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Let $p(x)$ be a polynomial of degree $3$ with real coefficient.

Then which of the following is possible $?$

$A. p(x)$ has no real root. WRONG.

$B.p(x)$ has exactly two real roots. WRONG AGAIN.

$C.p(1)=-1,p(2)=1,p(3)=11\ and\ p(4)=35.$

$D.i-1\ \ and\ \ i+1$ are roots of $p(x)$. As these are NOT conjugates, so cannot be roots of the same polynomial so AGAIN WRONG .

So, that leaves out only option $C$. Now , what I was thinking was that , I could say only $C$ is possible because I had ways of eliminating the rest . But what if the eliminations were not this easy , was there any way to say that , a polynomial of degree $3$ actually exists satisfying what is given in $C$ $?$ How can I actually find out such a polynomial $?$

Hope I could convey my question properly.

Thanks for any help.

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  • $\begingroup$ Please state the problem clearly. Is the problem asking which of a,b,c,d is correct? $\endgroup$ – user223391 Dec 19 '15 at 6:58
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    $\begingroup$ See if you can solve for the polynomial with linear algebra. Also seems like you didn't write out the full question. Shouldn't there be some kind of "which of the following are possible?" thing written in the problem statement? $\endgroup$ – anon Dec 19 '15 at 6:58
  • $\begingroup$ @anon : Yes . There is. I'll edit . $\endgroup$ – user118494 Dec 19 '15 at 7:00
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    $\begingroup$ For four points, one cannot always find a polynomial of degree exactly $3$. But $\le 3$ follows from general considerations. A nice way to do it is Lagrange interpolation. $\endgroup$ – André Nicolas Dec 19 '15 at 7:18
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Option $B$ is quite ambiguous since double roots can exist.

Concerning $C$, write the polynomial as $$P(x)=a+bx+cx^2+dx^3$$ so $$P(1)=a+b+c+d=-1$$ $$P(2)=a+2 b+4 c+8 d=1$$ $$P(3)=a+3 b+9 c+27 d=11$$ $$P(4)=a+4 b+16 c+64 d=35$$ Solve for $a,b,c,d$ using the method of your choice.

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  • $\begingroup$ I solved the system of equations and found values of $a,b,c,d.$ (Provided , I've not miscalculated any step) . So , does that mean the equation is found and option $C$ is a possibility $?$ $\endgroup$ – user118494 Dec 19 '15 at 7:47
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    $\begingroup$ @user118494. If all coefficients exist, $C$ is then true. $\endgroup$ – Claude Leibovici Dec 19 '15 at 7:53
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    $\begingroup$ @user118494. For sure, $d\neq 0$ is a requirement. $\endgroup$ – Claude Leibovici Dec 19 '15 at 10:07
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    $\begingroup$ @user118494. So, I am afraid that you have a mistake; $d\neq 0$ but $d\neq 3$ . $\endgroup$ – Claude Leibovici Dec 19 '15 at 10:23
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    $\begingroup$ You just face four simple linear equations for four unknowns. Please do it ! Cheers :-) $\endgroup$ – Claude Leibovici Dec 19 '15 at 12:28
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From the analytical point of view ,by Weierstrass approximation theorem, given the four points, we can construct an interpolating polynomial (recall from the numerical analysis classes )of degree 3 which assumes the same value at the given points.

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A real polynomial of degree $\leq 3$ can be fit to any given real values at $3+1$ points. The only thing requiring a calculation is to test that the $4$ given values are not fit by a polynomial of degree $2$ or lower. Since the points are at consecutive integer $x$ coordinates this exceptional case is easily tested by taking differences of the sequence.

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