Factor: $5 b ^ { 2 } - 27 b + 10$
A) $( b + 5 ) ( 5 b - 2 )$
B) $( b + 2 ) ( 5 b - 5 )$
C) $( b - 5 ) ( 5 b - 2 )$
D) $( b + 2 ) ( 5 b + 5 )$
E) Nonfactorable over the integers

Question

Quizplus · Accepted Answer

To factor $5b^2 - 27b + 10$, we look for two numbers that multiply to $5 \times 10 = 50$ and add to $-27$. The numbers $-25$ and $-2$ work, since $-25 \times -2 = 50$ and $-25 - 2 = -27$. We can rewrite the middle term using these numbers: $5b^2 - 25b - 2b + 10$, and then factor by grouping: $(5b^2 - 25b) - (2b - 10) = 5b(b - 5) - 2(b - 5) = (b - 5)(5b - 2)$.

Factor 5b2−27b+105 b ^ { 2 } - 27 b + 105b2−27b+10

Factor $5 b ^ { 2 } - 27 b + 10$