How do you divide #(a+3)/(a^2+a-12)div (a^2-9)/(a^2+7a+12)#?

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Answer 1 · 2016-11-29T20:29:24

#(a+3)/(a-3)^2#

(1) Using the rules of fractions, turn the second upside down ad change to multiply.

#(a+3)/(a^2+a-12)xx(a^2+7a+12)/(a^2-9)#

(2) Now factorise where possible

#(a+3)/((a+4)(a-3))xx((a+4)(a+3))/((a+3)(a-3))#

(3) Cancel where possible

#cancel((a+3))/((a+4)(a-3))xx((a+4)(a+3))/(cancel((a+3))(a-3))#

#1/((a+4)(a-3))xx((a+4)(a+3))/(a-3)#

#1/(cancel((a+4))(a-3))xx(cancel((a+4))(a+3))/(a-3)#

#=(a+3)/(a-3)^2#