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(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? ( :
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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
Thank you ( :
yw
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.
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