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I am having problems factoring the following quartic with quadratic roots.

$$3x^4 + x^3 + 2x^2 - x + 1 = \left(x^2 + x + 1 \right) \left( 3x^2 - 2x + 1 \right)$$

All the ways I've found online till now only show to factor quartics including linear factors. None of them show the process to factor the equation with pure quadratic factors. Could you please suggest a way to do so?

Rodrigo de Azevedo
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VD-Flash
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  • Just write $(x^2+ax+b)(3x^2+cx+d)$, compute the product and compare coefficients. This yields an easy system of equations in $a,b,c,d$. We obtain $a=b=d=1$ and $c=-2$. – Dietrich Burde Apr 30 '23 at 15:53
  • The problem is that neither one of the quadratics is factorable with real coefficients. Therefore the quartic can’t be factorable further (with real coefficients). – WindSoul Apr 30 '23 at 17:31
  • @DietrichBurde Thanks for suggesting that post. But I was looking for a formal procedure which can be taught to students. Something like grouping the terms and then factoring them. I'm not sure if such operation is possible on these types of equations – VD-Flash Apr 30 '23 at 17:35
  • Of course it is possible, but with comparison of coefficients. See my first comment. – Dietrich Burde Apr 30 '23 at 18:06
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    Can you explain what you want from this formal procedure? I'm not sure I see how any additional rigor can be introduced, so are you looking for something like the quadratic formula, but for quartic expressions? Are you only interested in factors over the rational quadratics? – Brian Tung Apr 30 '23 at 18:07
  • In general I think you pretty much have to do something like @Dietrich Burde suggested. For examples and references, see "3. Quadratic Factors of $;x^4 + 10x^2 - ; 96x - 71$" in this MSE answer, where in this example we are a bit more generous with the possible coefficients of the quadratic factors by using a method that always produces real-number coefficients of the quadratic factors (modulo solving at-most cubic equations). – Dave L. Renfro Apr 30 '23 at 19:11

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