Question

Find the illegal values of b in the fraction

Answer 1

Answer:

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

Step-by-step explanation:  

We have been given a rational expression: $\frac{2b^2+3b-10}{b^2-2b-8}$ 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.