did I solve these problems correctly stating the exclued values

Question

 

 

simplify the expression if possible and state the excluded values.

 

1.

 

5x - 5/10x^2 - 25x + 15

 

= 5(x - 1)/

   5(2x^2 -5x + 3)

 

= 5(x-1)  (1)/

  -5(x-1)  (2x -3)

 

=  1/2x-3

 

exclued values are 5,-5, -1, 1

  

 

2.

 

 w^2 + 6w + 9/

 2w^2 - 18

 

= ( w + 3) (w + 3)/

 -2( w + 3) (w - 3)

 

= w + 3/

   2w -3

 

 -3, 3 are exclued values

 

 

Did I solve both these problems correctly

Answer 1

1.

(5x - 5) / (10x^2 - 25x + 15) =

 

5(x - 1) / (5(2x^2 - 5x + 3)) =

 

(x - 1) / ((2x - 3)(x - 1)), note that (x-1)/(x-1) is 1 for all x ≠ 1, but if x = 1 then (x-1)/(x-1) = 0/0. Since we can't divide by 0 we have to exclude x = 1.

 

= 1 / (2x - 3), x ≠ 1 AND x ≠ 3/2 so 2x - 3 isn't 0.

 

2.

 

(w^2 + 6w + 9) / (2w^2 - 18) =

 

(w + 3)^2 / (2(w^2 - 9)) =

 

(w + 3)^2 / (2(w - 3)(w + 3)), x ≠ -3 AND x ≠ 3, then

 

= (w + 3) / (2(w - 3)), x ≠ -3 AND x ≠ 3

Answer 2

Not Really:

 5x  - 5/10 X^2 - 25X + 15 =

 

  (- 1/2)X^2  - 20 X + 15 =

 

  - 1/2 ( X^2 -40X - 30) =

 

    -1/2[ ( X- 20) ^2 - 430]=

 

      -1/2 ( x -20 +√430 ) ( X -20 - √430)

 

 

2.   W^2 + 6W + 9 = ( w +3 )^2

 

       2 W^2 - 18 = 2 ( W^2 - 9) = 2 ( w + 3) ( W - 3)

Answer 3

#2 is correct,

 

#1 is close.  You factor out a positive 5 first in the denominator, not -5.  Re-evaluate what excluded value's imply.  They are any values of x in the denominator that result in a factor of 0.  Therefore, solve for x:

 

(2x-3) = 0

 

(x-1) = 0

 

*EDIT*

This is based on the assumption that you left out parentheses denoting a numerator of 5x-5, and a denominator of 10x^2 - 25x + 15, which would indicate values of x in the denominator which result in a 0 denominator, or so called 'excluded values.'