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Question Number 82265 by Ali Yousafzai last updated on 19/Feb/20

$$f\mathit{a}\mathit{c}\mathit{t}\mathit{o}\mathit{r}\mathit{i}\mathit{z}\mathit{e}:$$$$(x+1)(x+2)(x+3)(x+6)-3x^2$$$$$$

Answered by MJS last updated on 19/Feb/20

$$=x^4+12x^3+44x^2+72x+36=$$$$[x=t-3]$$$$=t^4-10t^2+24t-27=$$$$=(t^2-\alpha t-\beta )(t^2+\alpha t-\gamma )=$$$$$$$$\mathrm{comparing}\mathrm{factors}\mathrm{leads}\mathrm{to}$$$$\{\begin{array}{l}{\alpha }^2+\beta +\gamma -10=0\\ \alpha \beta -\alpha \gamma +24=0\\ \beta \gamma +27=0\end{array}$$$$\Rightarrow \alpha =2\wedge \beta =-3\wedge \gamma =9$$$$$$$$=(t^2-2t+3)(t^2+2t-9)=$$$$=(x^2+4x+6)(x^2+8x+6)=$$$$=(x+4-\sqrt{10})(x+4+\sqrt{10})(x+2-\mathrm{i}\sqrt{2})(x+2+\mathrm{i}\sqrt{2})$$

Answered by behi83417@gmail.com last updated on 19/Feb/20

$$(\mathrm{x}+1)(\mathrm{x}+6)={\mathrm{x}}^2+7\mathrm{x}+6$$$$(\mathrm{x}+2)(\mathrm{x}+3)={\mathrm{x}}^2+5\mathrm{x}+6$$$$\stackrel{{\mathrm{x}}^2+6=\mathrm{t}}{\Rightarrow }(\mathrm{t}+5\mathrm{x})(\mathrm{t}+7\mathrm{x})={\mathrm{t}}^2+12\mathrm{tx}+{35\mathrm{x}}^2$$$$\Rightarrow \mathrm{p}={\mathrm{t}}^2+12\mathrm{tx}+{32\mathrm{x}}^2$$$$△={144\mathrm{x}}^2-4\times {32\mathrm{x}}^2={16\mathrm{x}}^2$$$$\Rightarrow \mathrm{t}=\frac{-12\mathrm{x}\pm 4\mathrm{x}}{2}=-8\mathrm{x},-4\mathrm{x}$$$$\Rightarrow \mathrm{p}=({\mathrm{x}}^2+4\mathrm{x}+6)({\mathrm{x}}^2+8\mathrm{x}+6)$$$$\Rightarrow \mathrm{p}=[({\mathrm{x}}^2+4\mathrm{x}+4)+2][({\mathrm{x}}^2+8\mathrm{x}+16)-10]=$$$$=[{(\mathrm{x}+2)}^2-{2\mathrm{i}}^2][{(\mathrm{x}+4)}^2-10]=$$$$=(\mathrm{x}+2+\mathrm{i}\sqrt{2})(\mathrm{x}+2-\mathrm{i}\sqrt{2})(\mathrm{x}+4+\sqrt{10})(\mathrm{x}+4-\sqrt{10}).$$

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