Can u Simplify: b*2+14b+49 over b*2-14b+49 devided by (b-7)*-2 over (b+7)*-2??

Question

This is a fraction problem. *2 means squared and *-2 mean to the negative squared.

Answer 1

Equals 1
Second Part= {(b-7)^-2=1/(b-7)^2} / {(B+7)^-2=1/(b+7)^2}
All that comes out to be {1/(b^2-14b+49)} / {1/b^2+14b+49)}
That equals (b^2-14b+49)/(b^2+14B+49) because the 1s cancel out.
Putting the 1st part along with the 2nd part=
{(b^2+14b+49)/(b^2-14B+49)} divided by {(b^2+14b+49)/(b^2-14b+49)}
That becomes
the 1st part multiplied by the reciprocal of 2nd part becoming
(b^4+28b^3+294b^2+1372b+2401)/(b^4-28b^3+294b^2-1372b+2401)
equals 1

Answer 2

ok, here's how you solve it

b*2 + 14b + 49 = (b+7)*2
b*2 - 14b + 49 = (b-7)*2
(b-7)*-2=1/(b-7)*2
and
(b+7)*-2 = 1/(b+7)*2

so

((b+7)*2/(b-7)*2)/((b-7)*2/(b+7)*2)

=1

hope it helps.

Answer 3

equals 1.