Constructing “nice” cubic polynomials for curve sketching.
(where
f
and f
all have integer zeros.)
(Taken from Tom Bruggeman, Xavier University and Tom Gush SUNY at Stony Brook)
Let
be a cubic polynomial with zeros 0, a and b (note: if one of the roots is NOT zero a simple translation can make it so).
x
3
2 abx then (1) f
3 x 2
2
ab and (2) f
6 x
2
.
Now if f
is to have nice integer roots (say c and d), then
(3) f
( x d )
3 x
2
3
c
3 cd
Comparing (1) and (3) yields
(4) 2
a b
c d
and (5) ab
3 cd .
From (5), one of a or b is divisible by 3 and because of symmetry suppose it is a.
From (4) (a + b) is divisible by 3, and so b must be divisible by 3 also.
Let a = 3A and b = 3B and (4) becomes
(6) 2( A + B ) = c + d and (5) becomes
(7) 3AB = cd
Eliminating A from the system (6) (7) yields..
2
Cd
B
B
3 C
d or
(8) 2 B 2
3 C
2 Cd
0
If B is to be an integer, the discriminant in the quadratic (8) above must be a perfect square
3 C
d
2
16 Cd
H
2 which can be rearranged to
(9) H
2
4 C
d
5 C
2
.
We know how to find all of the Pythagorean triples satisfying (9)
(10) 4 C
2 kst , H
2 t
2
, d
5 C
2 t
2
for integers s, t and any k.
From (8) B
3 C
H
Taking the plus sign (it turns out that choosing the minus sign gives the same result
4 with s and t interchanged) (10) gives… B
3 C
5 C
2 t
2
4
4 kst
2 ks 2
4
.
Putting k = 2r we get
From (10) C
rst and d
5 rst
2
2 t
2
r
2
s
2 t
Finally, substituting back we get a
3 rt
2 s
t b
3 rs s
2 t
c
3 rst d
r
2 s
t
s
2 t
The root of the second derivative is
3
2
4 st
t 2
The table below gives some examples of these “nice” cubics (I just used the third one down on a quiz I gave today)
Polynomial Roots of f Roots of f
Roots of f
x
3
3 x
2
4
1, 2, 2 0, 2 1 x
3
33 x
2
216 x
0,9,24 4,18 11 x
3
6 x
2
135 x
9, 0,15
5, 9 2 x 3
147 x
286
13, 2,11
7, 7 0 x
3
3 x
2
144 x
140 x
3
3 x
2
144 x
432
10, 1,14
12, 3,12
6,8
6,8
1
1
One thing worth mentioning…..don’t expect the function values at the relative extrema and P.O.I to be close to the x – axis.
For x
3
6 x
2
135 x , the relative max is at
5, 400
the relative min is at
and the P.O.I is at
I guess you can’t ask for everything to be “nice”.
Example: Let r = 1, t = 2 and s = 3. Then a = 48, b = 63, c = 18 and d = 56. This will yield
x 3
111 x
3024 x f
3 x 2
222 x
3024 f
6 x
222 roots of f
0, 48, 63 roots of f
18,56 and the root of f
37