# Can I find a non-trivial cubic polynomial with rational zeroes *and* rational turning points?

Maybe it's staring me in the face and I can't see it, but I wanted a problem for my students to be able to sketch a polynomial function, intercepts, turning points, and all. Preferably none of the zeroes are turning points, to make for a more interesting sketch.

• Please show us what you can, then we may guess if you can find such a cubic polynomial. Joke aside now, well, please provide context! math.meta.stackexchange.com/questions/9959/… This is a must, and must be so because you have students, so let us apply the same rules... Feb 28 at 1:34
• Is $a(a-3)(a-8)$ good for instance? Feb 28 at 1:39
• By "turning points," do you mean the local minima/maxima? Feb 28 at 7:48
• Take any rational $p, q$, then $(x-p)(x-q)^2$ would give you a monic example... Mar 1 at 14:35

## 3 Answers

$$a,b,c \in \mathbb Q.$$

$$(x-a)(x-b)(x-c)=0$$

$$x^3-(a+b+c)x^2+(ab+ac+bc)x-abc=0$$

$$3x^2-2(a+b+c)x+(ab+ac+bc)=0$$

$$x=\frac{(a+b+c)\pm \sqrt{(a+b+c)^2-3(ab+ac+bc)}}{3}$$

$$w=a+b+c$$

$$z=ab+ac+bc$$

$$x=\frac{w \pm \sqrt{w^2-3z}}{3}$$

So find $$w,z \in \mathbb Q$$.

$$p,q,m,n \in \mathbb N .w=p/q. z=m/n.$$

$$\frac{p^2}{q^2}-3\frac{m}{n}=r^2/s^2$$

$$ns^2p^2-3mq^2s^2=nr^2q^2$$

Let $$n=q=1$$

$$s^2p^2-3ms^2=r^2$$

Let $$r=ks, k \in \mathbb N$$.

$$p^2-3m=k^2$$

$$(p-k)(p+k)=3m$$

Let $$p=k+3, p+k=m\implies m=2k+3$$

So it looks like setting a value of $$k$$ can be used to give you what you seek.

Keep it simple, let $$k=1$$.

Then $$p=4$$ and $$m=5$$ $$\implies w=4, z=5$$.

Now find $$a,b,c$$ so that

$$a+b+c=4$$

$$ab+ac+bc=5$$

$$c=4-(a+b)$$

$$ab+(a+b)(4-a-b)=5$$

$$ab+4a-a^2-ab+4b-ab-b^2=5$$

$$-a^2+4a+4b-ab-b^2-5=0$$

$$-a^2+a(4-b)+(4b-b^2-5)=0$$

$$a=\frac{(b-4)\pm\sqrt{(4-b)^2+4(4b-b^2-5)}}{-2}$$

$$16+b^2-8b +16b-4b^2-20=-3b^2+8b+16$$

$$-3b^2+8b+16=j^2/l^2$$

$$-3b^2+8bl^2+(16l^2-j^2)=0$$

$$b=\frac{-8l^2\pm \sqrt{64l^4+12(16l^2-j^2)}}{-6}$$

$$64l^4+192l^2-12j^2=4(16l^4+48l^2-3j^2)$$

$$16(l^4+3l^2+9/4-9/4)-3j^2=16l^4+48l^2+36-36-3j^2$$

$$(4l^2+6)^2-3(12+j^2)$$

Now we need to select $$l,j \in \mathbb N$$ so that the expression produces a perfect square.

$$16l^2(l^2+3)-3j^2$$

Let $$l=1, j=4\implies b=8/3$$

$$-3b^2+8b+16=-3(64/9)+64/3+16=16$$

$$a=\frac{(b-4)\pm \sqrt{-3b^2+8b+16}}{-2}\in\{-4/3, 8/3 \}$$

Let's chose $$a=-4/3$$.

$$-4/3+8/3+c=4\implies c=8/3$$

So $$f(x)=(x+4/3)(x-8/3)(x-8/3)=(3x+4)(3x-8)^2/27$$

I think that might do it.

I suggest a frame change.

What you care about is that the students understand the relationships among intercepts (on both axes) and critical points. The algebra required to find those points matters less. An approximate sketch can show that knowledge. So ask questions where such an approximation can provide an answer. For example

What can you deduce about roots, turning points and inflection points of a cubic if you know there's a root at $$2$$ and a zero derivative at $$(-1, -1)$$? Can such a cubic have one, two or three real roots?

So, you want integers, $$a,b,c$$ such that $$a^2+b^2+c^2-ab-ac-bc$$ is a positive perfect square. There are lots of those,

0 3 8
0 6 16
1 4 9
1 6 22
1 25 10
4 25 20
8 24 14
...


As long as you divide those roots by a perfect square you, you will retain rational second derivatives.

• how do you get those if you are aiming for that perfect positive square to equal some $n^2$? Feb 28 at 2:46
• The formula given is just the discriminate of the second derivative. It is a perfect square if and only if the roots are rational. Dividing by a perfect square does not change the discriminate being a perfect square. I wrote a quick script to produce a bunch of answers for me and then posted some for the OP. The OP indicated they wanted some examples. Feb 28 at 8:14