Question Number 186292 by ajfour last updated on 03/Feb/23
$$\left({x}−{p}\right)\left({x}^{\mathrm{3}} −{x}−\frac{\mathrm{1}}{\mathrm{3}}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{4}} −{px}^{\mathrm{3}} −{x}^{\mathrm{2}} +\left({p}−\frac{\mathrm{1}}{\mathrm{3}}\right){x}+\frac{{p}}{\mathrm{3}}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} +{ax}+{h}\right)\left({x}^{\mathrm{2}} +{bx}+{k}\right)=\mathrm{0} \\ $$$${a}+{b}=−{p} \\ $$$${h}+{k}+{ab}=−\mathrm{1} \\ $$$${bh}+{ak}={p}−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${hk}=\frac{{p}}{\mathrm{3}} \\ $$$$−−−−−− \\ $$$${say}\:{ab}={t} \\ $$$$−−−−−− \\ $$$${ah}+{bk}+{p}−\frac{\mathrm{1}}{\mathrm{3}}={p}\left(\mathrm{1}+{t}\right) \\ $$$${bh}+{ak}−\left({p}−\frac{\mathrm{1}}{\mathrm{3}}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\left({a}−{b}\right)\left({h}−{k}\right)={pt}−{p}+\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${squaring} \\ $$$$\left({p}^{\mathrm{2}} −\mathrm{4}{t}\right)\left\{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} −\frac{\mathrm{4}{p}}{\mathrm{3}}\right\}=\left({pt}−{p}+\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$${say}\:\:{t}+\mathrm{1}={z} \\ $$$$\left({p}^{\mathrm{2}} +\mathrm{4}−\mathrm{4}{z}\right)\left({z}^{\mathrm{2}} −\frac{\mathrm{4}{p}}{\mathrm{3}}\right)=\left({pz}−\mathrm{2}{p}+\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$−\mathrm{4}{z}^{\mathrm{3}} +\left({p}^{\mathrm{2}} +\mathrm{4}\right){z}^{\mathrm{2}} +\frac{\mathrm{16}{pz}}{\mathrm{3}}−\frac{\mathrm{4}{p}}{\mathrm{3}}\left({p}^{\mathrm{2}} +\mathrm{4}\right) \\ $$$$\:\:\:\:={p}^{\mathrm{2}} {z}^{\mathrm{2}} −\mathrm{4}{p}\left({p}−\frac{\mathrm{1}}{\mathrm{3}}\right){z}+\mathrm{4}\left({p}−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:{z}^{\mathrm{3}} −{z}^{\mathrm{2}} −{p}\left({p}+\mathrm{1}\right){z}+\left({p}−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:+\frac{{p}}{\mathrm{3}}\left({p}^{\mathrm{2}} +\mathrm{4}\right)=\mathrm{0} \\ $$$$…..\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$