# how simplify the fraction (x^2+5ix-6)/(x^2+9)? what is 5i?

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You need to reduce the fraction to its lowest terms, hence you need to write the factored forms of both numerator and denominator.

Notice that numerator has complex coefficients (5i).

You need to find the roots of numerator and then you may write its factored form.

You need to use quadratic formula such that:

`x_(1,2) = (-5i +- sqrt((5i)^2-4*1*(-6)))/2`

`x_(1,2) = (-5i +- sqrt(-25+24))/2`

`x_(1,2) = (-5i +- sqrt(-1))/2`

You need to use complex number theory, hence `sqrt(-1) = i.`

`x_(1,2) = (-5i +- i)/2 =gt x_1 = -4i/2 =gt x_1 = -2i`

`x_2 = -6i/2 =gt x_2 = -3i`

You may write the factored form to numerator such that:

`(x^2+5ix-6)/(x^2+9) = ((x+2i)(x+3i))/(x^2+9)`

You may write denominator as difference of two squares such that:

`(x^2+9) = (x^2-(-9)) = (x-3i)(x+3i)`

`((x+2i)(x+3i))/(x^2+9) = ((x+2i)(x+3i))/((x-3i)(x+3i))`

Reducing like terms yields:

`(x+2i)/(x-3i)`

**Hence, reducing the fraction to its lowest terms yields `(x^2+5ix-6)/(x^2+9) = (x+2i)/(x-3i).` **