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Algebra Level 4

$f(x)$ is a $6^\text{th}$ degree polynomial satisfying $f(1-x)=f(x+1)$ for all real values of $x$ . If $f(x)$ has four distinct real roots and two real and equal roots, then find the sum of all roots of $f(x)=0$ .

\frac{1}{2}

4 6 2

3 solutions

Otto Bretscher
May 19, 2016

If $1+c$ is a root, then so is $1-c$ , with the same multiplicity. Thus the roots are $1\pm c_1, 1\pm c_2, 1,1$ , and the answer is $\boxed{6}$

Sir please can you clarify how the 2 equal roots came as 1

Priyamvad Tripathi - 5 years ago

Is it clear to you that the roots must take the form $1\pm c_1, 1\pm c_2, 1\pm c_3$ , and that we cannot have 1 as a non-repeated root? Assuming that that part is clear: One of the roots is repeated. Without loss of generality, suppose it is $1+c_1$ . Thus, one of the following must be true: $1+c_1=1-c_1,1+c_1=1+c_2,1+c_1=1-c_2,1+c_1=1+c_3,1+c_1=1-c_3$ . If the first one is true, then $c_1=0$ , and 1 is a repeated root. If any of the others are true, just label the variables so that we can say $1+c_1=1+c_2$ . This implies $c_1=c_2$ . But that would mean we have four repeated roots, which contradicts the hypothesis in the problem. Therefore, the two equal roots must be 1.

James Wilson - 3 years, 8 months ago

James Wilson
Oct 9, 2017

Since $f(1-x)=f(x+1)$ holds for all $x$ , $f(1-(x-1))=f((x-1)+1)$ holds for all $x$ , or simplified, $f(2-x)=f(x)$ holds for all $x$ . This leads to $(2-x-x_1)(2-x-x_2)(2-x-x_3)(2-x-x_4)(2-x-x_5)(2-x-x_5)$ $=(x+x_1-2)(x+x_2-2)(x+x_3-2)(x+x_4-2)(x+x_5-2)(x+x_5-2)$ $=(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)(x-x_5)$ for all $x$ . Equating the $x^5$ coefficients: $-(x_1+x_2+x_3+x_4+2x_5)=(x_1+x_2+x_3+x_4+2x_5)-12$ . Therefore, $x_1+x_2+x_3+x_4+2x_5=12/2=6$ . One could also note that it didn't matter that there was a repeated root.

I also like the Sammy Hagar reference.

James Wilson - 3 years, 7 months ago

You are right. No need for the last two conditions. Only the first suffices to get the sum of the roots.

A Former Brilliant Member - 2 years ago

Noe Blassel
Jun 6, 2019

Let $g(x)=f(x+1)$ is again a polynomial, $\deg{g}=6$ . Furthermore, $\rho$ is a root of $f \iff \rho-1$ is a root of $g$ , thus $\sum_{\rho:f(\rho)=0}\rho=\sum_{\rho:g(\rho)=0}\rho+6$ Also, $\forall x, f(1-x)=f(x+1) \iff \forall x, g(x)=g(-x)$ , i.e. $g$ is an even function. In particular, the coefficient of the power expansion of $g$ is $0$ for all odd powers. By Vieta's formula, the coefficient of the fifth power is minus the sum of all roots of $g$ . Hence this sum is $0$ , and the answer is $6$ . Note that in general, if $f$ is a real polynomial of degree $2n$ such that $f(\alpha-x)=f(\alpha+x)$ for some $\alpha \in \mathbb{R}$ , then the same reasoning shows that the sum of all roots of $f$ (counted with multiplicity, real or not) is $2n\alpha$ .

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