Question Number 214317 by muallimRiyoziyot last updated on 05/Dec/24
$${x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{11}{x}^{\mathrm{2}} +{x}−\mathrm{12}={f}\left({x}\right)×{g}\left({x}\right) \\ $$$${f}\left({x}\right)=?\:\:\:\:{g}\left({x}\right)=? \\ $$
Answered by A5T last updated on 05/Dec/24
![x^4 −11x^2 −12+x^3 +x =x^4 −12x^2 +x^2 −12+x(x^2 +1) =(x^2 −12)(x^2 +1)+x(x^2 +1)=(x^2 +1)(x^2 +x−12) =(x^2 +1)(x+4)(x−3) [f(x),g(x)]=[x^2 +1,x^2 +x−12] for instance.](Q214318.png)
$${x}^{\mathrm{4}} −\mathrm{11}{x}^{\mathrm{2}} −\mathrm{12}+{x}^{\mathrm{3}} +{x} \\ $$$$={x}^{\mathrm{4}} −\mathrm{12}{x}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{12}+{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$=\left({x}^{\mathrm{2}} −\mathrm{12}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)+{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right)=\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}−\mathrm{12}\right) \\ $$$$=\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}+\mathrm{4}\right)\left({x}−\mathrm{3}\right) \\ $$$$\left[{f}\left({x}\right),{g}\left({x}\right)\right]=\left[{x}^{\mathrm{2}} +\mathrm{1},{x}^{\mathrm{2}} +{x}−\mathrm{12}\right]\:{for}\:{instance}. \\ $$