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Question Number 175624 by infinityaction last updated on 04/Sep/22

$$\mathrm{let}\mathrm{p}\left(\mathrm{x}\right)={\mathrm{x}}^6+{\mathrm{ax}}^5+{\mathrm{bx}}^4+{\mathrm{cx}}^3+{\mathrm{dx}}^2+\mathrm{ex}+\mathrm{f}$$$$\mathrm{be}\mathrm{a}\mathrm{polynomial}\mathrm{function}\mathrm{such}$$$$\mathrm{that}\mathrm{p}\left(1\right)=1;\mathrm{p}\left(2\right)=2;\mathrm{p}\left(3\right)=3$$$$\mathrm{p}\left(4\right)=4;\mathrm{p}\left(5\right)=5;\mathrm{p}\left(6\right)=6\mathrm{then}$$$$\mathrm{find}\mathrm{p}\left(7\right)=?$$

Answered by cortano1 last updated on 04/Sep/22

$$\mathrm{p}\left(\mathrm{x}\right)=(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)(\mathrm{x}-4)(\mathrm{x}-5)(\mathrm{x}-6)+\mathrm{x}$$$$\mathrm{p}\left(7\right)=6.5.4.3.2.1+7=6!+7$$$$=727$$

Commented by BaliramKumar last updated on 05/Sep/22

$$\mathrm{p}\left(\mathrm{n}\right)=?[n>6]$$

Answered by leodera last updated on 04/Sep/22

$$$$$$p\left(x\right)=(x-1)(x-2)(x-3)(x-4)(x-5)(\mathit{x}-6)+\mathit{x}$$$$\mathit{p}\left(7\right)=6!+7=720+7=727$$

Answered by floor(10²Eta[1]) last updated on 04/Sep/22

$$\mathrm{p}\left(1\right)=1\Rightarrow \mathrm{p}\left(\mathrm{x}\right)=(\mathrm{x}-1)\mathrm{a}\left(\mathrm{x}\right)+1$$$$\mathrm{p}\left(2\right)=\mathrm{a}\left(2\right)+1=2\Rightarrow \mathrm{a}\left(2\right)=1\Rightarrow \mathrm{a}\left(\mathrm{x}\right)=(\mathrm{x}-2)\mathrm{b}\left(\mathrm{x}\right)+1$$$$\mathrm{p}\left(\mathrm{x}\right)=(\mathrm{x}-1)((\mathrm{x}-2)\mathrm{b}\left(\mathrm{x}\right)+1)+1$$$$\mathrm{p}\left(3\right)=2(\mathrm{b}\left(3\right)+1)+1=3\Rightarrow \mathrm{b}\left(3\right)=0\Rightarrow \mathrm{b}\left(\mathrm{x}\right)=(\mathrm{x}-3)\mathrm{c}\left(\mathrm{x}\right)$$$$\mathrm{p}\left(\mathrm{x}\right)=(\mathrm{x}-1)((\mathrm{x}-2)(\mathrm{x}-3)\mathrm{c}\left(\mathrm{x}\right)+1)+1$$$$\mathrm{p}\left(4\right)=3(2\mathrm{c}\left(4\right)+1)+1=4\Rightarrow \mathrm{c}\left(4\right)=0\Rightarrow \mathrm{c}\left(\mathrm{x}\right)=(\mathrm{x}-4)\mathrm{d}\left(\mathrm{x}\right)$$$$\Rightarrow \mathrm{p}\left(\mathrm{x}\right)=(\mathrm{x}-1)[(\mathrm{x}-2)(\mathrm{x}-3)(\mathrm{x}-4)\mathrm{d}\left(\mathrm{x}\right)+1)+1$$$$\mathrm{p}\left(5\right)=4[3.2\mathrm{d}\left(5\right)+1]+1=5\Rightarrow \mathrm{d}\left(5\right)=0\Rightarrow \mathrm{d}\left(\mathrm{x}\right)=(\mathrm{x}-5)\mathrm{e}\left(\mathrm{x}\right)$$$$\mathrm{p}\left(\mathrm{x}\right)=(\mathrm{x}-1)[(\mathrm{x}-2)(\mathrm{x}-3)(\mathrm{x}-4)(\mathrm{x}-5)\mathrm{e}\left(\mathrm{x}\right)+1]+1$$$$\mathrm{p}\left(6\right)=5[4.3.2\mathrm{e}\left(6\right)+1]+1=6\Rightarrow \mathrm{e}\left(6\right)=0\Rightarrow \mathrm{e}\left(\mathrm{x}\right)=(\mathrm{x}-6)\mathrm{C}$$$$\Rightarrow \mathrm{p}\left(\mathrm{x}\right)=(\mathrm{x}-1)[(\mathrm{x}-2)(\mathrm{x}-3)(\mathrm{x}-4)(\mathrm{x}-5)(\mathrm{x}-6)\mathrm{C}+1]+1$$$$=(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)(\mathrm{x}-4)(\mathrm{x}-5)(\mathrm{x}-6)\mathrm{C}+\mathrm{x}$$$$\mathrm{p}\left(7\right)=6!\mathrm{C}+7=720\mathrm{C}+7,\mathrm{C}\in {\mathbb{R}}^{\ast }$$$$\mathrm{but}\mathrm{C}=1\mathrm{because}\mathrm{p}\left(\mathrm{x}\right)={\mathrm{x}}^6+{\mathrm{ax}}^5+...$$$$\Rightarrow \mathrm{p}\left(7\right)=727$$

Commented by infinityaction last updated on 04/Sep/22

$$thankssir$$

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