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

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

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

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

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

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

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

$rArr 1/(d+4)$

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