what is the factorization of $121b^{4}-49$?\n(11b - 7)(11b - 7)\n(11b + 7)(11b - 7)\n(11b^{2}-7)(11b^{2}-7)\n…

Explanation:

Step1: Recognize difference - of - squares

The expression $121b^{4}-49$ is in the form $a^{2}-b^{2}$, where $a = 11b^{2}$ ($(11b^{2})^{2}=121b^{4}$) and $b = 7$ ($7^{2}=49$).

Step2: Apply difference - of - squares formula

The difference - of - squares formula is $a^{2}-b^{2}=(a + b)(a - b)$. Substituting $a = 11b^{2}$ and $b = 7$ into the formula, we get $121b^{4}-49=(11b^{2}+7)(11b^{2}-7)$.

Answer:

$(11b^{2}+7)(11b^{2}-7)$