How do you combine $(3n + 8)/(n^2 + 6n + 8) - ( 4n + 2)/(n^2 + n - 12)$ ?

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Answer 1 · 2018-04-17T01:36:03

$(3n+8)/(n^2+6n+8)-(4n-2)/(n^2+n-12)=(n^2+7n+20)/((n+4)(n+2)(3-n))$

$(3n+8)/(n^2+6n+8)-(4n-2)/(n^2+n-12)$

First factor the denominators.

$(3n+8)/((n+4)(n+2))-(4n-2)/((n+4)(n-3))$

The common denominator is $(n+4)(n+2)(n-3)$

Multiply each quotient by the appropriate factor to display the common denominator

$((3n+8))/((n+4)(n+2))((n-3))/((n-3))-((4n-2))/((n+4)(n-3))((n+2))/((n+2))$

Expand the numerators using the distributive property (or FOIL if you like).

$(3n^2-n-24)/((n+4)(n+2)(n-3))-(4n^2+6n-4)/((n+4)(n+2)(n-3))$

We can combine the quotients because they have a common denominator.

$(3n^2-n-24-4n^2-6n+4)/((n+4)(n+2)(n-3))$

Combine like terms.

$(-n^2-7n-20)/((n+4)(n+2)(n-3))=(n^2+7n+20)/((n+4)(n+2)(3-n))$

How do you combine (3n + 8)/(n^2 + 6n + 8) - ( 4n + 2)/(n^2 + n - 12)(3n + 8)/(n^2 + 6n + 8) - ( 4n + 2)/(n^2 + n - 12)?