How do you add $\frac { 4} { ( x + 3) ( x + 2) } + \frac { 16} { ( x + 2) ( x - 2) }$ ?

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Answer 1 · 2018-05-19T05:13:07

$(20)/((x+3)(x-2))$

or

$(20)/(x^2 + x -6)$

$4/((x+3)(x+2))+16/((x+2)(x-2))$

First find a common denominator, it must have all the factors of both denominators:

$(x+3)(x+2)(x-2)$

Then we have to multiply the missing factor times each term:

$(4(x-2))/((x+3)(x+2)(x-2))+(16(x+3))/((x+3)(x+2)(x-2))$

$(4x-8 +16x +48)/((x+3)(x+2)(x-2))$

$(20x+40)/((x+3)(x+2)(x-2))$

$(20(x+2))/((x+3)(x+2)(x-2))$

$(20)/((x+3)(x-2))$

How do you add \frac { 4} { ( x + 3) ( x + 2) } + \frac { 16} { ( x + 2) ( x - 2) }\frac { 4} { ( x + 3) ( x + 2) } + \frac { 16} { ( x + 2) ( x - 2) }?