Find the sum or difference.
-$$n ^ { 2 } \left( 3 n - \frac { 1 } { 4 } \right) \left( 11 n + \frac { 1 } { 2 } \right)$$
A) $33 n ^ { 4 } - \frac { 17 } { 4 } n ^ { 3 } - \frac { 1 } { 8 } n ^ { 2 }$
B) $33 n ^ { 4 } - \frac { 5 } { 4 } n ^ { 3 } - \frac { 1 } { 8 } n ^ { 2 }$
C) $33 n ^ { 4 } + \frac { 5 } { 4 } n ^ { 3 } - \frac { 1 } { 8 } n ^ { 2 }$
D) $33 n ^ { 4 } - \frac { 5 } { 4 } n ^ { 3 } + \frac { 1 } { 8 } n ^ { 2 }$

Question

Quizplus · Accepted Answer

To find the product, first multiply the two binomials: $(3n - \frac{1}{4})(11n + \frac{1}{2}) = 33n^2 + \frac{3}{2}n - \frac{11}{4}n - \frac{1}{8}$. Simplifying the middle terms gives $33n^2 + \frac{5}{4}n - \frac{1}{8}$. Then, multiply this result by $n^2$ to get $33n^4 + \frac{5}{4}n^3 - \frac{1}{8}n^2$, which matches option B.

Find the Sum or Difference n2(3n−14)(11n+12)n ^ { 2 } \left( 3 n - \frac { 1 } { 4 } \right) \left( 11 n + \frac { 1 } { 2 } \right)n2(3n−41​)(11n+21​)

Find the Sum or Difference $n ^ { 2 } \left( 3 n - \frac { 1 } { 4 } \right) \left( 11 n + \frac { 1 } { 2 } \right)$