Question Number 213624 by hardmath last updated on 10/Nov/24
$$\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{5x}\:−\:\mathrm{42}\:=\:\left(\mathrm{x}\:−\:\mathrm{3}\right)\centerdot\mathrm{P}\left(\mathrm{x}\right) \\ $$$$\mathrm{Find}:\:\:\:\mathrm{P}\left(\mathrm{3}\right)\:=\:? \\ $$
Answered by A5T last updated on 10/Nov/24
![(x−3)(x^2 +3x+14)−(x−3)P(x)=0 ⇒(x−3)[x^2 +3x+14−P(x)]=0 ⇒x=3 or P(x)=x^2 +3x+14⇒P(3)=32](https://www.tinkutara.com/question/Q213625.png)
$$\left({x}−\mathrm{3}\right)\left({x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{14}\right)−\left({x}−\mathrm{3}\right){P}\left({x}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({x}−\mathrm{3}\right)\left[{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{14}−{P}\left({x}\right)\right]=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{3}\:{or}\:{P}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{14}\Rightarrow{P}\left(\mathrm{3}\right)=\mathrm{32} \\ $$