How do you divide (20x^4+(5-12i)x^3-(25+3i)x^2+15ix)/(5x-3i)20x4+(512i)x3(25+3i)x2+15ix5x3i ?

2 Answers
Nov 15, 2015

(20x^4+(5-12i)x^3-(25+3i)x^2+15ix)/(5x-3i) = 4x^3+x^2-5x20x4+(512i)x3(25+3i)x2+15ix5x3i=4x3+x25x

Explanation:

The first term in the quotient is color(blue)(4x^3)4x3

Then 4x^3(5x-3i) = 20x^4-12ix^34x3(5x3i)=20x412ix3

Subtract this from the dividend:

(20x^4+(5-12i)x^3-(25+3i)x^2+15ix)-(20x^4-12ix^3)(20x4+(512i)x3(25+3i)x2+15ix)(20x412ix3)

=5x^3-(25+3i)x^2+15ix=5x3(25+3i)x2+15ix

The next term in the quotient is color(blue)(x^2)x2

Then x^2(5x-3i) = 5x^3-3ix^2x2(5x3i)=5x33ix2

Subtract this from the remainder:

(5x^3-(25+3i)x^2+15ix)-(5x^3-3ix^2)=-25x^2+15ix(5x3(25+3i)x2+15ix)(5x33ix2)=25x2+15ix

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

Then -5x(5x-3i) = -25x^2+15ix5x(5x3i)=25x2+15ix

...equalling the remainder.

So:

(20x^4+(5-12i)x^3-(25+3i)x^2+15ix)/(5x-3i)20x4+(512i)x3(25+3i)x2+15ix5x3i

= 4x^3+x^2-5x=4x3+x25x

Nov 15, 2015

4x^34x3 + x^2x2 -5x5x

Explanation:

(20x^4+(5−12i)x^3−(25+3i)x^2+15ix)/"5x-3i"20x4+(512i)x3(25+3i)x2+15ix5x-3i

(20x^4+5x^3−12ix^3−25x^2-3ix^2+15ix)/"5x-3i"20x4+5x312ix325x23ix2+15ix5x-3i

(20x^4−12ix^3+5x^3-3ix^2−25x^2+15ix)/"5x-3i"20x412ix3+5x33ix225x2+15ix5x-3i

(4x^3(5x-3i)+x^2(5x-3i)-5x(5x-3i))/"5x-3i"4x3(5x3i)+x2(5x3i)5x(5x3i)5x-3i

(5x-3i)(4x^3+x^2-5x)/"5x-3i"(5x3i)4x3+x25x5x-3i

(4x^3+x^2-5x)(4x3+x25x)