3x^(2)+2x-1)`
Now ,`(3x^(2)+2x-1) =3x^(2)+3x-x-1`[by splitting middle term]
`=3x(x+1)-1(x+1)=(x+1)(3x-1)`
`therefore 3x^(3)-x^(2) - 3x+1=(x-1)(x-1)(x+1)(3x-1)`