How do I solve (x-(2+i))(x-(2-i))(x-4)
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How do I solve (x-(2+i))(x-(2-i))(x-4)

[From: ] [author: ] [Date: 11-11-02] [Hit: ]
. If anybody could walk through this step by step I would be grateful...........
I get x^3 - 4x^2 - 5x - 20 , but the book says I am supposed to be getting x^3 - 8x^2 + 21x - 20 .... If anybody could walk through this step by step I would be grateful.

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[x-(2+i)][x-(2-i)](x-4)=(x-2-i)(x-2+i)(x…
=[(x-2)-i)][(x-2)+i)](x-4)

........ now you should use the formula

"a^2-b^2=(a-b)(a+b)" where "a" is "(x-2)"
and "i" is "b" (i^2 is equal to -1)

[(x-2)-i)][(x-2)+i)](x-4)=[(x-2)^2 - i^2](x-4)=
=(x^2-4x+4+1)(x-4)=(x^2-4x+5)(x-4)=
=x^3-4x^2-4x^2+16x+4x-16+x-4=
=x^3 - 8x^2 + 21x - 20
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