Question

How do you factor `x^3-9x^2+25x-21`, given that `x=3` is a zero ?

Answer 1

`x^3-9x^2+25x-21 = (x-3)(x-3-sqrt(2))(x-3+sqrt(2))`

Explanation:

`h(x) = x^3-9x^2+25x-21`

The difference of squares identity can be written:

`a^2-b^2 = (a-b)(a+b)`

We use this later with `a = (x-3)` and `b=sqrt(2)`

We are told that `3` is a zero, so `(x-3)` must be a factor:

`x^3-9x^2+25x-21 = (x-3)(x^2-6x+7)`

Then, completing the square we find:

`x^2-6x+7 = x^2-6x+9-2`

`color(white)(x^2-6x+7) = (x-3)^2-(sqrt(2))^2`

`color(white)(x^2-6x+7) = ((x-3)-sqrt(2))((x-3)+sqrt(2))`

`color(white)(x^2-6x+7) = (x-3-sqrt(2))(x-3+sqrt(2))`

Putting it all together:

`x^3-9x^2+25x-21 = (x-3)(x-3-sqrt(2))(x-3+sqrt(2))`