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\left(11y^{4}-7x\right)\left(11y^{4}+7x\right)
Rewrite 121y^{8}-49x^{2} as \left(11y^{4}\right)^{2}-\left(7x\right)^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).
\left(-7x+11y^{4}\right)\left(7x+11y^{4}\right)
Reorder the terms.