How do you add #\frac { 4} { ( x + 8) ^ { 2} } + \frac { 7} { x ^ { 2} + x - 56}#?

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Answer 1 · 2017-09-08T07:20:02

#4/(x+8)^2+7/(x^2+x-56)=(11x+28)/((x-7)(x+8)^2)#

#x^2+x-56#

= #x^2+8x-7x-56#

= #x(x+8)-7(x+8)#

= #(x-7)(x+8)#

Hence #4/(x+8)^2+7/(x^2+x-56)#

= #4/(x+8)^2+7/((x-7)(x+8))#

= #(4(x-7)+7(x+8))/((x-7)(x+8)^2)#

= #(4x-28+7x+56)/((x-7)(x+8)^2)#

= #(11x+28)/((x-7)(x+8)^2)#