Test 7 Section 3 #15

If you recognize that \dfrac{1}{3}(x+k)(x-k) contains a factored difference of two squares, then you can work from there to get this one very quickly.

    \begin{align*}&\dfrac{1}{3}(x+k)(x-k)\\=&\dfrac{1}{3}\left(x^2-k^2\right)\\=&\dfrac{1}{3}x^2-\dfrac{1}{3}k^2\end{align*}

If you know that has to equal \dfrac{1}{3}x^2-2, then you know that \dfrac{1}{3}k^2=2. Which answer choice makes that so? Yep, k=\sqrt{6} does:

    \begin{align*}&\dfrac{1}{3}x^2-\dfrac{1}{3}k^2\\=&\dfrac{1}{3}x^2-\dfrac{1}{3}\left(\sqrt{6}\right)^2\\=&\dfrac{1}{3}x^2-\dfrac{1}{3}(6)\\=&\dfrac{1}{3}x^2-2\end{align*}

 

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