Find the illegal values of b in the fraction

Question

asked Jan 13, 2019 195k views

1 Answer

Maaw · Answer 1 · 2019-01-20T00:58:27+0000

Answer:

$b=4\text { or }b=-2$

Step-by-step explanation:

We have been given a rational expression:
(2b^2+3b-10)/(b^2-2b-8) and we are asked to find the illegal values of b.

Illegal values of b will be those values, which will make our denominator 0.

To solve for illegal or excluded values of b, let us set our denominator equal to 0.

b^2-2b-8=0

We will factor out our equation by splitting the middle term.

b^2-4b+2b-8=0

b(b-4)+2(b-4)=0

(b-4)(b+2)=0

$b-4=0\text { or }b+2=0$

$b=4\text { or }b=-2$

Therefore,
$b=4\text { or }b=-2$ are illegal values of b in the given fraction.