Factor completely using the difference of squares along with factoring by grouping. $2 t - 2 u + t ^ { 2 } - u ^ { 2 }$
A) $( t - u ) ( 2 + t + u )$
B) $2 ( t - u ) ( t + u )$
C) $2 ( t - u ) + t ^ { 2 } + u ^ { 2 }$
D) Prime

Question

Quizplus · Accepted Answer

The expression can be rewritten as $t^2 - u^2 + 2t - 2u$. First, factor by grouping: $(t^2 - u^2) + (2t - 2u)$. The first group is a difference of squares: $(t-u)(t+u)$. The second group can be factored as $2(t-u)$. Factoring out $(t-u)$ from both groups gives $(t-u)(t+u+2)$.

Factor Completely Using the Difference of Squares Along with Factoring 2t−2u+t2−u22 t - 2 u + t ^ { 2 } - u ^ { 2 }2t−2u+t2−u2

Factor Completely Using the Difference of Squares Along with Factoring $2 t - 2 u + t ^ { 2 } - u ^ { 2 }$