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Question Number 103832 by bemath last updated on 17/Jul/20

$$p\left(x\right)=x^4+{ax}^3+{bx}^2+cx+d$$$$ifp\left(1\right)=10,p\left(2\right)=20and$$$$p\left(3\right)=30.find\frac{p\left(12\right)+p(-8)}{10}$$

Answered by floor(10²Eta[1]) last updated on 17/Jul/20

$$\mathrm{p}\left(\mathrm{x}\right)=(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)(\mathrm{x}-{\mathrm{x}}_4)+10\mathrm{x}$$$$\mathrm{p}\left(12\right)=11.10.9(12-{\mathrm{x}}_4)+120$$$$\mathrm{p}(-8)=(-9)(-10)(-11)(-8-{\mathrm{x}}_4)-80$$$$\frac{\mathrm{p}\left(12\right)+\mathrm{p}(-8)}{10}=99(12-{\mathrm{x}}_4)+12-99(-8-{\mathrm{x}}_4)-8$$$$=99[(12-{\mathrm{x}}_4)-(-8-{\mathrm{x}}_4)]+4=99.20+4=1984$$$$$$

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