Easy 10 points. What is (are) the non-permissible value(s) of x^2-3x+2/x-1

A)1
B)2
C)1&2
D)0

2) What is (are) the non-permissible value(s) of x^2-x-2/x^2+3x+2?
A)-1,2
B)2,1
C)1,-2
D)-1,-2

Please explain how you found the answers. Thank you.

So, we have x-1 in the denominator, and because dividing by zero is undefined, if we set x-1=0, and solve for x we get x=1, so when x=1 the denominator is zero. Also, if we factor out the top we get
(x-1)x(x-2)/(x-1), and as we can see x-1 is in both the numerator and denominator, meaning that x=1 is a non permissible value. So, the answer is A

For the second problem, we factor the denominator into two separate factors. so we have:
x^2-x-2/(x+2)x(x+1), and as we said in the last problem, the denominator can't equal zero, so we have to find where the denominator is equal to zero in order to determine the non-permissible values of x.
So:
x+2=0
x=-2
and
x+1=0,
x=-1
So the answer is D. x cannot be -1 or -2 because that would make the denominator zero, and you cannot divide by zero.

What is (are) the non-permissible value(s) of x^2-3x+2/x-1?

A)1 .................Ans
B)2
C)1&2
D)0

2) What is (are) the non-permissible value(s) of x^2-x-2/x^2+3x+2?
A)-1,2
B)2,1
C)1,-2
D)-1,-2 ..............Ans

1 is A

2 is D

just remember you cant divide by 0 and solve the second part for zero. so for (1) x-1=0 so x is 1
for (2) (-2)^2 + 3(-2) + 2 = 0
4+ (-6) + 2 =0
than try it with (-1)

here's a hint- non permissible values occur when the value of x will make the denominator zero. For the second question you will need to factor the denominator.

1) (x - 1)(x - 2)/(x - 1) = x - 2

x ≠ 1, x ≠ 2

C

2) (x - 2)(x + 1)/[(x + 1)(x + 2)] = (x - 2)/(x + 2)

x ≠ -2, x ≠ -1

D