Saturday, 13 June 2015

Factor by completing the square for` f(x)= x^2 -4x +9`

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.

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