# Could someone explain the solution to the problem in the screenshot?

This is from a past-years'-questions PDF for an Indian secondary school olympiad. Could someone explain the answer to question no. 6 shown in the picture?

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There are four roots so $a \neq 0$. Saying that the $\alpha_i$'s are the roots means that $$P(x) = a x^4 + b x^3 + x^2 + x + 1 = a(x - \alpha_1)(x - \alpha_2)(x- \alpha_3) ( x- \alpha_4).$$ Expand the right side so you can deduce by identification $$\prod_{i=1}^4 \alpha_i = \frac 1 a.$$ Now take the polynomial $P(x) = a x^4 + b x^3 + x^2 + x + 1$, divide by $x^4$ and set $y = 1/x$, you get the polynomial $$Q(y) = a + b y + y^2 + y^3 + y^4$$ and the roots of $Q$ are the inverse of the roots of $P$ by construction. So you know that $$Q(y) = (y - \beta_1) (y - \beta_2) (y - \beta_3) (y - \beta_4)$$ and when you expand you can deduce $$\sum_{i=1}^4 -\beta_i = 1 \ \ \text{ and } \ \sum_{1 \leq i < j \leq 4}^4 \beta_i \beta_j = 1$$ which correspond respectively to the coefficients of $y^3$ and $y^2$. Now with the last line of the proof, you know that $$\sum_{i=1}^4 \beta_i^2 < 0$$ so some of the $\beta_i$'s must be complex numbers.

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(+1) because this is the simplest/easiest to understand among all the answers. – achille hui Apr 13 '13 at 10:51
It took me a bit of time to see that "develop" means "expand" :) Nice, thanks - that was helpful. – Soham Chowdhury Apr 15 '13 at 6:00
Thanks. I used the french word for expand, I thought it was the same one in english :). – roger Apr 15 '13 at 11:47

The proof in the displayed solution has already been explained several times. I would just like to add a slightly more general statement (showing amongst others that the assumption $a\neq0$ is superfluous) and a (only very) slightly different point of view for the proof.

Proposition. No non constant real polynomial $P$ in $x$ whose lowest degree terms are $c_2x^2+c_1x+1$ with $2c_2\geq c_1^2$ can be factored as a product of real polynomials of degree$~1$.

Proof. In such a factorisation the product of all constant terms of the factors is the constant term $1$ of $P$, so none of those constant terms is $0$, and dividing every factor by its constant term does not change their product $P$, and makes all the constant terms equal to $1$. It will therefore suffice to show there is no decomposition of the form $$P=(a_1x+1)\ldots(a_dx+1)\quad\text{with}\quad a_1,\ldots,a_d\in\mathbf R.$$ A simple expansion of the product gives $a_1+\cdots+a_d=c_1$ and $\sum_{1\leq i<j\leq d}a_ia_j=c_2$. Then $$a_1^2+\cdots+a_d^2=c_1^2-2c_2\leq0,$$ contradicting that all $a_i$ are real and at least one is nonzero.

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Dear Marc, I prefer your Proposition to the screenshot solution: +1. – Georges Elencwajg Apr 13 '13 at 10:21
(+1) for the removal of superfluous assumption $a \ne 0$. – achille hui Apr 13 '13 at 10:52
This is a clean, simple proof. +1! – Soham Chowdhury Apr 15 '13 at 6:00

Since its zeroes are $\alpha_1,\alpha_2,\alpha_3$, and $\alpha_4$, the polynomial $ax^4+bx^3+x^+x+1$ is a constant multiple of $(x-\alpha_1)(x-\alpha_2)(x-\alpha_3)(x-\alpha_4)$. The leading term of this last product is $x^4$, so

$$ax^4+bx^3+x^2+x+1=a(x-\alpha_1)(x-\alpha_2)(x-\alpha_3)(x-\alpha_4)\;.\tag{1}$$

The constant term on the righthand side of $(1)$ is $a\alpha_1\alpha_2\alpha_3\alpha_4$; it must be equal to the constant term on the lefthand side, which is $1$, so $\alpha_1\alpha_2\alpha_3\alpha_4=\frac1a$. Now

\begin{align*} \left(x-\frac1{\alpha_1}\right)\left(x-\frac1{\alpha_2}\right)\left(x-\frac1{\alpha_3}\right)\left(x-\frac1{\alpha_4}\right)&=\frac{(\alpha_1x-1)(\alpha_2x-1)(\alpha_3x-1)(\alpha_4x-1)}{\alpha_1\alpha_2\alpha_3\alpha_4}\\\\ &=a(\alpha_1x-1)(\alpha_2x-1)(\alpha_3x-1)(\alpha_4x-1)\\\\ &=a(1-\alpha_1x)(1-\alpha_2x)(1-\alpha_3x)(1-\alpha_4x)\;, \end{align*}

and the factors of the product $(1-\alpha_1x)(1-\alpha_2x)(1-\alpha_3x)(1-\alpha_4x)$ have exactly the same coefficients as those of the product $(x-\alpha_1)(x-\alpha_2)(x-\alpha_3)(x-\alpha_4)$, but with the rôles of constant and $x$ term interchanged. The same must then be true of the products, so

$$a(1-\alpha_1x)(1-\alpha_2x)(1-\alpha_3x)(1-\alpha_4x)=a+bx+x^2+x^3+x^4\;.$$

Thus, with $\beta_k=\frac1{\alpha_k}$ for $k=1,2,3,4$ we have

$$(x-\beta_1)(x-\beta_2)(x-\beta_3)(x-\beta_4)=a+bx+x^2+x^3+x^4\;.\tag{2}$$

The coefficient of $x^3$ on the lefthand side of $(2)$ is $-\beta_1-\beta_2-\beta_3-\beta_4=-\sum_{k=1}^4\beta_k$, which must be equal to $1$, the coefficient of $x^3$ on the righthand side, so $$\sum_{k=1}^4\beta_k=-1\tag{3}\;.$$ The $x^2$ term on the lefthand side is the sum of all products of the form $\beta_i\beta_kx^2$ with $i\ne k$, so its coefficient is

$$\sum_{1\le i<k\le 4}\beta_i\beta_k=1\tag{4}$$

to match the righthand side of $(2)$.

Finally, $$\left(\sum_{k=1}^4\beta_k\right)^2=\sum_{k=1}^4\beta_k^2+2\sum_{1\le i<k\le 4}\beta_i\beta_k\;;\tag{5}$$

the factor of $2$ in the last term is to account for the fact that if $i\ne k$, the square on the left, when multiplied out, has both a $\beta_i\beta_k$ and a $\beta_k\beta_i$ term. Now rearrange $(5)$ and substitute the known values from $(3)$ and $(4)$:

$$\sum_{k=1}^4\beta_k^2=\left(\sum_{k=1}^4\beta_k\right)^2-2\sum_{1\le i<k\le 4}\beta_i\beta_k=1-2=-1\;.$$

But a sum of squares of real numbers cannot be negative, so the $\beta_k$ cannot all be real, and hence their reciprocals, the $\alpha_k$ that are the zeroes of the original polynomial, cannot all be real either.

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There are other answers here covering the entire solution. I will just elaborate on this important result being used here multiple times. Given a general fourth degree polynomial $ax^4+bx^3+cx^2+dx+e$, it must have four (complex) roots. Let's call them $r_1,r_2,r_3,r_4$. Then it must be the case that

$$x^4+\frac{b}{a}x^3+\frac{c}{a}x^2+\frac{d}{a}x+\frac{e}{a}=(x-r_1)(x-r_2)(x-r_3)(x-r_4)$$

we can do this because $a\neq 0$. Then when you expand this polynomial on the right hand side, we get

\begin{eqnarray*} x^4+\frac{b}{a}x^3+\frac{c}{a}x^2+\frac{d}{a}x+\frac{e}{a}&=&x^4\\ &-& (r_1+r_2+r_3+r_4)x^3\\ &+& (r_1r_2+r_1r_3+r_2r_3+r_1r_4+r_2r_4+r_3r_4)x^2\\ &-& (r_1r_2r_3+r_1r_2r_4+r_1r_3r_4+r_2r_3r_4)x\\ &+& r_1r_2r_3r_4. \end{eqnarray*}

Which then means that you can equate the coefficients and get

\begin{eqnarray*} \frac{b}{a}&=& -(r_1+r_2+r_3+r_4)\\ \frac{c}{a}&=& (r_1r_2+r_1r_3+r_2r_3+r_1r_4+r_2r_4+r_3r_4)\\ \frac{d}{a}&=& -(r_1r_2r_3+r_1r_2r_4+r_1r_3r_4+r_2r_3r_4)\\ \frac{e}{a}&=& r_1r_2r_3r_4. \end{eqnarray*}

The patterns are easy to remember here. The first is the negative of the sum of all the roots. The second is the sum of all possible two-combinations without any repetition. The third is the negative of the sum of all possible three-combinations without any repetition. The fourth is the sum of all the possible four-combinations but there is only one such combination because there are only four roots so that last is just the product of all the roots. Also note that the signs alternate.

This happens for any polynomial by the way. The second coefficient in a monic polynomial (when the first coefficient is zero after division by $a$) is ALWAYS the negative of the sum of all the roots and the last coefficient is always the product of the roots with the sign always being $(-1)^{\textrm{degree of the polynomial}}$. In this case the degree is even so the last coefficient is the product of the roots with a plus sign.

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And these are known as Viète's formulas (And he studied (law) at my university!) – Marc van Leeuwen Apr 15 '13 at 12:08