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

2 Answers
Feb 1, 2017

(x+4)/(x-8)

Explanation:

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)

Feb 1, 2017

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

Explanation:

(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)