Question

16 | vietas theorem.
20.f
write each binomial as a product.
-9 + 121m⁸n⁸
answer

Explanation:

Step1: Rearrange the binomial

Rewrite \(-9 + 121m^{8}n^{8}\) as \(121m^{8}n^{8}-9\).

Step2: Identify as difference of squares

Notice that \(121m^{8}n^{8}=(11m^{4}n^{4})^{2}\) and \(9 = 3^{2}\), so it's a difference of squares \(a^{2}-b^{2}\) where \(a = 11m^{4}n^{4}\) and \(b = 3\).

Step3: Apply difference of squares formula

The formula for difference of squares is \(a^{2}-b^{2}=(a + b)(a - b)\). Substituting the values, we get \((11m^{4}n^{4}+3)(11m^{4}n^{4}-3)\).

Answer:

\((11m^{4}n^{4}+3)(11m^{4}n^{4}-3)\)