How do you combine #(4a)/(a^2-ab-2b^2) -( 6b)/(a^2+4ab+3b)#?

Question

Nathan Axel · Accepted Answer

#(4a)/(a^2-ab-2b^2)-(6b)/(a^2+4ab+color(red)(3b^2))#

#=(2(2a^2+3ab+6b^2))/(a^3+2a^2b-5ab^2-6b^3)# This problem makes more sense if the expression is: #(4a)/(a^2-ab-2b^2)-(6b)/(a^2+4ab+color(red)(3b^2))# #=(4a)/((a+b)(a-2b))-(6b)/((a+b)(a+3b))# #=((4a)(a+3b)-(6b)(a-2b))/((a+b)(a-2b)(a+3b))# #=(4a^2+12ab-6ab+12b^2)/((a+b)(a-2b)(a+3b))# #=(2(2a^2+3ab+6b^2))/((a+b)(a-2b)(a+3b))# #=(2(2a^2+3ab+6b^2))/((a^2-ab-2b^2)(a+3b))# #=(2(2a^2+3ab+6b^2))/(a^3+2a^2b-5ab^2-6b^3)# Note that I did not try to factor #(2a^2+3ab+6b^2)# since it has a negative discriminant #Delta = 3^2-(4*2*6) = 9-48 = -39#, so no linear factors with Real coefficients.

Hazel Anglin · Answer

undefined To combine the given expressions, we need to find a common denominator. The common denominator for the two expressions is (a^2 - ab - 2b^2)(a^2 + 4ab + 3b).

Multiplying the first fraction by (a^2 + 4ab + 3b)/(a^2 + 4ab + 3b) and the second fraction by (a^2 - ab - 2b^2)/(a^2 - ab - 2b^2), we get:

(4a)/(a^2 - ab - 2b^2) * (a^2 + 4ab + 3b)/(a^2 + 4ab + 3b) - (6b)/(a^2 + 4ab + 3b) * (a^2 - ab - 2b^2)/(a^2 - ab - 2b^2)

Expanding and simplifying the numerators, we have:

(4a)(a^2 + 4ab + 3b) - (6b)(a^2 - ab - 2b^2)

Simplifying further, we get:

4a^3 + 16a^2b + 12ab - 6a^2b + 6ab^2 + 12b^3

Combining like terms, we have:

4a^3 + 10a^2b + 18ab^2 + 12b^3

Therefore, the combined expression is (4a^3 + 10a^2b + 18ab^2 + 12b^3)/(a^2 - ab - 2b^2).