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Question Number 107428 by abony1303 last updated on 10/Aug/20

$$\mathrm{The}\mathrm{polynomial}P\left(x\right)={\mathrm{x}}^3+{\mathrm{ax}}^2-4\mathrm{x}+\mathrm{b},$$$$\mathrm{where}\mathrm{a}\mathrm{and}\mathrm{b}\mathrm{are}\mathrm{constants}.\mathrm{Given}\mathrm{that}$$$$\mathrm{x}-2\mathrm{is}\mathrm{a}\mathrm{factor}\mathrm{of}P\left(x\right)\mathrm{and}\mathrm{that}\mathrm{a}\mathrm{remainder}$$$$\mathrm{of}6\mathrm{is}\mathrm{obtained}\mathrm{when}P\left(x\right)\mathrm{is}\mathrm{divided}\mathrm{by}$$$$(\mathrm{x}+1),\mathrm{find}\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{a}\mathrm{and}\mathrm{b}.$$

Commented by abony1303 last updated on 10/Aug/20

$$\mathrm{pls}\mathrm{help}$$

Answered by mr W last updated on 10/Aug/20

$$P\left(x\right)=x^3+{ax}^2-4x+b$$$$P\left(x\right)=(x-2)Q\left(x\right)$$$$P\left(2\right)=0$$$$2^3+a\times 2^2-4\times 2+b=0$$$$\Rightarrow 4a+b=0...\left(i\right)$$$$$$$$P\left(x\right)=(x+1)R\left(x\right)+6$$$$P(-1)=6$$$${(-1)}^3+a{(-1)}^2-4(-1)+b=6$$$$\Rightarrow a+b=3...\left(ii\right)$$$$\Rightarrow a=-1$$$$\Rightarrow b=4$$

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