How do you simplify #(v^2 + v - 12)/ (v^2 + 6v + 8) div (2v - 6)/( v +2)#?

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Answer 1 · 2016-07-06T11:58:54

#1/2#

the trynomial

#v^2+(v_1+v_2)v+(v_1*v_2)v#

is obtained by the product:

#(v+v_1)(v+v_2)#

so

#v^2+v-12=(v+4)(v-3)#

and

#v^2+6v+8=(v+4)(v+2)#

So you can rewrite

#(v^2+v-12)/(v^2+6v+8)-:(2v-6)/(v+2)=(cancel((v+4))(v-3))/(cancel((v+4))(v+2))-:(2(v-3))/(v+2)#

#=cancel(v-3)/cancel(v+2)*cancel(v+2)/(2cancel((v-3)))=1/2#