the sum of two polynomials is (10a^{2}b^{2}-8a^{2}b + 6ab^{2}-4ab + 2). if one addend is (-5a^{2}b^{2}+12a^{2…

the sum of two polynomials is (10a^{2}b^{2}-8a^{2}b + 6ab^{2}-4ab + 2). if one addend is (-5a^{2}b^{2}+12a^{2}b - 5), what is the other addend?\n(15a^{2}b^{2}-20a^{2}b + 6ab^{2}-4ab + 7)\n(5a^{2}b^{2}-20a^{2}b^{2}+7)\n(5a^{2}b^{2}+4a^{2}b^{2}+6ab - 4ab - 3)\n(-15a^{2}b^{2}+20a^{2}b^{2}-6ab + 4ab - 7)

the sum of two polynomials is (10a^{2}b^{2}-8a^{2}b + 6ab^{2}-4ab + 2). if one addend is (-5a^{2}b^{2}+12a^{2}b - 5), what is the other addend?\n(15a^{2}b^{2}-20a^{2}b + 6ab^{2}-4ab + 7)\n(5a^{2}b^{2}-20a^{2}b^{2}+7)\n(5a^{2}b^{2}+4a^{2}b^{2}+6ab - 4ab - 3)\n(-15a^{2}b^{2}+20a^{2}b^{2}-6ab + 4ab - 7)

Answer

Explanation:

Step1: Let the other addend be (x).

According to the relationship of polynomial addition: ((- 5a^{2}b^{2}+12a^{2}b - 5)+x=10a^{2}b^{2}-8a^{2}b + 6ab^{2}-4ab + 2). Then (x=(10a^{2}b^{2}-8a^{2}b + 6ab^{2}-4ab + 2)-(-5a^{2}b^{2}+12a^{2}b - 5)).

Step2: Remove the parentheses.

(x = 10a^{2}b^{2}-8a^{2}b + 6ab^{2}-4ab + 2 + 5a^{2}b^{2}-12a^{2}b + 5).

Step3: Combine like - terms.

For the (a^{2}b^{2}) terms: (10a^{2}b^{2}+5a^{2}b^{2}=(10 + 5)a^{2}b^{2}=15a^{2}b^{2}). For the (a^{2}b) terms: (-8a^{2}b-12a^{2}b=(-8-12)a^{2}b=-20a^{2}b). For the (ab^{2}) terms: (6ab^{2}). For the (ab) terms: (-4ab). For the constant terms: (2 + 5=7).

So (x=15a^{2}b^{2}-20a^{2}b + 6ab^{2}-4ab + 7).

Answer:

(15a^{2}b^{2}-20a^{2}b + 6ab^{2}-4ab + 7) (the first option)