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Let $P \in {\Bbb R} [x]$ be a cubic polynomial with real coefficients such that $$ |P(1)| = |P(2)| = |P(3)| = |P(5)| = |P(6)| = |P(7)| = 12 $$ Find the value of $\frac19 P(0)$

My approach so far:

Since $|P(x)| = 12$ for those six distinct points, $P(x)$ must take values $+12$ or $-12$ there. That means $P(x)^2 - 144 = 0$ for $x = 1, 2, 3, 5, 6, 7$. Hence, if I define $Q(x) = P(x)^2 - 144$, then $Q(x)$ is a degree-$6$ polynomial with these six roots. So it can be written as $$P(x)^2 - 144 = k(x-1) \cdot (x-2) \cdot (x-3) \cdot (x-5) \cdot (x-6) \cdot (x-7),$$ for some constant $k$. Now, since $P(x)$ is cubic, I can also write $$ P(x) = ax^3 + bx^2 + cx + d,$$ and substitute this into the above relation, but that quickly becomes messy. Could someone please help me formalize this step or show how to continue from here?

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  • $\begingroup$ You can write it as a system $Ax=b$ where $x$ contains the unknown coefficients of $P(x)^2$, and $b$ is the vector with all entries $144$ and $A$ is a Vamdernonde matrix. The matrix is non-singular (see the wikipedia on Vandermonde matrices) so there exists a solution. If you need expressions for the coefficients then solve the system some way. You could use a computer if it is too long by hand. In your case, you then need to check that this is actually the square of a polynomial, which is not clear to me. $\endgroup$ Commented Oct 7 at 11:12
  • $\begingroup$ Note that $-P$ satisfies the same hypothesis. So $P(0)/9$ is not unique. $\endgroup$ Commented Oct 7 at 11:25
  • $\begingroup$ It gets a little bit messy, but I don't think it's that bad. $\endgroup$ Commented Oct 7 at 11:35
  • $\begingroup$ @mukund Let me rephrase. If this comes from a book and someone who reads the same book asks the same question, there is a risk that people will invest some of their finite time answering a question that has already been answered. Does this misuse of human resources strike you as desirable? $\endgroup$ Commented Oct 9 at 11:38
  • $\begingroup$ i dont know the source saw it on yt took ss and it was in my archives $\endgroup$ Commented Oct 9 at 12:42

4 Answers 4

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Let's start from

$P^2=k(x-1)(x-2)(x-3)(x-5)(x-6)(x-7)+144.$

First off, define a translation $u=x-4$ in terms of which

$P^2=k(u+3)(u+2)(u+1)(u-1)(u-2)(u-3)+144=k(u^2-1)(u^2-4)(u^2-9)+144.$

Then we can more easily do the polynomial multiplication:

$P^2=k(\color{blue}{u^6-14u^4+49u^2})-(36k-144),$

whereupon we note that the blue expression is $(u^3-7u)^2$ so we should zero out the degree-zero term. Thus $k=4$, and the square root then gives

$P=\pm2(u^3-7u).$

We are to evaluate this at $x=0$, which corresponds to $u=x-4=-4$.

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  • $\begingroup$ I'm having trouble seeing why we need to zero out $36k-144$. Could you explain that? $\endgroup$ Commented Oct 8 at 21:47
  • $\begingroup$ To get just the part that's a square. Only for $a=0$ is $x^2-a$ invariably the square of another polyomial; same thing here. $\endgroup$ Commented Oct 8 at 21:54
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Check that $P(x)=+12$ can have maximum $3$ roots.
Similarly $P(x)=-12$ can have maximum $3$ roots.
Hence , together we can have maximum $3+3=6$ values if $x$ where $|P(x)|=12$
Naturally , we want the case where $|P(x)|=12$ does indeed have $6$ $x$ values.
Now that means , $P(1)=-12$ , then $P(2)=+12$ , $P(3)=+12$ , $P(5)=-12$ , $P(6)=-12$ , $P(7)=+12$ is the only way to get that (I am ignoring $-P(x)$ , which is essentially the same except for sign reversal)

Hence we try $y=\left(x-1\right)\left(x-5\right)\left(x-6\right)-12$ which satisfies $P(1)=-12$ $P(5)=-12$ , $P(6)=-12$ , though we do not yet get $+12$ values.
We must then try $y=A\left(x-1\right)\left(x-5\right)\left(x-6\right)-12$ for some $A$ value.
Now $A=2$ will work out fine !

Hence the Solution is :
$y=2\left(x-1\right)\left(x-5\right)\left(x-6\right)-12$

P(X)

Here $y=-2\left(x-1\right)\left(x-5\right)\left(x-6\right)+12$ will give same graph except for sign reversal.

Due to Cubic Polynomial Symmetry here , $P(4)=0$ , which is conveniently left out in the given list.

Depending on which we choose , $P(0)/9=\pm72/9=\pm8$

ADDENDUM :

Initially , to get the order $-12,+12,+12,-12,-12,+12$ , we can consider that Cubic Curves are generally like this Black Curve :
INITIAL

Here , the horizontal lines (Purple + Light Gray + Dark Gray) are Constant $y$ values.

Now , we can see that when-ever we want 3 intersections , then :

  • It can not be like the Light Gray lines , which have 1 intersection
  • It can not even be the Dark Gray lines , which have 2 intersection
  • It must be something like the middle Purple lines.

Eventually , which-ever 2 Purple lines we have to use , the order is always $-12,+12,+12,-12,-12,+12$ (ignoring the $-P$ Sign Exchange)

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    $\begingroup$ There are twenty possible sign patterns with three positive and three negative. How do you pick the two correct ones? $\endgroup$ Commented Oct 8 at 11:55
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    $\begingroup$ @OscarLanzi every time it goes from -12 to +12, the function needs to go through zero, so that can only happen three times for a cubic polynomial. At the same time, if it goes from +12 to +12 or from -12 to -12, there needs to be a local extremum between the points, which can only happen twice for a cubic polynomial. This should be enough to eliminate the other sign patterns, I think. $\endgroup$ Commented Oct 8 at 14:44
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    $\begingroup$ I think it can be simpler than that, @DanielSmania. For a cubic polynomial to have any repeating values, it must be divisible into three strict monotonic regions joined by two local extrema, either (increasing, decreasing, increasing) or (decreasing, increasing, decreasing). The values at the extrema cannot repeat more than twice, and there are, of course, no repeats within any individual strictly monotonic region. That gives us one +12 and one -12 in each monotonic region, ordered according to the monotonicity. Which yields either - + + - - + or the reverse. $\endgroup$ Commented Oct 9 at 1:46
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    $\begingroup$ @DanielSmania, I guess the ultimate length would depend on what parts were further detailed, to what extent, and also on how you measure (symbols? words? column space?). I don't think posting an additional solution merely to detail this approach to a sub-point would really add much to the answers already presented. I do think that deriving the needed pattern from the function shape is conceptually simpler and more satisfying than exhaustively analyzing all the possible cases. YMMV. $\endgroup$ Commented Oct 9 at 3:11
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    $\begingroup$ @JohnBollinger I incorporated your argument in my answer (with attribution). $\endgroup$ Commented Oct 9 at 15:41
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Claim: If $x_1 < x_2< x_3< x_4< x_5 < x_6$ are real numbers, $c >0$, $|P(x_i)|=c$ for $i\in \{1,2,\dots,6\}$, where $P$ is a real cubic polynomial, then the unique possible configurations of the sign of $P(x_i)$ are

\begin{array}{c|cccccc} x & x_1 & x_2 & x_3 & x_4 & x_5 & x_6 \\ \hline \operatorname{sign} P(x) & + & - & - & + & + & - \\ P(x) & c & -c & -c & c & c & -c \end{array}

\begin{array}{c|cccccc} x & x_1 & x_2 & x_3 & x_4 & x_5 & x_6 \\ \hline \operatorname{sign} P(x) & - & + & + & - & - & + \\ P(x) & -c & c & c & -c & -c & c \end{array}

The proof of the claim is the hard part. We provide in the end two proofs of this claim, one following an idea by JohnBollinger (given in a comment) and my original proof.

The claim implies that either $P(1)=P(5)=P(6)=12$ or $P(1)=P(5)=P(6)=-12$.

Note that the cubic polynomials that satisfies $P(1)=P(5)=P(6)=12$ are

$$P(x)= k(x-1)(x-5)(x-6)+12, \ k\neq 0.$$

Since we need $P(2)=P(3)=P(7)=-12$ we conclude that

$$-12= k(2-1)(2-5)(2-6)+12 = 12k+12.$$ $$-12= k(3-1)(3-5)(3-6)+12 = 12k+12.$$ $$-12= k(7-1)(7-5)(7-6)+12 = 12k+12.$$

so there is a unique solution $k = -2$. So

$$P(x)= -2(x-1)(x-5)(x-6)+12,$$

and the other solution for the case $P(1)=P(5)=P(6)=-12$ is $-P$.

in particular $P(0)= \pm 72$ and $P(0)/9= \pm 8$.

First Proof of the Claim (following JohnBollinger)

Here we follow the idea given by JohnBollinger in the comments. My original proof is in the end.

Let $S=\{x_1,x_2,x_3,x_4,x_5,x_6\}.$ We have that $P(x)=c$ has at most $3$ solutions and $P(x)=-c$ has at most 3 solutions. Since we need $|P(x)|=c$ for $x\in S$ and there are six elements in $S$ the only option is to have exactly $3$ solutions for each equation in $S$.

Let $a_1< a_2 < a_3$ and $b_1< b_2 < b_3$, with $S=\{a_1,a_2,a_3,b_1,b_2,b_3\}$, such that $P(a_i)=c$ and $P(b_i)=-c$. By Rolle's Theorem, there are two distinct critical points $d_1\in (a_1,a_2)$ and $d_2\in (a_2,a_3)$. In particular $d_1< d_2$. For the same reason there are two critical points $\hat{d}_1\in (b_1,b_2)$ and $\hat{d}_2\in (b_2,b_3)$, with $\hat{d}_1< \hat{d}_2$.

Since $P'$ is a quadratic polynomial, we conclude that

-$P$ has exactly two critical points, and in particular $\hat{d}_1=d_1$ and $\hat{d}_2=d_2$,

-$P$ is strictly monotone on each one of the intervals $I_1=(-\infty,d_1)$, $I_2=(d_1,d_2)$ and $I_3=(d_2,+\infty)$, and the signal of the derivative changes when one cross a critical point.

-$a_1,b_1\in I_1$, $a_2,b_2\in I_2$ and $a_3,b_3\in I_3$.

There are two cases

Case I. We have

\begin{array}{c|ccc} interval & I_1 & I_2 & I_3 \\ \hline P' \ is & positive & negative & positive \\ \hline P \ is \ monotone & increasing & decreasing & increasing \end{array}

Since $P(a_i) > P(b_i)$ we conclude that $$b_1 < a_1< a_2 < b_2< b_3 < a_3,$$

so

\begin{array}{c|cccccc} x & x_1 & x_2 & x_3 & x_4 & x_5 & x_6 \\ \hline \operatorname{sign} P(x) & - & + & + & - & - & + \\ P(x) & -c & c & c & -c & -c & c \end{array}

Case II. We have

\begin{array}{c|ccc} interval & I_1 & I_2 & I_3 \\ \hline P' \ is & negative & positive & negative \\ \hline P \ is \ monotone & decreasing & increasing & decreasing \end{array}

Since $P(a_i) > P(b_i)$ we conclude that $$a_1 < b_1< b_2 < a_2< a_3 < b_3,$$

so

\begin{array}{c|cccccc} x & x_1 & x_2 & x_3 & x_4 & x_5 & x_6 \\ \hline \operatorname{sign} P(x) & + & - & - & + & + & - \\ P(x) & c & -c & -c & c & c & -c \end{array}

Second proof of the Claim (original proof)

I will write down the proof of the claim for our case $x_1=1, x_2=2, x_3=3,x_4=5, x_5=6, x_6=7$, and $c=12$. The generalization is rather obvious.

As already observed $P(x)=12$ has at most $3$ solutions and $P(x)=-12$ has at most 3 solutions. Since we need $|P(x)|=12$ for $x\in S=\{1,2,3,5,6,7\}$ the only option is to have exactly $3$ solutions for each equation in $S$.

Note that all points in $S$ are simple roots of $P(x)\pm 12$, that is, $P'(x)\neq 0$ for $x\in S$.

We now look for the possible signs of $P$ at $S$: there are $20$ configurations of $3$ signs $+$ and $3$ signs $-$ in the $6$ positions of $S$, but we are going to see that only $2$ are possible for a cubic polynomial.

Crucial observations are that

-the patterns $++$, $--$ imply, by Rolle's Theorem, the existence of a critical point of $P$ in the corresponding interval.

-the patterns $+-$, $-+$ imply, by the Intermediary Value Theorem, the existence of a root of $P$ in the corresponding interval.

-As a consequence the patterns $+-+$, $-+-$ imply the existence of two roots in corresponding interval. By Rolle's Theorem there is also a critical point in the corresponding interval.

Case I. If the signs are \begin{array}{c|cccccc} x & 1 & 2 & 3 & 5 & 6 & 7 \\ \hline \text{sign}(P(x)) & + & + & + & - & - & - \end{array} \begin{array}{c|cccccc} x & 1 & 2 & 3 & 5 & 6 & 7 \\ \hline \text{sign}(P(x)) & - & - & - & + & + & + \end{array} \begin{array}{c|cccccc} x & 1 & 2 & 3 & 5 & 6 & 7 \\ \hline \text{sign}(P(x)) & - & + & + & +& - & - \end{array} \begin{array}{c|cccccc} x & 1 & 2 & 3 & 5 & 6 & 7 \\ \hline \text{sign}(P(x)) & + & - & - & -& + & + \end{array} \begin{array}{c|cccccc} x & 1 & 2 & 3 & 5 & 6 & 7 \\ \hline \text{sign}(P(x)) & - & - & + & +& + & - \end{array} \begin{array}{c|cccccc} x & 1 & 2 & 3 & 5 & 6 & 7 \\ \hline \text{sign}(P(x)) &+ & + & - & -& - & + \end{array} $P$ would have at least $3$ critical points. Impossible.

Case II. If the signs are \begin{array}{c|cccccc} x & 1 & 2 & 3 & 5 & 6 & 7 \\ \hline \text{sign}(P(x)) & + & - & + & - & + & - \end{array}

\begin{array}{c|cccccc} x & 1 & 2 & 3 & 5 & 6 & 7 \\ \hline \text{sign}(P(x)) & - & + & - & + & - & + \end{array}

\begin{array}{c|cccccc} x & 1 & 2 & 3 & 5 & 6 & 7 \\ \hline \text{sign}(P(x)) & - & + & + & - & + & - \end{array} \begin{array}{c|cccccc} x & 1 & 2 & 3 & 5 & 6 & 7 \\ \hline \text{sign}(P(x)) & + & - & - & + & - & + \end{array}

\begin{array}{c|cccccc} x & 1 & 2 & 3 & 5 & 6 & 7 \\ \hline \text{sign}(P(x)) & - & + & - & + & + & - \end{array} \begin{array}{c|cccccc} x & 1 & 2 & 3 & 5 & 6 & 7 \\ \hline \text{sign}(P(x)) & + & - & + & - & - & + \end{array}

then $P$ would have at least $4$ roots. That is not possible.

Case III. If the signs are \begin{array}{c|cccccc} x & 1 & 2 & 3 & 5 & 6 & 7 \\ \hline \text{sign}(P(x)) & + & + & - & + & - & - \end{array} \begin{array}{c|cccccc} x & 1 & 2 & 3 & 5 & 6 & 7 \\ \hline \text{sign}(P(x)) & - & - & + & - & + & + \end{array}

then $P$ would have 3 critical points. One in each interval $(1,2)$ and $(6,7)$ and, since $P$ would have roots in $(3,5)$ and $(5,6)$, we would have a critical point in $(3,6)$. Impossible.

Case IV. If the signs are (thanks to tkf for fixing the first two configurations)

\begin{array}{c|cccccc} x & 1 & 2 & 3 & 5 & 6 & 7 \\ \hline \text{sign}(P(x)) & + & + & - & - & + & - \end{array} \begin{array}{c|cccccc} x & 1 & 2 & 3 & 5 & 6 & 7 \\ \hline \text{sign}(P(x)) & - & - & + & + & - & + \end{array} \begin{array}{c|cccccc} x & 1 & 2 & 3 & 5 & 6 & 7 \\ \hline \text{sign}(P(x)) & + & - & + & + & - & - \end{array} \begin{array}{c|cccccc} x & 1 & 2 & 3 & 5 & 6 & 7 \\ \hline \text{sign}(P(x)) & - & + & - & - & + & + \end{array}

$P$ would have at $3$ critical points, using an argument similar to case $III$. Impossible.

Case V. So the remaining cases are

\begin{array}{c|cccccc} x & 1 & 2 & 3 & 5 & 6 & 7 \\ \hline \operatorname{sign} P(x) & + & - & - & + & + & - \\ P(x) & 12 & -12 & -12 & 12 & 12 & -12 \end{array}

\begin{array}{c|cccccc} x & 1 & 2 & 3 & 5 & 6 & 7 \\ \hline \operatorname{sign} P(x) & - & + & + & - & - & + \\ P(x) & -12 & 12 & 12 & -12 & -12 & 12 \end{array}

This completes the proof of the claim in our case.

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Another solution, starting exactly as the first solution does, i.e., using that the values at $1,2,3,5,6,7$ are $-12,12,12,-12,-12,12$, is Lagrange interpolation. One can use an online calculator, or even do it by hand. It gives $$ P(x)=2x^3-24x^2+82x-72. $$

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    $\begingroup$ Please explain more $\endgroup$ Commented Oct 7 at 12:34
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    $\begingroup$ You might state that with a cubic the only possible sign sequences are $-++--+$ or $+--++-$. $\endgroup$ Commented Oct 7 at 14:26
  • $\begingroup$ @mukund See Lagrange interpolation for the formula. $\endgroup$ Commented Oct 7 at 15:03
  • $\begingroup$ Daniel's answer goes to a lot of trouble to figure out the sign pattern, and then a bit of work to find the polynomial. Sweeping that away with "starting exactly as the first" and "Lagrange interpolation" make this short, but doesn't really give any insight into how to solve the problem. $\endgroup$ Commented Oct 8 at 22:15
  • $\begingroup$ @Teepeemm This answer, too long for a comment, is just a response to Prem's answer, improving on the step of "trying several different constants" ( We must then try $y=A(x−1)(x−5)(x−6)−12$ for some $A$ value.) to find the polynomial. So far, nobody has mentioned that one has the "famous" Lagrange interpolation" for this. $\endgroup$ Commented Oct 9 at 7:55

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