Please help me..

I have come across this type of problems many times in brilliant(sure you would to)

A monic polynomial of degree NN leaves the remainder:-

AA when divided by (x±n1)(x \pm n_{1})

BB when divided by (x±n2)(x \pm n_{2})

CC when divided by (x±n3)(x \pm n_{3})

DD when divided by (x±n4)(x \pm n_{4})

EE when divided by (x±n5)(x \pm n_{5})

FF when divided by (x±n6)(x \pm n_{6})

and so on.....

Find out the value of P(Z)P(Z)

So, I just wanted to know a more or less shortcut method to solve this type of problem rather than just to plug the values and solve the equations(this can sometimes become a hilarious job)..

Any type of help will be appreciated.

Thank you\color{limegreen}{\text{Thank you}}.

Note by Anik Mandal
4 years, 5 months ago

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Anik Mandal - 4 years, 5 months ago

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Yes. Satvik is right. Method of differences is easy to employ too.

Krishna Ar - 4 years, 5 months ago

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If all the nn's are consecutive, you can use the Method of Differences. Check it's Wiki.

Satvik Golechha - 4 years, 5 months ago

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Thanks...

Anik Mandal - 4 years, 5 months ago

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Anik, because the polynomial problems are linear, we can always solve it using matrix. For example for the problem: Am I cubic?, my solution is as follows. I use an Microsoft Excel spreadsheet to do the matrix calculations.

Assuming that f(x)f(x) is a cubic polynomial, we can write in matrix form:

XA=B[13121110232221203332313043424140][a3a2a1a0]=[4347]XA=B\quad \Rightarrow \begin{bmatrix} 1^3 & 1^2 & 1^1 & 1^0 \\ 2^3 & 2^2 & 2^1 & 2^0 \\ 3^3 & 3^2 & 3^1 & 3^0 \\ 4^3 & 4^2 & 4^1 & 4^0 \end{bmatrix} \begin{bmatrix} a_3 \\ a_2 \\ a_1 \\ a_0 \end{bmatrix} = \begin{bmatrix} 4 \\ 3 \\ 4 \\ 7 \end{bmatrix}

We can find AA as follows:

A=X1B=[1612121632472113321271164641][4347]=[0147]A = X^{-1}B = \begin{bmatrix} -\frac{1}{6} & \frac {1}{2} &-\frac {1}{2} & \frac {1}{6} \\ \frac {3}{2} & -4 & \frac {7}{2} &-1 \\ -\frac {13}{3} & \frac {21}{2} & -7 & \frac{11}{6} \\ 4 & -6 & 4 & -1 \end{bmatrix} \begin{bmatrix} 4 \\ 3 \\ 4 \\ 7 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ -4 \\ 7 \end{bmatrix}

It is shown that: f(x)=x24x+7f(x) = x^2-4x+7 showing that: No\boxed{No}, it is not a cubic polynomial.

The spreadsheet.

Chew-Seong Cheong - 4 years, 5 months ago

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Thanks@Chew-Seong Cheong

Anik Mandal - 4 years, 5 months ago

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Sir how did we conclude that it is quadratic polynomial

Gaurav Jain - 4 years, 3 months ago

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Gaurav, A={0,1,4,7}A = \{0, 1, -4, 7\} means that f(x)=(0)x3+x24x+7f(x) = (0)x^3 + x^2-4x+7.

Chew-Seong Cheong - 4 years, 3 months ago

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@Chew-Seong Cheong Oh! I got , Thank you Sir

Gaurav Jain - 4 years, 3 months ago

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