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

A monic polynomial of degree $N$ leaves the remainder:-

$A$ when divided by $(x \pm n_{1})$

$B$ when divided by $(x \pm n_{2})$

$C$ when divided by $(x \pm n_{3})$

$D$ when divided by $(x \pm n_{4})$

$E$ when divided by $(x \pm n_{5})$

$F$ when divided by $(x \pm n_{6})$

and so on.....

Find out the value of $P(Z)$

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.

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

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## Comments

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

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

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Thanks...

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

$XA=B\quad \Rightarrow \begin{bmatrix} 1^3 & 1^2 & 1^1 & 1^0 \\ 2^3 & 2^2 & 2^1 & 2^0 \\ 3^3 & 3^2 & 3^1 & 3^0 \\ 4^3 & 4^2 & 4^1 & 4^0 \end{bmatrix} \begin{bmatrix} a_3 \\ a_2 \\ a_1 \\ a_0 \end{bmatrix} = \begin{bmatrix} 4 \\ 3 \\ 4 \\ 7 \end{bmatrix}$

We can find $A$ as follows:

$A = X^{-1}B = \begin{bmatrix} -\frac{1}{6} & \frac {1}{2} &-\frac {1}{2} & \frac {1}{6} \\ \frac {3}{2} & -4 & \frac {7}{2} &-1 \\ -\frac {13}{3} & \frac {21}{2} & -7 & \frac{11}{6} \\ 4 & -6 & 4 & -1 \end{bmatrix} \begin{bmatrix} 4 \\ 3 \\ 4 \\ 7 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ -4 \\ 7 \end{bmatrix}$

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

The spreadsheet.

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Thanks@Chew-Seong Cheong

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Sir how did we conclude that it is quadratic polynomial

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

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