How do you put (x+5)(x-3)[x-(4-i)][x-(4+i)] into standard form

1 answer:

6 0

$(x+5)(x-3)[x-(4-i)][x-(4+i)]
 \\ 
(x^{2}+5x-3x-15)(x-4+i)(x-4-i)
\\
(x^{2}+5x-3x-15)(x^{2} -4x-xi-4x+16+4i+xi-4i- i^{2} )
\\
(x^{2}+5x-3x-15)( x^{2} -8x+16- i^{2})
\\
(x^{2}+5x-3x-15)( x^{2} -8x+16-(-1))
\\
(x^{2}+5x-3x-15)( x^{2} -8x+16+1)
\\
(x^{2}+5x-3x-15)( x^{2} -8x+17)$
From there just distribute and simplify.

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