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Question Number 96904 by pticantor last updated on 05/Jun/20

Commented by pticantor last updated on 05/Jun/20

$$\mathit{s}\mathit{h}\mathit{o}\mathit{w}\mathit{t}\mathit{h}\mathit{a}\mathit{t}\mathit{t}\mathit{h}\mathit{e}\mathit{r}\mathit{e}\mathit{a}\mathit{r}\mathit{e}4\mathit{r}\mathit{e}\mathit{a}\mathit{l}\mathit{n}\mathit{u}\mathit{m}\mathit{b}\mathit{e}\mathit{r}\mathit{s}\mathit{a},\mathit{b},\mathit{c}\mathit{a}\mathit{n}\mathit{d}\mathit{d}$$$$\mathit{w}\mathit{h}\mathit{i}\mathit{c}\mathit{h}\mathit{w}\mathit{i}\mathit{l}\mathit{l}\mathit{b}\mathit{e}\mathit{s}\mathit{p}\mathit{e}\mathit{c}\mathit{i}\mathit{f}\mathit{i}\mathit{e}\mathit{d}\mathit{s}\mathit{u}\mathit{c}\mathit{h}\mathit{t}\mathit{h}\mathit{a}\mathit{t}$$$${({\mathit{x}}^2-4\mathit{x}+5)}^2+{(\mathit{x}+1)}^2=({\mathit{x}}^2+\mathit{a}\mathit{x}+\mathit{b})({\mathit{x}}^2+\mathit{c}\mathit{x}+\mathit{d})$$

Commented by john santu last updated on 05/Jun/20

$$\Rightarrow x^4-8x^3+27x^2-38x+26=$$$$(x^2-2x+2)(x^2-6x+13)$$$$\{\begin{array}{l}a=-2,b=2,c=-6,d=13or\\ a=-6,b=13,c=-2,d=2\end{array}$$

Commented by pticantor last updated on 05/Jun/20

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{y}\mathit{o}\mathit{u}\mathit{s}\mathit{i}\mathit{r}$$

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