Question Number 82265 by Ali Yousafzai last updated on 19/Feb/20
$${f}\boldsymbol{{actorize}}: \\ $$$$\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)\left({x}+\mathrm{6}\right)−\mathrm{3}{x}^{\mathrm{2}} \\ $$$$ \\ $$
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 −αt−β)(t^2 +αt−γ)= comparing factors leads to { ((α^2 +β+γ−10=0)),((αβ−αγ+24=0)),((βγ+27=0)) :} ⇒ α=2∧β=−3∧γ=9 =(t^2 −2t+3)(t^2 +2t−9)= =(x^2 +4x+6)(x^2 +8x+6)= =(x+4−(√(10)))(x+4+(√(10)))(x+2−i(√2))(x+2+i(√2))](https://www.tinkutara.com/question/Q82266.png)
$$={x}^{\mathrm{4}} +\mathrm{12}{x}^{\mathrm{3}} +\mathrm{44}{x}^{\mathrm{2}} +\mathrm{72}{x}+\mathrm{36}= \\ $$$$\:\:\:\:\:\left[{x}={t}−\mathrm{3}\right] \\ $$$$={t}^{\mathrm{4}} −\mathrm{10}{t}^{\mathrm{2}} +\mathrm{24}{t}−\mathrm{27}= \\ $$$$=\left({t}^{\mathrm{2}} −\alpha{t}−\beta\right)\left({t}^{\mathrm{2}} +\alpha{t}−\gamma\right)= \\ $$$$ \\ $$$$\:\:\:\:\:\mathrm{comparing}\:\mathrm{factors}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\:\:\:\:\:\begin{cases}{\alpha^{\mathrm{2}} +\beta+\gamma−\mathrm{10}=\mathrm{0}}\\{\alpha\beta−\alpha\gamma+\mathrm{24}=\mathrm{0}}\\{\beta\gamma+\mathrm{27}=\mathrm{0}}\end{cases} \\ $$$$\:\:\:\:\:\Rightarrow\:\alpha=\mathrm{2}\wedge\beta=−\mathrm{3}\wedge\gamma=\mathrm{9} \\ $$$$ \\ $$$$=\left({t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{3}\right)\left({t}^{\mathrm{2}} +\mathrm{2}{t}−\mathrm{9}\right)= \\ $$$$=\left({x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{6}\right)\left({x}^{\mathrm{2}} +\mathrm{8}{x}+\mathrm{6}\right)= \\ $$$$=\left({x}+\mathrm{4}−\sqrt{\mathrm{10}}\right)\left({x}+\mathrm{4}+\sqrt{\mathrm{10}}\right)\left({x}+\mathrm{2}−\mathrm{i}\sqrt{\mathrm{2}}\right)\left({x}+\mathrm{2}+\mathrm{i}\sqrt{\mathrm{2}}\right) \\ $$
Answered by behi83417@gmail.com last updated on 19/Feb/20
![(x+1)(x+6)=x^2 +7x+6 (x+2)(x+3)=x^2 +5x+6 ⇒^(x^2 +6=t) (t+5x)(t+7x)=t^2 +12tx+35x^2 ⇒p=t^2 +12tx+32x^2 △=144x^2 −4×32x^2 =16x^2 ⇒t=((−12x±4x)/2)=−8x,−4x ⇒p=(x^2 +4x+6)(x^2 +8x+6) ⇒p=[(x^2 +4x+4)+2][(x^2 +8x+16)−10]= =[(x+2)^2 −2i^2 ][(x+4)^2 −10]= =(x+2+i(√2))(x+2−i(√2))(x+4+(√(10)))(x+4−(√(10))) .](https://www.tinkutara.com/question/Q82282.png)
$$\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}+\mathrm{6}\right)=\mathrm{x}^{\mathrm{2}} +\mathrm{7x}+\mathrm{6} \\ $$$$\left(\mathrm{x}+\mathrm{2}\right)\left(\mathrm{x}+\mathrm{3}\right)=\mathrm{x}^{\mathrm{2}} +\mathrm{5x}+\mathrm{6} \\ $$$$\overset{\mathrm{x}^{\mathrm{2}} +\mathrm{6}=\mathrm{t}} {\Rightarrow}\left(\mathrm{t}+\mathrm{5x}\right)\left(\mathrm{t}+\mathrm{7x}\right)=\mathrm{t}^{\mathrm{2}} +\mathrm{12tx}+\mathrm{35x}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{p}=\mathrm{t}^{\mathrm{2}} +\mathrm{12tx}+\mathrm{32x}^{\mathrm{2}} \\ $$$$\bigtriangleup=\mathrm{144x}^{\mathrm{2}} −\mathrm{4}×\mathrm{32x}^{\mathrm{2}} =\mathrm{16x}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{t}=\frac{−\mathrm{12x}\pm\mathrm{4x}}{\mathrm{2}}=−\mathrm{8x},−\mathrm{4x} \\ $$$$\Rightarrow\mathrm{p}=\left(\mathrm{x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{6}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{8x}+\mathrm{6}\right) \\ $$$$\Rightarrow\mathrm{p}=\left[\left(\mathrm{x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{4}\right)+\mathrm{2}\right]\left[\left(\mathrm{x}^{\mathrm{2}} +\mathrm{8x}+\mathrm{16}\right)−\mathrm{10}\right]= \\ $$$$=\left[\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2i}^{\mathrm{2}} \right]\left[\left(\mathrm{x}+\mathrm{4}\right)^{\mathrm{2}} −\mathrm{10}\right]= \\ $$$$=\left(\mathrm{x}+\mathrm{2}+\mathrm{i}\sqrt{\mathrm{2}}\right)\left(\mathrm{x}+\mathrm{2}−\mathrm{i}\sqrt{\mathrm{2}}\right)\left(\mathrm{x}+\mathrm{4}+\sqrt{\mathrm{10}}\right)\left(\mathrm{x}+\mathrm{4}−\sqrt{\mathrm{10}}\right)\:. \\ $$