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Question Number 171456 by mathlove last updated on 15/Jun/22

Commented by mathlove last updated on 16/Jun/22

solution??

$${solution}?? \\ $$

Answered by Rasheed.Sindhi last updated on 16/Jun/22

Let P(x)=ax^2 +bx+c  • P(x) is divided by x−3 leaving     remainder 4. By synthetic division:   determinant (((3)),a,b,c),(,,(3a),(9a+3b)),(,a,(3a+b),(9a+3b+c=4^★ )))    Q(x)=ax+(3a+b)  •Q(x) is divided by x+3 leaving     remainder(R) 2.     R=Q(−3)=a(−3)+(3a+b)=2             b=2         ^★  9a+3(2)+c=4⇒9a+c=−2  •P(x) is divided by x^2 −9   x^2 −9)ax^2 +bx+c^(a)                _− ax^2         −_(+) 9a   Remainder: bx+9a+c=(2)x+(−2)                 =2x−2    ans

$${Let}\:{P}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c} \\ $$$$\bullet\:{P}\left({x}\right)\:{is}\:{divided}\:{by}\:{x}−\mathrm{3}\:{leaving} \\ $$$$\:\:\:{remainder}\:\mathrm{4}.\:{By}\:{synthetic}\:{division}: \\ $$$$\begin{array}{|c|c|c|}{\left.\mathrm{3}\right)}&\hline{{a}}&\hline{{b}}&\hline{{c}}\\{}&\hline{}&\hline{\mathrm{3}{a}}&\hline{\mathrm{9}{a}+\mathrm{3}{b}}\\{}&\hline{{a}}&\hline{\mathrm{3}{a}+{b}}&\hline{\mathrm{9}{a}+\mathrm{3}{b}+{c}=\mathrm{4}^{\bigstar} }\\\hline\end{array} \\ $$$$\:\:{Q}\left({x}\right)={ax}+\left(\mathrm{3}{a}+{b}\right) \\ $$$$\bullet{Q}\left({x}\right)\:{is}\:{divided}\:{by}\:{x}+\mathrm{3}\:{leaving} \\ $$$$\:\:\:{remainder}\left({R}\right)\:\mathrm{2}. \\ $$$$\:\:\:{R}={Q}\left(−\mathrm{3}\right)={a}\left(−\mathrm{3}\right)+\left(\mathrm{3}{a}+{b}\right)=\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{b}=\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:^{\bigstar} \:\mathrm{9}{a}+\mathrm{3}\left(\mathrm{2}\right)+{c}=\mathrm{4}\Rightarrow\mathrm{9}{a}+{c}=−\mathrm{2} \\ $$$$\bullet{P}\left({x}\right)\:{is}\:{divided}\:{by}\:{x}^{\mathrm{2}} −\mathrm{9} \\ $$$$\left.\:{x}^{\mathrm{2}} −\mathrm{9}\right)\overset{{a}} {{ax}^{\mathrm{2}} +{bx}+{c}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:_{−} {ax}^{\mathrm{2}} \:\:\:\:\:\:\:\:\underset{+} {−}\mathrm{9}{a} \\ $$$$\:{Remainder}:\:{bx}+\mathrm{9}{a}+{c}=\left(\mathrm{2}\right){x}+\left(−\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}{x}−\mathrm{2}\:\:\:\:{ans} \\ $$

Commented by mathlove last updated on 16/Jun/22

thanks

$${thanks} \\ $$

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