How do you evaluate #\frac { 2( x ^ { 2} + 32) } { ( x + 8) ( x - 8) } - \frac { x - 4} { x + 8}#?

Question

979 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-01T18:34:00

#(x+4)/(x-8)#

Take out #1/(x+8)# as a common factor

#1/(x+8) [ (2(x^2+32))/(x-8) - (x-4)]#

#1/(x+8) [(2(x^2 +32)- (x-4)(x-8))/(x-8) ]#

#1/(x+8) [(2x^2 +64- x^2+12x-32)/(x-8)]#

#(x^2 +12x+32)/((x+8)(x-8))#

Now factorise the numerator

#((x+8)(x+4))/((x+8)(x-8))#

#(x+4)/(x-8)#

Answer 2 · 2017-02-01T18:41:10

#=(x+4)/(x-8)#

#(2(x^2+32))/((x+8)(x-8))-(x-4)/(x+8)#

#=(2(x^2+32)-(x-4)(x-8))/((x+8)(x-8))#

#=(2(x^2+32)-(x^2-8x-4x+32))/((x+8)(x-8))#

#=(2x^2+64-x^2+12x-32)/((x+8)(x-8))#

#=(x^2+12x+32)/((x+8)(x-8))#

#=(x^2+8x+4x+32)/((x+8)(x-8))#

#=(x(x+8)+4(x+8))/((x+8)(x-8))#

#=((x+8)(x+4))/((x+8)(x-8))#

#=(x+4)/(x-8)#