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Factor $9x^4-37x^2+4$ using difference of cubes, sum of cubes, difference of squares, substitution, grouping or common factoring, whichever is simplest method.

Usually factoring isn't to difficult for me, but for some reason I'm stuck on this one. I would usually use the factor theorem but the question specifically states to use one of the methods listed. And the only one this equation kind of works with is substitution, at least to my eyes, but I'm having trouble with doing it that way and I can't figure any other way to do it besides the factor theorem.

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  • $\begingroup$ Presumably "usually factoring isn't too difficult for me, [but I can't quite see how to do this one]" was a standard line of would-be court mathematicians in the early 16th century, when solving cubic and quartic equations was an unsolved problem. I imagine that Cardano, Tartaglia, Ferrari and their contemporaris would not have had much trouble with your example (a quartic that is actually quadratic in $x^2$). $\endgroup$ – Rob Arthan Feb 22 at 1:55
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You have an $x^4$, an $x^2$ and a constant term. Therefore substitute $u = x^2$ to get the quadratic $9u^2-37u+4$, which factors as $(9u - 1)(u - 4) = (9x^2 - 1)(x^2 - 4)$.

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As Toby said, you can factor it down to $(9x^2 - 1)(x^2 - 4)$. However, note that both of these factors are themselves differences of squares. So, you may factor further to get $(3x-1)(3x+1)(x-2)(x+2)$ over the integers. If you allow rational numbers, you can pull out a factor of $3$ from the first $2$ factors to get $9(x-\frac13)(x+\frac13)(x-2)(x+2).$

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