(x-(4-i))(x-(4+i)) = x^2-8x+17
I don't understand where the 17 comes from :o can someone explain this to me? ( :

Question

(x-(4-i))(x-(4+i)) = x^2-8x+17 
I don't understand where the 17 comes from :o can someone explain this to me? ( :

hero · Answer

Well, let's do this:

((x - 4) - i))((x - 4) + i) = (x - 4)^2 - i^2
                                   = (x - 4)(x - 4) - i^2
                                   = x(x - 4) - 4(x - 4) - i^2
                                   = x^2 - 4x - 4x + 16 -(-1)
                                   = x^2 - 8x + 16 + 1
                                   = x^2 - 8x + 17

anonymous · Answer

Thank you ( :

hero · Answer

yw

whpalmer4 · Answer

that's a nifty way to cut down on the effort of evaluating that multiplication!

But, if you aren't so observant, 
$$(x-4-i)(x-4+i) = x^2 -4x + ix - 4x + 16 -4i -ix + 4i - i^2$$$$\qquad = x^2 -8x + 16 - (-1)$$$$\qquad = x^2-8x+17$$
just the same.