Question

Factorize `x^3-7x+6`?

Answer 1

Factors of `x^3-7x+6=0` are `(x-1)(x+2)(x-3)=0`

Explanation:

As the coefficients (`1-7+6=0`) in the given equation add up to zero, it is apparent that `x=1` is a solution of `x^3-7x+6=0` and `(x-1)` is a factor of `x^3-7x+6`.

Dividing `x^3-7x+6` by `(x-1)`

`x^3-7x+6=x^2(x-1)+x(x-1)-6(x-1)`

= `(x-1)(x^2+x-6)` - now splitting the middle term

= `(x-1)(x^2+2x-3x-6)`

= `(x-1)(x(x+2)-3(x+2))`

= `(x-1)(x+2)(x-3)`

Hence factors of `x^3-7x+6=0` are `(x-1)(x+2)(x-3)=0`

Answer 2

The answer is `=(x-1)(x+3)(x-2)`

Explanation:

Let `f(x)=x^3-7x+6`

`f(1)=1-7+6=0`

Therefore,

`(x-1)` is a factor

To find the other factors, we do a long division

`color(white)(aaaa)` `x^3` `color(white)(aaaaa)` `-7x+6` `color(white)(aaaaaa)` `∣` `x-1`

`color(white)(aaaa)` `x^3-x^2` `color(white)(aaaa)` ###color(white)(aaaaaaaaaa)#`∣` `x^2+x-6`

`color(white)(aaaaa)` `0+x^2-7x`

`color(white)(aaaaaaa)` `+x^2-x`

`color(white)(aaaaaaaa)` `+0-6x+6`

`color(white)(aaaaaaaaaaaa)` `-6x+6`

`color(white)(aaaaaaaaaaaaa)` `-0+0`

Therefore,

`(x^3-7x+6)/(x-1)=x^2+x-6=(x+3)(x-2)`

So,

`(x^3-7x+6)=(x-1)(x+3)(x-2)`