Question

How do you factor completely `y=2x^3 - 11x^2 + 14x - 3`?

Answer 1

`2x^3-11x^2+14x-3 = (2x-3)(x-2-sqrt(3))(x-2+sqrt(3))`

Explanation:

`f(x) = 2x^3-11x^2+14x-3`

By the rational root theorem, any zeros of `f(x)` must be expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-3` and `q` a divisor of the coefficient `2` of the leading term.

That means that the only possible rational zeros are:

`+-1/2`, `+-1`, `+-3/2`, `+-3`

Trying each in turn, we eventually find:

`f(3/2) = 27/4-99/4+21-3 = (27-99+84-12)/4 = 0`

So `x=3/2` is a zero and `(2x-3)` a factor:

`2x^3-11x^2+14x-3 = (2x-3)(x^2-4x+1)`

The remaining quadratic factor can be factored by completing the square and using the difference of squares identity:

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

with `a=(x-2)` and `b=sqrt(3)` as follows:

`x^2-4x+1 = x^2-4x+4-3`

`= (x-2)^2-(sqrt(3))^2`

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

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

Putting it all together:

`2x^3-11x^2+14x-3 = (2x-3)(x-2-sqrt(3))(x-2+sqrt(3))`