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rokotherodent

  • 3 years ago

Factor completely? y^2x - y^x - 20?

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  1. lgbasallote
    • 3 years ago
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    hmm you can let a = y^x first so \[y^{2x} - y^x - 20\] will become\[a^2 - a - 20\] i suppose you can solve for a there?

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