Factor completely \( y^{4}-81 \) \( \left(y^{2}-9\right)^{2} \) \( \left(y^{2}-9\right)\left(y^{2}+9\right) \) \( (y-9)^{4} \) \( \left(y^{2}+9\right)(y-3)(y+3) \)

To factor \( y^{4}-81 \) completely, we can recognize it as a difference of squares. It can be expressed as \( (y^2)^2 - 9^2 \), which factors to \( (y^2 - 9)(y^2 + 9) \). The \( y^2 - 9 \) part can be further factored, as it is also a difference of squares: \( (y - 3)(y + 3) \). So, the complete factorization is: \[ (y - 3)(y + 3)(y^2 + 9) \] Remember, \( y^2 + 9 \) does not factor further over the real numbers, maintaining its importance in the complex number realm! When factoring polynomials, a common mistake is to overlook the power of squares or the possibility of further factoring, such as missing that \( y^2 - 9 \) can break down into \( (y - 3)(y + 3) \). Always double-check for additional steps that can simplify your solution further!