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how-to-show-f-x-x-4-2x-3-5x-2-16x-20-in-the-form-of-x-2-x-a-2-4-x-b-2-

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Question Number 162351 by ZiYangLee last updated on 29/Dec/21
how to show f(x)=x^4 +2x^3 +5x^2 −16x−20 in the form of (x^2 +x+a)^2 −4(x+b)^2 .
$$\mathrm{how}\:\mathrm{to}\:\mathrm{show}\: \\ $$$${f}\left({x}\right)={x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{2}} −\mathrm{16}{x}−\mathrm{20}\: \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{form}\:\mathrm{of}\:\left({x}^{\mathrm{2}} +{x}+{a}\right)^{\mathrm{2}} −\mathrm{4}\left({x}+{b}\right)^{\mathrm{2}} . \\ $$
Answered by Rasheed.Sindhi last updated on 29/Dec/21
f(x)=x^4 +2x^3 +5x^2 −16x−20 g(x)=(x^2 +x+a)^2 −4(x+b)^2 (x^2 +x+a)^2 −4(x+b)^2 =x^4 +x^2 +a^2 +2x^3 +2ax+2ax^2 −4x^2 −8bx−b^2 =x^4 +2x^3 +2ax^2 −4x^2 +x^2 +a^2 +2ax−8bx−4b^2 =x^4 +2x^3 +(2a−3)x^2 +(2a−8b)x+a^2 −4b^2 Comparing coefficients of f & g 2a−3=5 ∧ 2a−8b=−16 ∧ a^2 −4b^2 =−20 a=4 ∧b=3 ∧ 4^2 −4.3^2 =−20✓ g(x)=(x^2 +x+4)^2 −4(x+3)^2
$${f}\left({x}\right)={x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{2}} −\mathrm{16}{x}−\mathrm{20}\: \\ $$$${g}\left({x}\right)=\left({x}^{\mathrm{2}} +{x}+{a}\right)^{\mathrm{2}} −\mathrm{4}\left({x}+{b}\right)^{\mathrm{2}} \\ $$$$ \\ $$$$\left({x}^{\mathrm{2}} +{x}+{a}\right)^{\mathrm{2}} −\mathrm{4}\left({x}+{b}\right)^{\mathrm{2}} \\ $$$$={x}^{\mathrm{4}} +{x}^{\mathrm{2}} +{a}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{3}} +\mathrm{2}{ax}+\mathrm{2}{ax}^{\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} −\mathrm{8}{bx}−{b}^{\mathrm{2}} \\ $$$$={x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} +\mathrm{2}{ax}^{\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} +{x}^{\mathrm{2}} +{a}^{\mathrm{2}} +\mathrm{2}{ax}−\mathrm{8}{bx}−\mathrm{4}{b}^{\mathrm{2}} \\ $$$$={x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} +\left(\mathrm{2}{a}−\mathrm{3}\right){x}^{\mathrm{2}} +\left(\mathrm{2}{a}−\mathrm{8}{b}\right){x}+{a}^{\mathrm{2}} −\mathrm{4}{b}^{\mathrm{2}} \\ $$$${Comparing}\:{coefficients}\:{of}\:{f}\:\&\:{g} \\ $$$$\mathrm{2}{a}−\mathrm{3}=\mathrm{5}\:\wedge\:\mathrm{2}{a}−\mathrm{8}{b}=−\mathrm{16}\:\wedge\:{a}^{\mathrm{2}} −\mathrm{4}{b}^{\mathrm{2}} =−\mathrm{20} \\ $$$${a}=\mathrm{4}\:\wedge{b}=\mathrm{3}\:\wedge\:\mathrm{4}^{\mathrm{2}} −\mathrm{4}.\mathrm{3}^{\mathrm{2}} =−\mathrm{20}\checkmark \\ $$$${g}\left({x}\right)=\left({x}^{\mathrm{2}} +{x}+\mathrm{4}\right)^{\mathrm{2}} −\mathrm{4}\left({x}+\mathrm{3}\right)^{\mathrm{2}} \\ $$
Commented by mr W last updated on 29/Dec/21
nice working!
$${nice}\:{working}! \\ $$
Commented by Rasheed.Sindhi last updated on 29/Dec/21
ThanX for mentioning my mistake Sir!
$$\mathbb{T}\mathrm{han}\mathbb{X}\:\mathrm{for}\:\mathrm{mentioning}\:\mathrm{my}\:\mathrm{mistake}\:\mathbb{S}\mathrm{ir}! \\ $$
Commented by Rasheed.Sindhi last updated on 29/Dec/21
Than_(sir!) ^(Lo_(f) t) X
$$\mathbb{T}\underset{\boldsymbol{\mathrm{sir}}!} {\overset{\mathbb{L}\underset{\boldsymbol{\mathrm{f}}} {\boldsymbol{\mathrm{o}}}\mathrm{t}} {\mathrm{han}}}\mathbb{X} \\ $$

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