Factor completely.
-$121 \mathrm { k } ^ { 2 } - 100 \mathrm {~m} ^ { 2 }$
A) $( 11 \mathrm { k } + 10 \mathrm {~m} ) ( 11 \mathrm { k } - 10 \mathrm {~m} )$
B) $( 11 \mathrm { k } + 10 \mathrm {~m} ) ^ { 2 }$
C) $( 11 \mathrm { k } - 10 \mathrm {~m} ) ^ { 2 }$
D) $( 121 \mathrm { k } + \mathrm { m } ) ( \mathrm { k } - 100 \mathrm {~m} )$

Question

Quizplus · Accepted Answer

This is a difference of squares problem. Using the formula: 
$$
a^2-b^2=(a+b)(a-b)
$$
where $a = 11k$ and $b=10m$, we get
\begin{align*}
121 k ^ { 2 } - 100 m ^ { 2 } &= (11k)^2-(10m)^2\
&=(11k+10m)(11k-10m)
\end{align*}
Therefore, the answer is (A).

Factor Completely 121k2−100 m2121 \mathrm { k } ^ { 2 } - 100 \mathrm {~m} ^ { 2 }121k2−100 m2

Factor Completely $121 \mathrm { k } ^ { 2 } - 100 \mathrm {~m} ^ { 2 }$