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Question Number 98796 by Lekhraj last updated on 16/Jun/20

Answered by MJS last updated on 16/Jun/20

$$-\frac{267}{2}$$

Commented by Lekhraj last updated on 16/Jun/20

$$\mathrm{How}?$$

Commented by MJS last updated on 16/Jun/20

$$p\left(x\right)=c_3{x}^3+c_2{x}^2+c_{1}x+c_0$$$$\mathrm{insert}\mathrm{the}\mathrm{given}\mathrm{values}\mathrm{and}\mathrm{solve}\mathrm{the}\mathrm{linear}$$$$\mathrm{system}\mathrm{for}{c}_j$$

Commented by Lekhraj last updated on 16/Jun/20

$$\mathrm{Calculatins}\mathrm{are}\mathrm{very}\mathrm{lengthy}.$$$$\mathrm{Is}\mathrm{there}\mathrm{any}\mathrm{short}\mathrm{methld}\mathrm{or}\mathrm{else}$$$$\mathrm{you}\mathrm{may}\mathrm{please}\mathrm{show}\mathrm{the}\mathrm{whole}$$$$\mathrm{solution}.$$$$\mathrm{Thanks}$$$$$$

Answered by mr W last updated on 17/Jun/20

$$alternativemethod:$$$$p\left(1\right)=p\left(\frac{1}{4}\right)=3$$$$\Rightarrow p\left(x\right)=a\left(x\right)(x-1)(x-\frac{1}{4})+3$$$$p\left(\frac{1}{3}\right)=a\left(\frac{1}{3}\right)(\frac{1}{3}-1)(\frac{1}{3}-\frac{1}{4})+3=4$$$$\Rightarrow a\left(\frac{1}{3}\right)=-18$$$$\Rightarrow a\left(x\right)=b(x-\frac{1}{3})-18$$$$\Rightarrow p\left(x\right)=(b(x-\frac{1}{3})-18)(x-1)(x-\frac{1}{4})+3$$$$p\left(\frac{1}{2}\right)=(b(\frac{1}{2}-\frac{1}{3})-18)(\frac{1}{2}-1)(\frac{1}{2}-\frac{1}{4})+3=6$$$$\Rightarrow b=-36$$$$\Rightarrow p\left(x\right)=(-36(x-\frac{1}{3})-18)(x-1)(x-\frac{1}{4})+3$$$$\Rightarrow p\left(x\right)=-18(2x+\frac{1}{3})(x-1)(x-\frac{1}{4})+3$$$$$$$$p\left(2\right)=-18(2\times 2+\frac{1}{3})(2-1)(2-\frac{1}{4})+3$$$$=-3\times 13\times \frac{7}{2}+3$$$$=-\frac{267}{2}$$

Commented by Lekhraj last updated on 17/Jun/20

$$\mathrm{Thanks}.$$

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