Question

Simplify.

Answer 1

\[5(a2+5a+6)/3(a^2-49)(a-7) \div 61(a+3)/\]

Answer 2

\[\frac{5(a^2+5a+6}{3(a^2-49)(a-7)} \div \frac{61a+3}{missing}\]Something like that?

Answer 3

factorise the quadratic term in the numerator of the first fraction
\[(a^2+5a+6)\rightarrow(a+2)(a+3)\],

Answer 4

oh my bad on the missing part it's [div] 61(a+3) / 31(a+7)

Answer 5

\[\frac{5(a^2+5a+6)}{3(a^2-49)(a-7)} \div \frac{61(a+3)}{31(a+7)}\]\[=\frac{5(a+2)(a+3)}{3(a+7)(a-7)(a-7)} \times \frac{31(a+7)}{61(a+3)}\]\[=\]