How do you simplify #\frac { 5a ^ { 2} - 45} { 3a ^ { 2} - 12a } \div \frac { a ^ { 3} + 3a ^ { 2} } { 4a ^ { 2} - 16a } \cdot \frac { 21a ^ { 3} + 6a ^ { 2} } { 4a ^ { 2} - 12a }#?

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Answer 1 · 2018-01-14T13:33:07

#(5(7a+2))/a#

#(5a^2-45)/(3a^2-12a) * (4a^2-16a)/(a^3+3a^2)*(21a^3+6a^2)/(4a^2-12a)#

(dividing by a fraction means multiplying by its reciprocal)

#3a^2-12a = 3a(a-4)#
#4a^2-16a = 4a(a-4)#

#(5a^2-45)/(3acancel((a-4))) * (4acancel((a-4)))/(a^3+3a^2)*(21a^3+6a^2)/(4a^2-12a)#

#=(5a^2-45)/(3a) * (4a)/(a^3+3a^2)*(21a^3+6a^2)/(4a^2-12a)#

#=(5a^2-45)/(3) * (4)/(a^3+3a^2)*(21a^3+6a^2)/(4a^2-12a)#

#a^3+3a^2 = a^2(a+3)#
#5a^2-45 = 5(a^2-9) = 5(a+3)(a-3)#

#(5cancel((a+3))(a-3))/(3) * (4)/(a^2cancel((a+3)))*(21a^3+6a^2)/(4a^2-12a)#

#=(5(a-3))/(3) * (4)/(a^2)*(21a^3+6a^2)/(4a^2-12a)#

#4a^2-12a = 4a(a-3)#

#(5cancel((a-3)))/(3) * (4)/(a^2)*(21a^3+6a^2)/(4acancel((a-3))#

#=(5)/(3) * (4)/(a^2)*(21a^3+6a^2)/(4a)#

#21a^3+6a^2 = 3a^2(7a+2)#

#(5)/(cancel3) * (4)/(cancel(a^2))*(cancel3cancel(a^2)(7a+2))/(4a)#

#=5/1 * 4/1 * (7a+2)/(4a)#

#5/1 * cancel4/1 * (7a+2)/(cancel4a) = (5*(7a+2))/a#

#=(5(7a+2))/a#