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

2 Answers
Nov 15, 2015

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

Explanation:

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

Nov 15, 2015

4x^3 + x^2 -5x

Explanation:

(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)