As – 16a,
as – 4a3,
2a3 + 8a - 4a2 - 16

1) as-16a

taknig a as common

=a(s-16)

2)as-4a³

taking a as common

=a(s-4a²)

3)2a³+8a-4a²-16

taking common terms

=2a(a²+4)-4(a²+4)

=(2a-4)(a²+4)

again taking common

=2(a-2)(a²+4)

i hope this will help you :)

answer rather than having to add and subtract the measurements of the semi-circle, notice that the semicircle to the left of ad is actually identical in area to the blank semicircle-shaped space to the left of bc. so if we transferred the red semicircle into the blank space, we would end up with a square of side length 24 cm. the area of a square is s^2 = 24^2 = 676 cm^2.

u also answered this so i think u would know but i am here to

put in or use mine