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

Answer:

A. $15a^{2}b^{2}-20a^{2}b + 6ab^{2}-4ab + 7$

Explanation:

Step1: Let the unknown addend be $x$.

Let the sum of the two polynomials be $S = 10a^{2}b^{2}-8a^{2}b + 6ab^{2}-4ab + 2$, and one addend be $y=-5a^{2}b^{2}+12a^{2}b - 5$. Then $S=x + y$, so $x=S - y$.

Step2: Substitute the polynomials.

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

Step3: Remove the parentheses.

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

Step4: Combine like - terms.

For the $a^{2}b^{2}$ terms: $10a^{2}b^{2}+5a^{2}b^{2}=15a^{2}b^{2}$; for the $a^{2}b$ terms: $-8a^{2}b-12a^{2}b=-20a^{2}b$; the $ab^{2}$ term remains $6ab^{2}$; the $ab$ term remains $-4ab$; and the constant terms $2 + 5=7$. So $x = 15a^{2}b^{2}-20a^{2}b + 6ab^{2}-4ab + 7$.