How do you simplify #(4x^2)/(4x^2-9)#?

2 Answers
Jun 1, 2017

Answer:

#(4x^2)/((2x+3)(2x-3)#

This expression does not simplify.

Explanation:

#(4x^2)/(4x^2-9)#

Do not be tempted to cancel the #4x^2# in the numerator and denominator. The fact that there are 2 terms in the denominator means you cannot cancel.

Perhaps factorising the denominator will help?

#(4x^2)/((2x+3)(2x-3)#

This expression does not simplify.

Jun 12, 2017

Answer:

Another way of writing this which might be considered "simple" is to decompose this into partial fractions.

This gives the answer #3/(2(2x-3))-3/(2(2x+3))#.

Explanation:

(Note that for all intents and purposes, #(4x^2)/((2x-3)(2x+3)# is perfectly fine. This method just shows another way to simplify a rational function.)

The separation through partial fractions will look like this:

#(4x^2)/((2x+3)(2x-3)) = A/(2x+3) + B/(2x-3)#

#4x^2 = A(2x-3) + B(2x+3)#

Let #x=3/2#:

#4(3/2)^2 = A(2(3/2)-3) + B(2(3/2)+3)#

#4(9/4) = 0A+6B#

#9 = 6B#

#3/2 = B#

Let #x=-3/2#

#4(-3/2)^2 = A(2(-3/2)-3)+B(2(-3/2)+3)#

#4(9/4) = -6A + 0B#

#9 = -6A#

#-3/2 = A#

Therefore:

#(4x^2)/((2x+3)(2x-3)) = (-3/2)/(2x+3) + (3/2)/(2x-3)#

#3/(2(2x-3))-3/(2(2x+3))#

Final Answer