Factor completely.
-$4 \mathrm{k}^{2}-121 \mathrm{~m}^{2}$
A) $(4 k+m)(k-121 m)$
B) $(2 \mathrm{k}+11 \mathrm{~m})(2 \mathrm{k}-11 \mathrm{~m})$
C) $(2 k-11 m)^{2}$
D) $(2 \mathrm{k}+11 \mathrm{~m})^{2}$

Question

Quizplus · Accepted Answer

This expression is a difference of squares, which can be factored as $(a+b)(a-b)$. Here, $a^2 = 4k^2$ (so $a = 2k$) and $b^2 = 121m^2$ (so $b = 11m$). Therefore, the factored form is $(2k+11m)(2k-11m)$.

Factor Completely 4k2−121 m24 \mathrm{k}^{2}-121 \mathrm{~m}^{2}4k2−121 m2 A) (4k+m)(k−121m)(4 k+m)(k-121 m)(4k+m)(k−121m) B) (2k+11 m)(2k−11 m)(2 \mathrm{k}+11 \mathrm{~m})(2 \mathrm{k}-11 \mathrm{~m})(2k+11 m)(2k−11 m)

Factor Completely $4 \mathrm{k}^{2}-121 \mathrm{~m}^{2}$ A) $(4 k+m)(k-121 m)$ B) $(2 \mathrm{k}+11 \mathrm{~m})(2 \mathrm{k}-11 \mathrm{~m})$