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

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Answer 1 · 2018-05-19T05:13:07

$\frac{20}{(x + 3) (x - 2)}$ $(20)/((x+3)(x-2))$

or

$\frac{20}{x^{2} + x - 6}$ $(20)/(x^2 + x -6)$

$\frac{4}{(x + 3) (x + 2)} + \frac{16}{(x + 2) (x - 2)}$ $4/((x+3)(x+2))+16/((x+2)(x-2))$

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

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

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

$\frac{4 (x - 2)}{(x + 3) (x + 2) (x - 2)} + \frac{16 (x + 3)}{(x + 3) (x + 2) (x - 2)}$ $(4(x-2))/((x+3)(x+2)(x-2))+(16(x+3))/((x+3)(x+2)(x-2))$

$\frac{4 x - 8 + 16 x + 48}{(x + 3) (x + 2) (x - 2)}$ $(4x-8 +16x +48)/((x+3)(x+2)(x-2))$

$\frac{20 x + 40}{(x + 3) (x + 2) (x - 2)}$ $(20x+40)/((x+3)(x+2)(x-2))$

$\frac{20 (x + 2)}{(x + 3) (x + 2) (x - 2)}$ $(20(x+2))/((x+3)(x+2)(x-2))$

$\frac{20}{(x + 3) (x - 2)}$ $(20)/((x+3)(x-2))$

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