How did we know we can factor the following polynomial?

Question

Here is the equation I have $x^4 - 2 x^3 - 3x^2 + 4x + 2 = 0,$ the book said we can factor it to obtain $(x^2 - 2)(x^2 - 2x -1) = 0$ but I do not know how is this true, the rational root test does not say that $\sqrt{2}$ is a root. Can someone explain to me which theorem say this please?

Answer 1

If the rational root theorem fails, one potential route is to "conclude" (perhaps falsely, but nonetheless proceed as if you have, since you might as well try something) that you can only factor into two "nice" quadratics. Since the original polynomial is monic, we can go ahead and say, for some $a,b,c,d$, $$ x^4 - 2 x^3 - 3x^2 + 4x + 2 = (x^2 + ax + b)(x^2 + cx + d) $$ Expand the right-hand side, and compare coefficients. You'll get a system of equations: $$\begin{cases} a+c &= -2 \\ ac+b+d &= -3 \\ ad+bc &= 4 \\ bd &= 2 \end{cases}$$ From here, you should try to solve the system of equations. Of course, this is easier said than done, as you have no reason to suspect, a priori, that $a,b,c,d$ are "nice". However, again assuming they are nice (say, integers) should lead to the desired conclusion in this case.

For instance, if $a,b,c,d$ are integers, then $(b,d) \in \{(1,2),(2,1),(-1,-2),(-2,-1)\}$, and you can try to solve the system from there. Or you could use the fact that $a+c=-2$ to derive a substitution and perhaps clean the system up a bit. However you prefer to solve, you should arrive at the factorization with some trial and error.

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Other routes than assuming the factorization into quadratics may exist, of course. Such as:

• You could try to see if this polynomial matches any familiar forms. This doesn't seem to, though, and this is a skill more honed by experience than anything. (In general, for instance, you might easily notice the binomial coefficients arising - not in this case though.)

• The quartic formula, but this is painful. You could try to find roots via any other method, too, that might seem plausible: numerical methods like Newton's method, for example. (They can at least provide guesses for roots that are easily checked.) From there, you could divide by $x-(\text{that root})$ and get a perhaps-simpler cubic, or iterate that method.

• More obscure tricks/tools of note may be Ferrari's method for quartics, or Descartes' rule of signs and Sturm's theorem for discerning the types or locations of roots, if you're interested.

Answer 2

Any polynomial can be reduced to the depressed form with a linear substitution. In this case, the substitution $\,x = z + \dfrac{1}{2}\,$ results in a simple biquadratic, which is straightforward to factor:

$$ \begin{align} x^4 - 2 x^3 - 3x^2 + 4x + 2 &= \left(z+\frac{1}{2}\right)^4 - 2 \left(z+\frac{1}{2}\right)^3 - 3\left(z+\frac{1}{2}\right)^2 + 4\left(z+\frac{1}{2}\right) + 2 \\ &= z^4 - \frac{9}{2} z^2 + \frac{49}{16} \\ &= \frac{1}{16} \left(4 z^2 - 9 - 4 \sqrt{2}\right) \left(4 z^2 - 9 + 4 \sqrt{2}\right) \end{align} $$

Recognizing that $\,9 \pm 4 \sqrt{2} = \left(2 \sqrt{2} \pm 1\right)^2\,$, each factor can be written as a difference of squares:

$$ \begin{align} \frac{1}{16} \left(4 z^2 - 9 - 4 \sqrt{2}\right) \left(4 z^2 - 9 + 4 \sqrt{2}\right) &= \frac{1}{16}\left((2z)^2 - \left(2 \sqrt{2} + 1\right)^2\right)\left((2z)^2 - \left(2 \sqrt{2} - 1\right)^2\right) \\ &= \frac{1}{16} \color{blue}{\left(2z-2\sqrt{2}-1\right)}\color{brown}{\left(2z+2\sqrt{2}+1\right)} \\ &\qquad\;\;\; \color{brown}{\left(2z-2\sqrt{2}+1\right)}\color{blue}{\left(2z+2\sqrt{2}-1\right)} \end{align} $$

Substituting back $\,2z = 2x - 1\,$ and grouping the colored pairs produces the given factorization.

Answer 3

I guess there's a trick specifically for seeking roots of integer-coefficient quartics of the form $\sqrt m$ where $m$ is a nonsquare rational number. Note that if $\sqrt m$ is a root of $ax^4+bx^3+cx^2+dx+e$, then it must separately be a root of both $ax^4+cx^2+e$ and $bx^3+dx$. (This is because $$ a\sqrt m^4+b\sqrt m^3+c\sqrt m^2+d\sqrt m+e = (am^2+cm+e) + (bm+d)\sqrt m $$ can equal $0$ only if both $am^2+cm+e=0$ and $bm+d=0$, since $\sqrt m$ is irrational.) In particular, the roots of $bx^3+dx$ (other than $x=0$) are $\pm\sqrt{-d/b}$, and so one can plug this into $ax^4+cx^2+e$ to see if it happens to be a root as well. For the polynomial in the OP, this works because $\pm\sqrt2$ are the nonzero roots of $-2x^3+4x$ and just happen to be roots of $x^4-3x^2+2$ as well.

To phrase this as a one-line test for such roots of integer-coefficient quartics: $$ \text{If } bd\ne0 \text{ and } ad^2-cbd+eb^2=0 \text{, then } \pm\sqrt{-d/b} \text{ are roots of } ax^4+bx^3+cx^2+dx+e. $$