# Polynomial Graph Question Whose Solution Does Not Make Sense

## The question is:

How many roots of the equation $$3x^4 +6x^3 + x^2 +6x +3$$ are real?

## The textbook's solution is:

Let $$f(x) = 3x^4 +6x^3 + x^2 +6x +3$$ .
Now $$f'(x) =12x^3 +18x^2 +2x +6\\ \implies f''(x) = 36x^2 +36x +2 ≠ 0 \; \forall x \in \mathbb{R}$$ So, graph of $$f'(x)$$ intersects $$x$$-axis only once.
Hence, $$f(x)$$ has only one turning point.
So $$f(x)=0$$ has maximum two real roots.
Now $$f(0)=3>0$$ and $$f(-1)=-5<0$$
So graph cuts $$x$$-axis between $$-1$$ and $$1$$.
Hence $$f(x)=0$$ has two real roots.

## My Question:

How on earth does $$f''(x)$$ have no real solutions? Using the quadratic formula I get $$2$$ real negative roots. Am I missing something here? Also how does the fact that the graph cuts the $$x$$-axis between $$1$$ and $$-1$$ imply that there are $$2$$ real roots? I just started reading the portion in the textbook on Graphs Of Polynomial Functions and I am in $$12$$th grade so if you could, please explain in a little simple way thank you. Also sorry if the formatting is weird this is the first time I have asked a question here :) .

• Well, $f''(x)$ certainly has real roots, so that's wrong. Which means several of the subsequent implications made are wrong, even though some conclusions happens to be correct. As $f\to \infty$ on either side of the $x$ axis, it will have an even number of roots (counting multiplicities), so if it cuts the axis anywhere once, it must have a minimum of two real roots. Commented Jun 10, 2021 at 8:00
• Alright thanks a lot! Was breaking my head over this question (I hate leaving questions unsolved). Thanks again. Commented Jun 10, 2021 at 8:37
• @Macavity what do you mean by counting multiplicities? I was aware of the fact that the end behaviour of a polynomial depends on the degree and the nature of the coefficient of the highest term. Does it depend on the multiplicity of the roots as well? What I know is that multiplicity of a root is the power of that root in the factorization of the polynomial? Is my definition wrong? Commented Jun 10, 2021 at 8:53
• Yes. However if you look at the polynomial $(x-1)^4$, it has a root of multiplicity $4$ at $x=1$. So even though it "touches" the x-axis, there is only one distinct root, unless you count the multiplicity. Commented Jun 10, 2021 at 9:01
• Ah ok that makes sense. Thanks for helping me! Commented Jun 10, 2021 at 9:35

Since you have sorted out the source of misunderstanding, here I offer an alternative solution, which quite surprisingly* allows you to find the exact roots of the polynomial while avoiding calculus. The method is (much) longer, but I find it interesting and I hope you will too.

(*I say surprisingly because mostly when solving quartic equations we don't have a hope of finding the exact roots without using the quartic formula.)

## Alternative Solution

We begin with a key observation: when a polynomial's coefficients are mirror images of each other about the centre of the polynomial, then the roots of the polynomial come as pairs of reciprocals of each other. For example, the roots of the polynomial $$2x^2+5x+2$$ are $$-2$$ and $$-0.5$$ which are reciprocals of each other.

If you would like me to, I can prove this for you, but it may be a good exercise to try it for yourself.

Using this fact, we may now observe that your polynomial satisfies this property: the coefficients are unchanged if we 'reflect' them about the middle term.

Therefore, since the polynomial is a quartic it has $$4$$ roots and its roots must be of the form $$-\alpha,-\frac{1}{\alpha},-\beta, -\frac{1}{\beta}.$$ Using the Factor theorem, we can now see that your polynomial can be factorised in the following way: \begin{align}3x^4+6x^3+x^2+6x+3&\equiv 3(x+\alpha)\left(x+\frac{1}{\alpha}\right)(x+\beta)\left(x+\frac{1}{\beta}\right)\\ &\equiv 3\left(x^2+x\left(\alpha+\frac{1}{\alpha}\right)+1\right)\left(x^2+x\left(\beta+\frac{1}{\beta}\right)+1\right)\\ &\equiv 3x^4+3x^3\left(\alpha+\frac{1}{\alpha}+\beta+\frac{1}{\beta}\right)+3x^2\left(2+\left(\alpha+\frac{1}{\alpha}\right)\left(\beta+\frac{1}{\beta}\right)\right)\\ &\quad+3x\left(\alpha+\frac{1}{\alpha}+\beta+\frac{1}{\beta}\right)+3 \end{align} Divide both sides by $$3$$ and for simplicity's sake let $$a=\alpha+\frac{1}{\alpha},\quad b=\beta+\frac{1}{\beta}$$ and we find that $$x^4+2x^3+\frac{1}{3}x^2+2x+1\equiv x^4+x^3(a+b)+x^2(2+ab)+x(a+b)+1.$$ What we really have here is a pair of simultaneous equations: \begin{align} a+b&=2\iff a=2-b\\ 2+ab&=\frac{1}{3} \end{align}

Replacing the $$a$$ in the $$2$$nd equation by the $$a$$ in the $$1$$st equation, we are left with a quadratic in $$b$$:

\begin{align} 2+b(2-b)=\frac{1}{3}&\implies b^2-2b-\frac{5}{3}=0\\ &\implies b=\frac{3\pm 2\sqrt6}{3} \end{align} The nice thing about this solution is that we then find that $$a=\frac{3\mp 2\sqrt6}{3}.$$ Therefore, without loss of generality we can let $$a=\frac{3+2\sqrt6}{3},\quad b=\frac{3- 2\sqrt6}{3}.$$ We now have $$2$$ quadratic equations that allow us to find the exact roots of the quartic polynomial!!

\begin{align}&\frac{3+2\sqrt6}{2}=\alpha+\frac{1}{\alpha}\implies \alpha^2-\alpha\left(\frac{3+ 2\sqrt6}{2}\right)+1=0\\ &\frac{3-2\sqrt6}{2}=\beta+\frac{1}{\beta}\implies \beta^2-\beta\left(\frac{3-2\sqrt6}{2}\right)+1=0 \end{align} Now all you need to find the values of $$-\alpha,-\frac{1}{\alpha},-\beta, -\frac{1}{\beta}$$ is to solve these two quadratic equations, and indeed only one of them has real roots (which you can find exactly using the quadratic formula).

I hope that was helpful, or you at least found it interesting. Sorry for the length, I've just tried to make it is clear as possible. If you have any questions please don't hesitate to ask :)

• That was amazing! I really can't thank you enough for providing such an amazing solution to this problem! Just one question, what do you mean by without loss of generality? Commented Jun 10, 2021 at 12:22
• @Anili My pleasure, I'm so glad you liked it! :) It's always lovely when someone appreciates your work. 'Without of loss of generality' means that we're not losing any validity for our solution by scrapping an arbitrary assumption or condition. In this case, since $a$ or $b$ could be equal to $(3+2\sqrt6)/3$, we're not losing any validity in our solution by letting $a=(3+2\sqrt6)/3$ and letting $b$ equal something else. Is that any clearer? Wikipedia actually has an article on it that may help. en.wikipedia.org/wiki/Without_loss_of_generality Commented Jun 10, 2021 at 12:29
• Yes that helps! Thanks again for the really interesting solution! Commented Jun 10, 2021 at 12:32
• @Anili You are really welcome! If you found the initial observation about the 'mirroring' of the coefficients interesting you may want to check out this Wikipedia page too. en.wikipedia.org/wiki/Reciprocal_polynomial Very nice $1$st question by the way, I hope you continue to use and benefit from this site. Commented Jun 10, 2021 at 12:34
• Thanks for all the info! I'm preparing for IIT Advanced (an exam which about 200k students appear in and only 10k get selected), an exam to get into colleges in India and any and all info helps. I didn't even expect my question to be answered and instead it was answered in a mind blowing way. Thanks so much for everything! Commented Jun 10, 2021 at 12:45