Question

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

Answer 1

`(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`

Answer 2

`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)`