How do you solve #(10h^2 + 21h - 10) /( 12h^2 - 7h - 12 )div( 2h^2 + 9h + 10)/( 4h^2 + 11h + 6 )#?

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Answer 1 · 2016-08-30T01:53:35

The expression can be simplified to #(5h - 2)/(3h - 4)#, with the restrictions being #h!=-3/4, 4/3, -2, -5/2#

Make it into a multiplication:

#=> (10h^2 + 21h - 10)/(12h^2 - 7h - 12) xx (4h^2 + 11h + 6)/(2h^2 + 9h + 10)#

Factor everything:

#=> (10h^2 + 25h - 4h - 10)/(12h^2 - 16h + 9h - 12) xx (4h^2 + 8h + 3h + 6)/(2h^2 + 4h + 5h + 10)#

#=> ((5h(2h + 5) - 2(2h + 5)))/((4h(3h - 4) + 3(3h - 4))) xx ((4h(h + 2) + 3(h + 2)))/(2h(h + 2) + 5(h + 2))#

#=> ((5h - 2)(2h + 5))/((4h + 3)(3h - 4)) xx ((4h + 3)(h + 2))/((2h + 5)(h + 2))#

Cancel using the property #a/a = 1, a !=0#

#=>(5h - 2)/(3h - 4)#

Finally, state restrictions on the variable. They are: #h!= -3/4, 4/3, -2 and -5/2#.

Hopefully this helps!