## Q&Q – Vertex Form

Following up yesterday’s post, I’ve spent some time with factoring quartics as the composition of two quadratics, and looking at the quadratics in vertex form to get a better graphical understanding of the problem. Also, yesterday I mentioned the apparent dis-symmetry in the freedom allowed when choosing $a_2$ as one of the free parameters, versus choosing $a_1$. I realized that came from the assumption that the leading coefficient for the quartic was positive, and probably has no further bearing on the problem.

So, let’s get back to it, as simple as possible. Our overall goal is to write $f=x^4+bx^3+cx^2+dx$ as the composition $p_1\circ p_2$ where $p_1$ and $p_2$ are quadratics. Since we now know we have enough freedom in choosing some of the coefficients, without losing any power, let’s go ahead and assume that both $p_1$ and $p_2$ are monic. I think quadratics are easier to visualize in vertex form, so write $p_i=(x-h_i)^2+k_i$, $i=1,2$ (so the vertex for $p_i$ is at $(h_i,k_i)$).

Since we’re also allowed to freely choose the constant term for $p_2$, let’s choose it to be 0. This has several wonderful implications. The constant term for $p_2$ is $h_2^2+k_2$, so if this is 0, we can set $k_2=-h_2^2$ to eliminate a parameter. Now that the constant term for $p_2$ is 0, and since we’re assuming the constant term for the composite is also 0, we see that the constant term for $p_1$ is also 0. Thus, $k_1=-h_1^2$, and we have written $p_i=(x-h_i)^2-h_i^2$. This means the vertices for the two parabolas lie on the parabola $y=-x^2$, which is somehow visually appealing. We can picture two parabolas opening up with vertices travelling around on $y=-x^2$, and make a nice movie of corresponding quartic compositions. Perhaps I’ll see what I can rig up in Maple, or something.

With these formulas for $p_1$ and $p_2$, some things are quickly apparent. It’s comforting to check that the relation $8d=4bc-b^3$ (from yesterday) is satisfied by the coefficients of the composite $p_1\circ p_2$ (here, $b,c,d$ are the coefficients of $x^3,x^2,x$ in the composite). Next, notice that

$(p_1\circ p_2)(x+h_2)=((x+h_2-h_2)^2-h_2^2-h_1)^2-h_1^2=(x^2-(h_2^2+h_1))^2-h_1^2$

which is an even function (symmetric about the $y$-axis). Since $g(x+h_2)$ is just a horizontal shift of $g(x)$ (shifting left $h_2$ units), this means that $p_1\circ p_2$ is symmetric around $x=h_2$. Unless I’ve got some implications wrong somewhere, I’m pretty sure what we’ve just shown is that a quartic can be written as the composite of two quadratics precisely when it is symmetric around a vertical axis. But I’d like to double check that statement (or let you).

Suppose you had a fixed quartic $f=x^4+bx^3+cx^2+dx$ (with $8d=4bc-b^3$) and wanted to identify the $h_1$ and $h_2$ for the quadratics that we’re talking about. It’s easy to do, with the work from yesterday. Yesterday, we said that the linear coefficient of $p_2$ was equal to $b/2$. Thus, $-2h_2=b/2$, so $h_2=-b/4$. We also found that the linear coefficient of $p_1$ would be $c-b^2/4$ . So $-2h_1=c-b^2/4$, so $h_1=\frac{b^2}{8}-\frac{c}{2}$, which is $-d/b$ if $b\neq 0$. Anyway, the point is that the vertices for the quadratics can be easily picked out from the coefficients of the quartic you started with.

Finally, a nice thing about the constant terms being 0 for these two quadratics is that then $p_i=x^2-2h_ix=x(x-2h_i)$, so one of the roots is 0, and the other is $h_i$. So the composite, $p_1\circ p_2$ will be 0 when $p_2=0$ (so $x=h_2\pm h_2$) or when $p_2=2h_1$ (so $x=h_2\pm \sqrt{h_2^2+2h_1}$). It’s also not hard to find the zeros for the first and second derivatives for $p_1\circ p_2$. The first derivative is 0 at $h_2$ and $h_2\pm \sqrt{h_2^2+h_1}$, and the second derivative is 0 at $h_2\pm \sqrt{(h_2^2+h_1)/3}$. Of course, not all of these values in square roots need be positive, but whatever. The symmetry around $x=h_2$ is readily apparent.

So, I’m feeling good about factoring quartics as quadratics. I’ve got no real questions remaining, personally. If you do, please let me know. What’s next? Well, there’s no point trying to factor a quintic as a composite, because 5 is prime, so the next logical choice would be a sixth degree polynomial as a quadratic and a cubic. As fun as that sounds, I may leave it to somebody else. I’ve got another project in mind in relation to polynomials, but you’ll just have to come back and find out about that later.

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### One Response to “Q&Q – Vertex Form”

1. Finishing Quartics « Sumidiot’s Blog Says:

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