How do you divide (20x^4+(5-12i)x^3-(25+3i)x^2+15ix)/(5x-3i)20x4+(5−12i)x3−(25+3i)x2+15ix5x−3i ?
2 Answers
(20x^4+(5-12i)x^3-(25+3i)x^2+15ix)/(5x-3i) = 4x^3+x^2-5x20x4+(5−12i)x3−(25+3i)x2+15ix5x−3i=4x3+x2−5x
Explanation:
The first term in the quotient is
Then
Subtract this from the dividend:
(20x^4+(5-12i)x^3-(25+3i)x^2+15ix)-(20x^4-12ix^3)(20x4+(5−12i)x3−(25+3i)x2+15ix)−(20x4−12ix3)
=5x^3-(25+3i)x^2+15ix=5x3−(25+3i)x2+15ix
The next term in the quotient is
Then
Subtract this from the remainder:
(5x^3-(25+3i)x^2+15ix)-(5x^3-3ix^2)=-25x^2+15ix(5x3−(25+3i)x2+15ix)−(5x3−3ix2)=−25x2+15ix
The next term in the quotient is
Then
...equalling the remainder.
So:
(20x^4+(5-12i)x^3-(25+3i)x^2+15ix)/(5x-3i)20x4+(5−12i)x3−(25+3i)x2+15ix5x−3i
= 4x^3+x^2-5x=4x3+x2−5x