Question

How do you solve `4x^3-12x^2-3x+9=0`?

Answer 1

`x=3,-sqrt3/2` or `sqrt3/2`

Explanation:

According to factor theorem , if `f(a)=0`, then `(x–a)` is a factor of the polynomial `f(x)`. Converse of this theorem is also true i.e. if `(x-a)` is a factor of the polynomial `f(x)`, then `f(a)=0`.

Here for `x=3`, `f(x)=4x^3-12x^2-3x+9`, we have

`f(3)=4*3^3-12*3^2-3*3+9=108-108-9+9=0`

Hence `x-3` is a factor of `4x^3-12x^2-3x+9` and hence

`4x^3-12x^2-3x+9=0`

`=>4x^2(x-3)-3(x-3)=0`

or `(x-3)(4x^2-3)=0`

i.e. `(x-3)(2x+sqrt3)(2x-sqrt3)=0`

Hence `x=3,-sqrt3/2` or `sqrt3/2`