Factor the expression completely. If the expression is prime, so indicate. $7 ( x + y ) ^ { 2 } + ( x + y ) - 8$
A) $( 7 x + 7 y + 8 ) ( x + y - 1 )$
B) $( 7 x + 7 y - 8 ) ( x + y + 1 )$
C) $( 8 x + 8 y + 7 ) ( x + y - 1 )$
D) $( 8 x + 8 y - 7 ) ( x + y + 1 )$
E) prime

Question

Quizplus · Accepted Answer

We can start by factoring out the common factor of $(x+y)$: $$7(x+y)^2 + (x+y) - 8 = (x+y)\left(7(x+y) + 1ight) - 8.$$ Now, we can see that we have a difference of squares: $$\begin{aligned} (x+y)\left(7(x+y) + 1ight) - 8 &= (x+y)\left(7(x+y) + 1 - 56\cdot\frac{1}{8}ight) \ &= (x+y)\left(7(x+y) - 7\cdot 8\cdot\frac{1}{8} + \frac{1}{8}ight) \ &= (x+y)(7(x+y) - 7\cdot 8)(x+y) + \frac{1}{8}(x+y) \ &= (x+y)(7x+ 7y + 8)(x+y) + \frac{1}{8}(x+y). \end{aligned}$$ Finally, we factor out the common factor of $(x+y)$ again to get the fully factored form: $$(x+y)\left[(7x+7y+8)+\frac{1}{8}ight]=(7x+7y+8)(x+y-1).$$

Factor the Expression Completely 7(x+y)2+(x+y)−87 ( x + y ) ^ { 2 } + ( x + y ) - 87(x+y)2+(x+y)−8

Factor the Expression Completely $7 ( x + y ) ^ { 2 } + ( x + y ) - 8$