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Question Number 68351 by mhmd last updated on 09/Sep/19

$$\mathrm{If}2x^2+(2p-13)x+2=0\mathrm{is}\mathrm{exactly}\mathrm{divisible}$$$$\mathrm{by}x-3,\mathrm{then}\mathrm{the}\mathrm{value}\mathrm{of}p\mathrm{is}$$

Answered by $@ty@m123 last updated on 09/Sep/19

$$f\left(x\right)=2x^2+(2p-13)x+2$$$$ATQ,$$$$f\left(3\right)=0$$$$\Rightarrow 2\times 3^2+(2p-13)\times 3+2=0$$$$\Rightarrow 18+6p-39+2=0$$$$\Rightarrow 6p=39-20$$$$\Rightarrow p=\frac{19}{6}$$

Answered by Rasheed.Sindhi last updated on 10/Sep/19

$$\mathbb{A}\mathrm{n}\mathbb{O}\mathrm{ther}\mathbb{W}\mathrm{ay}$$$$(2x^2+(2p-13)x+2)\mathbf{÷}(\mathrm{x}-3)$$$$\mathrm{By}\mathrm{synthetic}\mathrm{division}:$$$$\left(\begin{array}{cccc}3)& 2& 2\mathrm{p}-13& 2\\ & & 6& 6\mathrm{p}-21\\ & 2& 2\mathrm{p}-7& 6\mathrm{p}-19\end{array}\right)$$$$\mathrm{As}\mathrm{x}-3\mathrm{is}\mathrm{factor}$$$$\therefore 6\mathrm{p}-19=0\Rightarrow \mathrm{p}=\frac{19}{6}$$

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