Question Number 38910 by Rio Mike last updated on 01/Jul/18
$${Given}\:{a}\:{polynomial}\:{f}\left({x}\right)\:{where} \\ $$$${f}\left({x}\right)\:=\:{x}^{\mathrm{3}} \:+\:{ax}\:^{\mathrm{2}} \:+\:\mathrm{3}{x}\:+\:{b} \\ $$$${and}\:\left(\:{x}−\:\mathrm{1}\right)\:{is}\:{a}\:{factor}\:{of}\:{f}\left({x}\right) \\ $$$${and}\:{f}\:'\left({x}\right)=\:\:\mathrm{12}\:{at}\:{point}\:\left(\:\mathrm{2},\mathrm{4}\right) \\ $$$${find}\:{the}\:{values}\:{of}\:{a}\:{and}\:{b} \\ $$$${hence}\:{factorise}\:{f}\left({x}\right)\:{completely} \\ $$$$ \\ $$
Answered by ajfour last updated on 01/Jul/18
$$\mathrm{1}+{a}+\mathrm{3}+{b}=\mathrm{0} \\ $$$${f}\:'\left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{ax}+\mathrm{3} \\ $$$${f}\:'\left(\mathrm{2}\right)=\mathrm{12}+\mathrm{4}{a}+\mathrm{3}=\mathrm{12} \\ $$$$\Rightarrow\:\:\:{a}=−\frac{\mathrm{3}}{\mathrm{4}}\:\:;\:\:{b}=\:−\frac{\mathrm{13}}{\mathrm{4}} \\ $$$${f}\left({x}\right)=\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{px}+{q}\right) \\ $$$$\Rightarrow\:\:\:\:\:−{q}\:=\:{b}\:=\:−\frac{\mathrm{13}}{\mathrm{4}} \\ $$$${coefficient}\:{of}\:{x}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:{p}−\mathrm{1}\:=\:{a}\:=\:−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${f}\left({x}\right)=\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +\frac{{x}}{\mathrm{4}}+\frac{\mathrm{13}}{\mathrm{4}}\right)\:. \\ $$