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.

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

Note by Anik Mandal
3 years, 8 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 $$f(x)$$ 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.

- 3 years, 7 months ago

Sir how did we conclude that it is quadratic polynomial

- 3 years, 6 months ago

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

- 3 years, 6 months ago

Oh! I got , Thank you Sir

- 3 years, 6 months ago

Thanks@Chew-Seong Cheong

- 3 years, 7 months ago

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

- 3 years, 8 months ago

Thanks...

- 3 years, 7 months ago

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

- 3 years, 8 months ago