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

Question

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

Eva Adamson · Accepted Answer

#(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))#

Kennedy Adams · Answer

undefined To combine the given expressions, we need to find a common denominator for both fractions. The denominators are (n^2 + 6n + 8) and (n^2 + n - 12).

The common denominator is (n^2 + 6n + 8)(n^2 + n - 12).

To combine the fractions, we multiply the numerator and denominator of the first fraction by (n^2 + n - 12), and the numerator and denominator of the second fraction by (n^2 + 6n + 8).

This gives us ((3n + 8)(n^2 + n - 12))/((n^2 + 6n + 8)(n^2 + n - 12)) - ((4n + 2)(n^2 + 6n + 8))/((n^2 + 6n + 8)(n^2 + n - 12)).

Next, we simplify the numerators by distributing and combining like terms.

The simplified expression is (3n^3 + 3n^2 - 36n + 8n^2 + 8n - 96 - 4n^3 - 24n^2 - 32n - 8)/(n^4 + 7n^3 - 4n^2 - 96n + 96).

Combining like terms in the numerator gives us (-n^3 - 13n^2 - 60n - 104)/(n^4 + 7n^3 - 4n^2 - 96n + 96).

Therefore, the combined expression is (-n^3 - 13n^2 - 60n - 104)/(n^4 + 7n^3 - 4n^2 - 96n + 96).