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Factor completely.2x^2-50

User Grrussel
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\begin{gathered} 2x^2-50=2(x^2-25) \\ \text{ Remember that a\textasciicircum{}2 - b\textasciicircum{}2 = (a-b)(a+b)} \\ =2(x^2-5^2) \\ =2(x-5)(x+5) \end{gathered}

Te polynomial factors completely as 2(x-5)(x+5)


\begin{gathered} 81-4x^2\text{, since 81 = }9^2\text{, we have} \\ 81-4x^2=9^2-4x^2\text{, but }4=2^2 \\ =9^2-2^2x^2=9^2-(2x)^2^{} \\ =(9-2x)(9+2x) \end{gathered}

User Conor Gallagher
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