Answer:
![x^(2) -1](https://img.qammunity.org/2020/formulas/mathematics/middle-school/sdn30c3yinrb5lsb4qhcpk9pk8qo43guji.png)
Explanation:
we know that
Every difference of squares problem can be factored as follows:
![a^(2)-b^(2)=(a+b)(a-b)](https://img.qammunity.org/2020/formulas/mathematics/high-school/1rvptfxczq4m33jszcjnw15bghp3jsmltg.png)
If the polynomial represent a difference of squares every number must be a perfect square (Remember that a number is a perfect square if its square root is an integer.)
Verify each case
case 1) we have
![x^(2) -1](https://img.qammunity.org/2020/formulas/mathematics/middle-school/sdn30c3yinrb5lsb4qhcpk9pk8qo43guji.png)
In this case both numbers are perfect square
so
![x^(2) -1=(x+1)(x-1)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/yiebfpicd0akyyobg64qgdsttvymk2stsh.png)
therefore
The polynomial represent a difference of squares
case 2) we have
![x^(2) -8](https://img.qammunity.org/2020/formulas/mathematics/middle-school/rfx7szifyc2hded6wy1qyj8pqgkpcnv32d.png)
In this case 8 is not a perfect square
therefore
The polynomial not represent a difference of squares
case 3) we have
![4x^(2) +16](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ui6clboul4immkqqz7m4rt0f50kl4iiuui.png)
![4x^(2)+16=4(x^(2)+4)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/y5mzupa5yv0lt33uyyfxkfp6t77gvt35m5.png)
In this case both numbers are perfect square
but is a sum of squares
therefore
The polynomial not represent a difference of squares
case 4) we have
![9x^(2)-18](https://img.qammunity.org/2020/formulas/mathematics/middle-school/1cieyizo9updh54rkoe3zewfsyzjgmjana.png)
![9x^(2)-18=9(x^(2)-2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/m0k1i6ef4c8k76yp0vpu8zgmzogd4jmqug.png)
In this case 2 is not a perfect square
therefore
The polynomial not represent a difference of squares