Answer:
The simplified form is
.
is the excluded value for the given expression.
Step-by-step explanation:
Given:
The expression given is:
![\dfrac{2a^2-4a+2}{3a^2-3}](https://tex.z-dn.net/?f=%5Cdfrac%7B2a%5E2-4a%2B2%7D%7B3a%5E2-3%7D)
Let us simplify the numerator and denominator separately.
The numerator is given as ![2a^2-4a+2](https://tex.z-dn.net/?f=2a%5E2-4a%2B2)
2 is a common factor in all the three terms. So, we factor it out. This gives,
![=2(a^2-2a+1)](https://tex.z-dn.net/?f=%3D2%28a%5E2-2a%2B1%29)
Now, ![a^2-2a+1=(a-1)(a-1)](https://tex.z-dn.net/?f=a%5E2-2a%2B1%3D%28a-1%29%28a-1%29)
Therefore, the numerator becomes ![2(a-1)(a-1)](https://tex.z-dn.net/?f=2%28a-1%29%28a-1%29)
The denominator is given as: ![3a^2-3](https://tex.z-dn.net/?f=3a%5E2-3)
Factoring out 3, we get
![3(a^2-1)](https://tex.z-dn.net/?f=3%28a%5E2-1%29)
Now,
is of the form ![a^2-b^2=(a-b)(a+b)](https://tex.z-dn.net/?f=a%5E2-b%5E2%3D%28a-b%29%28a%2Bb%29)
So, ![a^2-1=(a-1)(a+1)](https://tex.z-dn.net/?f=a%5E2-1%3D%28a-1%29%28a%2B1%29)
Therefore, the denominator becomes ![3(a-1)(a+1)](https://tex.z-dn.net/?f=3%28a-1%29%28a%2B1%29)
Now, the given expression is simplified to:
![\frac{2a^2-4a+2}{3a^2-3}=\frac{2(x-1)(x-1)}{3(x-1)(x+1)}](https://tex.z-dn.net/?f=%5Cfrac%7B2a%5E2-4a%2B2%7D%7B3a%5E2-3%7D%3D%5Cfrac%7B2%28x-1%29%28x-1%29%7D%7B3%28x-1%29%28x%2B1%29%7D)
There is
in the numerator and denominator. We can cancel them only if
as for
, the given expression is undefined.
Now, cancelling the like terms considering
, we get:
![\dfrac{2a^2-4a+2}{3a^2-3}=\dfrac{2(x-1)}{3(x+1)}](https://tex.z-dn.net/?f=%5Cdfrac%7B2a%5E2-4a%2B2%7D%7B3a%5E2-3%7D%3D%5Cdfrac%7B2%28x-1%29%7D%7B3%28x%2B1%29%7D)
Therefore, the simplified form is ![\dfrac{2(x-1)}{3(x+1)}](https://tex.z-dn.net/?f=%5Cdfrac%7B2%28x-1%29%7D%7B3%28x%2B1%29%7D)
The simplification is true only if
. So,
is the excluded value for the given expression.