Solve the problem.
-$$- 12 a ^ { 3 } b ^ { 7 } \left( - 11 a ^ { 4 } b ^ { 9 } + 4 b ^ { 7 } + 11 a \right)$$
A) $a ^ { 4 } b ^ { 9 } - 48 b ^ { 7 } - 132 a$
B) $\quad a ^ { 7 } b ^ { 16 } - 48 a ^ { 3 } b ^ { 14 } - 132 a ^ { 4 } b ^ { 7 }$
C) $a ^ { 7 } b ^ { 16 } - 48 a ^ { 3 } b ^ { 14 } + 11 a$
D) $\quad a ^ { 7 } b ^ { 16 } + 4 b ^ { 7 } + 11 a$

Question

Quizplus · Accepted Answer

When multiplying terms with exponents, you add the exponents of like bases. For the first term inside the parentheses, $-12a^3b^7$ multiplies with $-11a^4b^9$ to give $132a^{3+4}b^{7+9} = 132a^7b^{16}$. For the second term, $-12a^3b^7$ multiplies with $4b^7$ to give $-48a^3b^{7+7} = -48a^3b^{14}$. For the third term, $-12a^3b^7$ multiplies with $11a$ to give $-132a^{3+1}b^7 = -132a^4b^7$.

Solve the Problem −12a3b7(−11a4b9+4b7+11a)- 12 a ^ { 3 } b ^ { 7 } \left( - 11 a ^ { 4 } b ^ { 9 } + 4 b ^ { 7 } + 11 a \right)−12a3b7(−11a4b9+4b7+11a)

Solve the Problem $- 12 a ^ { 3 } b ^ { 7 } \left( - 11 a ^ { 4 } b ^ { 9 } + 4 b ^ { 7 } + 11 a \right)$