## Reducing Quartics to Cubics

```Suppose we are given the length of the hypotenuse of a right triangle,
and we also know the sides of a rectangle inscribed in the triangle
as shown below
__
|\
|  \
|    \
|_____ \  H
Y   |     |  \
|     |    \
|    b|      \
__   |_____|________\
a
|<-----X------>|

Given a, b, and H, how can we determine the remaining dimensions of
the right triangle?  If Y denotes the height of the triangle, then
the equation of the hypotenuse is y = Y - (Y/X)x.  We also know
Y = sqrt(H^2-X^2), and when we plug "a" into the equation for the
line we must get "b".  Therefore,

b  =  sqrt(H^2 - X^2) (1 - a/X)

Squaring both sides and rearranging terms gives the quartic

X^4 - (2a)X^3 + (a^2+b^2-H^2)X^2 + (2aH^2)X - (a^2 H^2) = 0

For example, if a=11, b=5, and H=25 we have

X^4 - 22 X^3 - 479 X^2 + 13750 X - 75625  =  0

The Galois group of this polynomial is the fully symmetric group S_4,
so there's no hope of reducing the general problem to anything simpler
than the general solution of the quartic.  The roots of the above
polynomial give the four solutions

X             Y
-----------   -----------
23.10702596    9.54281674
14.59544931   20.29711456
9.05666503  -23.30186298
-24.75914031    3.46193175

These correspond to the four lines that pass through the point (a,b)
and such that the distance between the x-intercept and the y-intercept
is H.  Of course, the first two are the solutions that match the
geometrical shown above.

Interestingly, if we define Z = (Y - b)/a, we have the quartic

Z^4 + (2b/a)Z^3 + (1+(b/a)^2-(H/a)^2)Z^2 + (2b/a)Z + (b/a)^2 = 0

which is palindromic if a=b.  In The Fundamental Theorem of Palindromic
Polynomials we showed how a palindromic quartic of this kind can be
reduced to a resolvant quadratic, which explains why the problem is
easy if the rectangle is a square.

Returning to the original quartic

X^4 - (2a)X^3 + (a^2+b^2-H^2)X^2 + (2aH^2)X - (a^2 H^2) = 0      (1)

it occurs to me that the palindromic procedure can be applied to ANY
quartic of this type, not just those with a=b, reducing it to, at most,
a cubic.  Thus, it can be added to the well known methods of Ferrari and
Descartes for reducing quartics to cubics.

Given any quartic

Ax^4 + Bx^3 + Cx^2 + Dx + E = 0                      (2)

it was shown in the previous article that this can be converted to
"palindromic" form if E = A(D/B)^2.  Making the substitution x=z+k
we have a quartic in z whose coefficients A',B'...,E' are a function
of the parameter k.  To ensure that this polynomial can be converted
to palindromic form we need only impose the requirement that
E' = A'(D'/B')^2, which gives a cubic for k

p k^3  +  q k^2  +  r k  +  s  =  0                  (3)
where
p  =  (B^3 - 4ABC + 8DA^2)
q  =  (2ABD + 16EA^2 + CB^2 - 4AC^2)
r  =  (8ABE - 4ACD + DB^2)

With k equal to any root of this cubic we can substitute into the
general quartic (2), reducing it to a palindromic form with a

Of course, if E=A(D/B)^2 then s=0, and we have k=0 is a root of (3),
meaning that no shifting of the roots is required to make the original
equation (2) palindromic.  Also, note that the expression for "p" is
the same as the "discriminant" in the note on palindromic polynomials.
The sign of "p" determines whether the roots are real or complex
conjugates, and p=0 implies two equal real roots. In that case the
equation for k reduces to a quadratic, so again the solution is easy.

Finally, what about the case of equation (1) with a=b?  In that case,
none of the coefficients of the resolvant cubic (3) vanish, so it might
seem strange that this case can be easily solved.  However, when a=b it
turns out that the resolvant cubic (3) factors as

4 (a^2 + H^2) (a-k) (2ak^2 - H^ k + aH^2) = 0

so this gives the simple transformation given previously.
```