Factor completely using grouping.
-
A) 7(2y + 1)(y - 5)
B) (14y - 7)(y + 5)
C) 7(2y - 1)(y + 5)
D) prime

Question

Factor completely using grouping.
-
A) 7(2y + 1)(y - 5) 
B) (14y - 7)(y + 5) 
C) 7(2y - 1)(y + 5) 
D) prime

Quizplus · Accepted Answer

First, we can group the first two terms and the last two terms:
11ebb17b_2a1e_afd6_afdc_5503b336bd0a_TB8591_12
Then we can factor out a 7 from the first group and a -1 from the second group:
11ebb17b_2a1e_afd6_afdc_5503b336bd0a_TB8591_13
Combining the terms inside the parentheses gives us:
7(2y - 1)(y + 5) First, we can factor out the greatest common factor of 2y - 1 from the first two terms and the greatest common factor of -5 from the last two terms, giving us:
(2y - 1)(eb + 17a) - 5(2y - 1)(a + 17b)
We can now factor out the common factor of (2y - 1) from both terms, giving us:
(2y - 1)(eb + 17a - 5a - 85b)
Simplifying the coefficients of a and b, we get:
(2y - 1)(12a - 68b)
We can further simplify this expression by factoring out the greatest common factor of 4 from both terms, giving us:
4(2y - 1)(3a - 17b)
Finally, we can factor out the common factor of 2y - 1, giving us the fully factored expression:
4(2y - 1)(3a - 17b)

Factor Completely Using Grouping