Factorise `x^(3)-7x+6`.

Explanation

Here, constant term=6
Factors of `6=+-1, +-2,+-3,+-6`
Let `p(x)=x^(3)-7x+6`
Put x=1
Remainder `=p(1)=1^(3)-7(1)+6=1-7+6=0`
`therefore (x-1)`, is factor of p(x).
Now, `p(x)=x^(3)-7x+6=x^(2)(x-1)+x^(2)-7x+6`
`=x^(2)(x-1)+x(x-1)-6x+6`
`=x^(2)(x-1)+x(x-1)-6(x-1)`
`=(x-1)(x^(2)+x-6)`
`=(x-1)[x^(2)+3x-2x-6]`
`=(x-1)[x(+3)-2(x+3)]`
=(x-1)(x+3)(x-2)