Question

factor completely? a^4-50a^2b^2+49b^4

Answer 1

Let \(\Large m = a^2\) and \(\Large n = b^2\)


\[\Large a^4-50a^2b^2+49b^4\]


\[\Large (a^2)^2-50a^2b^2+49(b^2)^2\]


\[\Large m^2-50mn+49n^2\]


\[\Large (m-49n)(m-n)\]


\[\Large (a^2-49b^2)(a^2-b^2)\]


\[\Large (a^2-(7b)^2)(a^2-b^2)\]


\[\Large (a-7b)(a+7b)(a^2-b^2)\]


\[\Large (a-7b)(a+7b)(a-b)(a+b)\]


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So \[\Large a^4-50a^2b^2+49b^4\] completely factors to \[\Large (a-7b)(a+7b)(a-b)(a+b)\]