the sum of two polynomials is (8d^{5}-3c^{3}d^{2}+5c^{2}d^{3}-4cd^{4}+9). if one addend is…

the sum of two polynomials is (8d^{5}-3c^{3}d^{2}+5c^{2}d^{3}-4cd^{4}+9). if one addend is (2d^{5}-c^{3}d^{2}+8cd^{4}+1), what is the other addend?\n(6d^{5}-2c^{3}d^{2}+5c^{2}d^{3}-12cd^{4}+8)\n(6d^{5}-4c^{3}d^{2}+3c^{2}d^{3}-4cd^{4}+8)\n(6d^{5}-4c^{3}d^{2}+5c^{2}d^{3}-12cd^{4}+8)\n(6d^{5}-2c^{3}d^{2}-3c^{2}d^{3}-4cd^{4}+8)

the sum of two polynomials is (8d^{5}-3c^{3}d^{2}+5c^{2}d^{3}-4cd^{4}+9). if one addend is (2d^{5}-c^{3}d^{2}+8cd^{4}+1), what is the other addend?\n(6d^{5}-2c^{3}d^{2}+5c^{2}d^{3}-12cd^{4}+8)\n(6d^{5}-4c^{3}d^{2}+3c^{2}d^{3}-4cd^{4}+8)\n(6d^{5}-4c^{3}d^{2}+5c^{2}d^{3}-12cd^{4}+8)\n(6d^{5}-2c^{3}d^{2}-3c^{2}d^{3}-4cd^{4}+8)

Answer

Explanation:

Step1: Let the other addend be (x).

According to the rule of polynomial addition: ((2d^{5}-c^{3}d^{2}+8cd^{4}+1)+x = 8d^{5}-3c^{3}d^{2}+5c^{2}d^{3}-4cd^{4}+9). Then (x=(8d^{5}-3c^{3}d^{2}+5c^{2}d^{3}-4cd^{4}+9)-(2d^{5}-c^{3}d^{2}+8cd^{4}+1)).

Step2: Remove the parentheses.

(x = 8d^{5}-3c^{3}d^{2}+5c^{2}d^{3}-4cd^{4}+9 - 2d^{5}+c^{3}d^{2}-8cd^{4}-1).

Step3: Combine like - terms.

For the (d^{5}) terms: (8d^{5}-2d^{5}=6d^{5}). For the (c^{3}d^{2}) terms: (-3c^{3}d^{2}+c^{3}d^{2}=-2c^{3}d^{2}). For the (c^{2}d^{3}) terms: (5c^{2}d^{3}) (no other (c^{2}d^{3}) terms to combine). For the (cd^{4}) terms: (-4cd^{4}-8cd^{4}=-12cd^{4}). For the constant terms: (9 - 1=8).

So (x = 6d^{5}-2c^{3}d^{2}+5c^{2}d^{3}-12cd^{4}+8).

Answer:

(6d^{5}-2c^{3}d^{2}+5c^{2}d^{3}-12cd^{4}+8) (the first option).