How do you divide #(20x^4+(5-12i)x^3-(25+3i)x^2+15ix)/(5x-3i)# ?

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Answer 1 · 2015-11-15T15:11:27

#(20x^4+(5-12i)x^3-(25+3i)x^2+15ix)/(5x-3i) = 4x^3+x^2-5x#

The first term in the quotient is #color(blue)(4x^3)#

Then #4x^3(5x-3i) = 20x^4-12ix^3#

Subtract this from the dividend:

#(20x^4+(5-12i)x^3-(25+3i)x^2+15ix)-(20x^4-12ix^3)#

#=5x^3-(25+3i)x^2+15ix#

The next term in the quotient is #color(blue)(x^2)#

Then #x^2(5x-3i) = 5x^3-3ix^2#

Subtract this from the remainder:

#(5x^3-(25+3i)x^2+15ix)-(5x^3-3ix^2)=-25x^2+15ix#

The next term in the quotient is #color(blue)(-5x)#

Then #-5x(5x-3i) = -25x^2+15ix#

...equalling the remainder.

So:

#(20x^4+(5-12i)x^3-(25+3i)x^2+15ix)/(5x-3i)#

#= 4x^3+x^2-5x#

Answer 2 · 2015-11-15T15:41:20

#4x^3# + #x^2# -#5x#

#(20x^4+(5−12i)x^3−(25+3i)x^2+15ix)/"5x-3i"#

#(20x^4+5x^3−12ix^3−25x^2-3ix^2+15ix)/"5x-3i"#

#(20x^4−12ix^3+5x^3-3ix^2−25x^2+15ix)/"5x-3i"#

#(4x^3(5x-3i)+x^2(5x-3i)-5x(5x-3i))/"5x-3i"#

#(5x-3i)(4x^3+x^2-5x)/"5x-3i"#

#(4x^3+x^2-5x)#