Please help me..

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.

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

#Polynomials

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

@Aditya Raut @Satvik Golechha @Calvin Lin @Krishna Ar @Sandeep Bhardwaj @Trevor Arashiro @Finn Hulse @Agnishom Chattopadhyay @brian charlesworth and all other $\color{#D61F06}{\text{Brilliant' ians}}$

Anik Mandal - 6 years, 4 months ago

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

Krishna Ar - 6 years, 4 months ago

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

Satvik Golechha - 6 years, 4 months ago

Thanks...

Anik Mandal - 6 years, 4 months ago

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.

Chew-Seong Cheong - 6 years, 4 months ago

Thanks@Chew-Seong Cheong

Anik Mandal - 6 years, 4 months ago

Sir how did we conclude that it is quadratic polynomial

Gaurav Jain - 6 years, 3 months ago

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

Chew-Seong Cheong - 6 years, 3 months ago

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

Gaurav Jain - 6 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$