Factor the expression by factoring out the common binomial factor. ( 2 x − 3 ) ( x + 5 ) + ( 7 x + 1 ) ( x + 5 ) ( 2 x - 3 ) ( x + 5 ) + ( 7 x + 1 ) ( x + 5 ) ( 2 x − 3 ) ( x + 5 ) + ( 7 x + 1 ) ( x + 5 ) A) ( 9 x − 2 ) ( 2 x − 3 ) ( 9 x - 2 ) ( 2 x - 3 ) ( 9 x − 2 ) ( 2 x − 3 ) B) ( 2 x − 3 ) ( 7 x + 1 ) ( 2 x - 3 ) ( 7 x + 1 ) ( 2 x − 3 ) ( 7 x + 1 ) C) ( 7 x + 1 ) ( x + 5 ) ( 7 x + 1 ) ( x + 5 ) ( 7 x + 1 ) ( x + 5 ) D) ( 9 x − 2 ) ( x + 5 ) ( 9 x - 2 ) ( x + 5 ) ( 9 x − 2 ) ( x + 5 ) E) ( 2 x − 3 ) ( x + 5 ) ( 2 x - 3 ) ( x + 5 ) ( 2 x − 3 ) ( x + 5 )

Question

Factor the expression by factoring out the common binomial factor.   ( 2 x − 3 )  ( x + 5 )  + ( 7 x + 1 )  ( x + 5 )  ( 2 x - 3 )  ( x + 5 )  + ( 7 x + 1 )  ( x + 5 )  ( 2 x − 3 )  ( x + 5 )  + ( 7 x + 1 )  ( x + 5 )  A)    ( 9 x − 2 )  ( 2 x − 3 )  ( 9 x - 2 )  ( 2 x - 3 )  ( 9 x − 2 )  ( 2 x − 3 )  B)    ( 2 x − 3 )  ( 7 x + 1 )  ( 2 x - 3 )  ( 7 x + 1 )  ( 2 x − 3 )  ( 7 x + 1 )  C)    ( 7 x + 1 )  ( x + 5 )  ( 7 x + 1 )  ( x + 5 )  ( 7 x + 1 )  ( x + 5 )  D)    ( 9 x − 2 )  ( x + 5 )  ( 9 x - 2 )  ( x + 5 )  ( 9 x − 2 )  ( x + 5 )  E)    ( 2 x − 3 )  ( x + 5 )  ( 2 x - 3 )  ( x + 5 )  ( 2 x − 3 )  ( x + 5 )

Shikha Maskey · Accepted Answer

Final Answer : D, Explanation :   The common binomial factor in both terms is   ( x + 5 ) (x + 5) ( x + 5 )  , so when factored out, the expression becomes   ( 2 x − 3 + 7 x + 1 ) ( x + 5 ) (2x - 3 + 7x + 1)(x + 5) ( 2 x − 3 + 7 x + 1 ) ( x + 5 )  , which simplifies to   ( 9 x − 2 ) ( x + 5 ) (9x - 2)(x + 5) ( 9 x − 2 ) ( x + 5 )  .

Factor the expression by factoring out the common binomial factor. $(2 x - 3) (x + 5) + (7 x + 1) (x + 5)$

A) $(9 x - 2) (2 x - 3)$
B) $(2 x - 3) (7 x + 1)$
C) $(7 x + 1) (x + 5)$
D) $(9 x - 2) (x + 5)$
E) $(2 x - 3) (x + 5)$

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