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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