What is #(4x^2-1)/(2x^2-5x-3) * (x^2-6x+9)/(2x^2+5x-3)#, simplified?

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Answer 1 · 2017-01-30T21:48:34

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

First, you would factor all the polynomials and get:

#4x^2-1=(2x-1)(2x+1)#

#x^2-6x+9=(x-3)^2#

Let's find the zeros of

1) #2x^2-5x-3# and 2) #2x^2+5x-3# by the quadratic formula:

#x=(5+-sqrt(25+24))/4=(5+-7)/4#

#x_1=-1/2; x_2=3#

Then

1) #2x^2-5x-3=2(x+1/2)(x-3)=(2x+1)(x-3)#

#x=(-5+-sqrt(25+24))/4=(-5+-7)/4#

#x_1=-3; x_2=1/2#

Then

2) #2x^2+5x-3=2(x+3)(x-1/2)=(x+3)(2x-1)#

Then the given expression is:

#(cancel((2x-1))cancel((2x+1)))/(cancel((2x+1))cancel((x-3)))*((x-3)^cancel2)/((x+3)cancel((2x-1)))#

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