Ask your own question, for FREE!
Pre-Algebra 23 Online
OpenStudy (anonymous):

2a(a-5)+4(a-5)

OpenStudy (kc_kennylau):

Hint: (a-5) is the common factor :)

OpenStudy (kc_kennylau):

Always take out the common factor :)

OpenStudy (shamil98):

\[\huge 2a(a-5) + 4(a-5)\] Distribute. \[\huge 2a^2 - 10a + 4a - 20\] Simplify. \[\huge 2a^2 - 6a - 20\] \[\huge 2(a^2 -3a + 10)\] \[\huge 2(a-5)(a+2)\]

OpenStudy (shamil98):

oops i changed the sign there

OpenStudy (shamil98):

\[\huge 2(a^2 -3a - 10)\] \[\huge 2(a-5)(a+2)\] is the correct factor though

OpenStudy (kc_kennylau):

lolz he's gone

OpenStudy (shamil98):

yep

OpenStudy (unklerhaukus):

Why did you distribute @shamil98 ? kc_kennylau's method is only two steps 2a(a-5)+4(a-5) (2a+4)(a-5) 2(a+2)(a-5)

OpenStudy (anonymous):

2a(a-5)+4(a-5) Multiply 2a by each term inside the parentheses. 2a^(2)-10a+4(a-5) Multiply 4 by each term inside the parentheses. 2a^(2)-10a+4a-20 Since -10a and 4a are like terms, subtract 4a from -10a to get -6a. 2a^(2)-6a-20

OpenStudy (anonymous):

|dw:1388976695466:dw|

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!