How do you evaluate $\frac { 4} { x ^ { 2} - 25} + \frac { 6} { x ^ { 2} + 6x + 5}$ ?

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Answer 1 · 2017-11-11T05:29:37

$(10x - 26)/((x-5)(x+5)(x+1) )$

$4/(x^2 -25) + 6/(x^2 + 6x +5)$

to factor $x^2-25$ you need to understand difference of squares where $(x-5)(x+5) = x^2 -25$

$4/((x-5)(x+5)) + 6/((x+1)(x+5)$

our LCD is $(x-5)(x+5)(x+1)$

$4/((x-5)(x+5)) * ((x+1))/((x+1)) = (4(x+1))/((x-5)(x+5)(x+1)$ and

$6/((x+1)(x+5)) * ((x-5))/((x-5)) = (6(x-5))/((x-5)(x+5)(x+1))$

$(4(x+1) + 6(x-5))/((x-5)(x+5)(x+1) )$

Distributive Property

$(4x + 4 + 6x - 30)/((x-5)(x+5)(x+1) )$

add like terms

$(10x - 26)/((x-5)(x+5)(x+1) )$ Fully simplified

How do you evaluate \frac { 4} { x ^ { 2} - 25} + \frac { 6} { x ^ { 2} + 6x + 5}\frac { 4} { x ^ { 2} - 25} + \frac { 6} { x ^ { 2} + 6x + 5}?