Express `f(x) = x^2 - 4x + 9` as `x^2 - 2*x*2 + 2^2 - 2^2 + 9 = (x - 2)^2 - 4 + 9 = (x - 2)^2 + 5.` We used the formula `(a - b)^2 = a^2 - 2ab + b^2` in the reverse direction.

We see that this function is always positive for real `x,` therefore it cannot be factored using real coefficients.

But it can using complex numbers: `-5 = (i sqrt(5))^2` and

`f(x) =(x - 2)^2 + 5 =(x - 2)^2 -(i sqrt(5))^2 = (x - 2 -i sqrt(5))(x - 2 + i sqrt(5)).`

Here we used the formula `a^2 - b^2 = (a - b)(a + b).`

The answer is **impossible** *for real coefficients* and
`(x - 2 -i sqrt(5))(x - 2 + i sqrt(5))` *for complex
coefficients*.

First, when we complete the square for a quadratic function of the form `f(x)=ax^2+bx+c` we get it into the form `(x+b/2)^2-(b/2)^2+c` and then simplify. To do this we must add and subtract (to not change the equation) factors of `(b/2)^2` .

In this case:

`(b/2)^2=(-4/2)^2=4`

And

`b/2=-4/2=-2`

Now add and subtract factors of `(b/2)^2`

`x^2 -4x +9+(4-4)`

`(x^2-4x+4)-4+9`

Factor the term in the parenthesis.

`(x-2)^2-4+9`

This is now in the form ` (x+b/2)^2-(b/2)^2+c` , now simplify.

`f(x)=(x-2)^2+5`

This form is convenient for reading off the horizontal and vertical shift of the parabola. The vertex is at `(2,5)` .

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