Let be a polynomial function $P\in \mathbb{R}[X]$.If I divide $P$ by $(x-1)(x-2)(x-3)(x-4)$ I get a remainder without "free term" ( like $ax^{3}+bx^{2}+cx$ )

I have to calculate the determinant:

$$\begin{vmatrix} P(1) & 1 & 1 &1 \\ P(2) & 2 & 4 &8 \\ P(3) & 3 & 9 &27 \\ P(4) & 4 & 16 &64 \end{vmatrix}$$

My try: I wrote that $P(x)=(x-1)(x-2)(x-3)(x-4)\cdot Q(x)+ax^{3}+bx^{2}+cx$

So for $x=1=> P(1)=a+b+c$

$x=2=> P(2)=8a+4b+2c$

$x=3=> P(3)=27a+9b+3c$

$x=4=> P(4)=64a+16b+4c$

And now I just have to replace the results in my determinant but it takes me a lot of time to solve the determinant.I'm wondering if there is a short way to solve this. Can you help me with some ideas?


Hint: Take column one and subtract $c$ from column $2,$ $b$ from column 3 and $a$ of column 4. Do you get a column full of $0's$? What is the determinant of that?

  • $\begingroup$ I also tried like this: I multiplied first column with (-1) then to the first column I added the second column,the third one the last one.So I get the first line full of 0.But I remain with a determinant which also difficult to calculate. $\endgroup$ – DaniVaja Apr 15 at 19:57
  • $\begingroup$ Your method is fine but multiply the second column by $c$ instead and the third one by $b$ and the fourth one by $a.$ You should get 0. And determinant is never changed by those operations so determinant should be 0. $\endgroup$ – Phicar Apr 15 at 19:59
  • $\begingroup$ So I got the first column full of 0 so determinant is 0, right ? $\endgroup$ – DaniVaja Apr 15 at 20:18
  • $\begingroup$ yes. A vector full of $0's$ gives you determinant $0$ $\endgroup$ – Phicar Apr 15 at 20:24
  • $\begingroup$ Thank you for your help :) $\endgroup$ – DaniVaja Apr 15 at 20:58

Hint: What is $a$ times the fourth column, plus $b$ times the third column, plus $c$ times the second?

  • $\begingroup$ I think I don't understand your question.Sorry for my English but I'm not native and I'm not so familiar with math terms. $\endgroup$ – DaniVaja Apr 15 at 19:58

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