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

If 2x^2 +(2p−13)x+2=0 is exactly divisible  by x−3, then the value of p is

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

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

f(x)=2x^2 +(2p−13)x+2  ATQ,  f(3)=0  ⇒2×3^2 +(2p−13)×3+2=0  ⇒18+6p−39+2=0  ⇒6p=39−20  ⇒p=((19)/6)

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

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

An Other Way  (2x^2 +(2p−13)x+2 )÷(x−3)  By synthetic division:   (((3)),2,(2p−13),(       2)),(,,(       6),(6p−21)),(,2,(2p−7),(6p−19)) )  As x−3 is factor  ∴   6p−19=0⇒p=((19)/6)

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

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