Question Number 162351 by ZiYangLee last updated on 29/Dec/21
$$\mathrm{how}\:\mathrm{to}\:\mathrm{show}\: \\ $$$${f}\left({x}\right)={x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{2}} −\mathrm{16}{x}−\mathrm{20}\: \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{form}\:\mathrm{of}\:\left({x}^{\mathrm{2}} +{x}+{a}\right)^{\mathrm{2}} −\mathrm{4}\left({x}+{b}\right)^{\mathrm{2}} . \\ $$
Answered by Rasheed.Sindhi last updated on 29/Dec/21
$${f}\left({x}\right)={x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{2}} −\mathrm{16}{x}−\mathrm{20}\: \\ $$$${g}\left({x}\right)=\left({x}^{\mathrm{2}} +{x}+{a}\right)^{\mathrm{2}} −\mathrm{4}\left({x}+{b}\right)^{\mathrm{2}} \\ $$$$ \\ $$$$\left({x}^{\mathrm{2}} +{x}+{a}\right)^{\mathrm{2}} −\mathrm{4}\left({x}+{b}\right)^{\mathrm{2}} \\ $$$$={x}^{\mathrm{4}} +{x}^{\mathrm{2}} +{a}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{3}} +\mathrm{2}{ax}+\mathrm{2}{ax}^{\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} −\mathrm{8}{bx}−{b}^{\mathrm{2}} \\ $$$$={x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} +\mathrm{2}{ax}^{\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} +{x}^{\mathrm{2}} +{a}^{\mathrm{2}} +\mathrm{2}{ax}−\mathrm{8}{bx}−\mathrm{4}{b}^{\mathrm{2}} \\ $$$$={x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} +\left(\mathrm{2}{a}−\mathrm{3}\right){x}^{\mathrm{2}} +\left(\mathrm{2}{a}−\mathrm{8}{b}\right){x}+{a}^{\mathrm{2}} −\mathrm{4}{b}^{\mathrm{2}} \\ $$$${Comparing}\:{coefficients}\:{of}\:{f}\:\&\:{g} \\ $$$$\mathrm{2}{a}−\mathrm{3}=\mathrm{5}\:\wedge\:\mathrm{2}{a}−\mathrm{8}{b}=−\mathrm{16}\:\wedge\:{a}^{\mathrm{2}} −\mathrm{4}{b}^{\mathrm{2}} =−\mathrm{20} \\ $$$${a}=\mathrm{4}\:\wedge{b}=\mathrm{3}\:\wedge\:\mathrm{4}^{\mathrm{2}} −\mathrm{4}.\mathrm{3}^{\mathrm{2}} =−\mathrm{20}\checkmark \\ $$$${g}\left({x}\right)=\left({x}^{\mathrm{2}} +{x}+\mathrm{4}\right)^{\mathrm{2}} −\mathrm{4}\left({x}+\mathrm{3}\right)^{\mathrm{2}} \\ $$
Commented by mr W last updated on 29/Dec/21
$${nice}\:{working}! \\ $$
Commented by Rasheed.Sindhi last updated on 29/Dec/21
$$\mathbb{T}\mathrm{han}\mathbb{X}\:\mathrm{for}\:\mathrm{mentioning}\:\mathrm{my}\:\mathrm{mistake}\:\mathbb{S}\mathrm{ir}! \\ $$
Commented by Rasheed.Sindhi last updated on 29/Dec/21
$$\mathbb{T}\underset{\boldsymbol{\mathrm{sir}}!} {\overset{\mathbb{L}\underset{\boldsymbol{\mathrm{f}}} {\boldsymbol{\mathrm{o}}}\mathrm{t}} {\mathrm{han}}}\mathbb{X} \\ $$