How do you simplify #(2d-7)/(2d^2+d-28)#?

Question

1478 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-17T11:00:08

The answer is #=1/(d+4)#

Let's factorise the denominator

#2d^2+d-28=(2d-7)(d+4)#

Therefore,

#(2d-7)/(2d^2+d-28)=cancel(2d-7)/(cancel(2d-7)(d+4))#

#=1/(d+4)#

Answer 2 · 2017-01-17T11:01:57

#1/(d+4)#

Given, #(2d - 7)/[2d^2 +d-28]#

#rArr(2d-7)/[2d^2+(8-7)d-28]#

#rArr (2d-7)/[2d^2+8d-7d-28]#

#rArr(2d-7)/[2d(d+4)-7(d+4)]#

#rArr(2d-7)/[(2d-7)(d+4)]#

#rArr cancel(2d-7)/[cancel{(2d-7)}(d+4)]#

#rArr 1/(d+4)#