Question Number 118184 by mathocean1 last updated on 15/Oct/20
$${factorise}\:{x}^{\mathrm{4}} +\mathrm{4} \\ $$
Commented by bemath last updated on 16/Oct/20
$$\left({x}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} \:=\:\left({x}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} \\ $$$$=\:\left({x}^{\mathrm{2}} +\mathrm{2}−\mathrm{2}{x}\right)\left({x}^{\mathrm{2}} +\mathrm{2}+\mathrm{2}{x}\right) \\ $$$$ \\ $$$${recall}\:{equality}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\:\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{2}{ab} \\ $$
Answered by Dwaipayan Shikari last updated on 15/Oct/20
$${x}^{\mathrm{4}} +\mathrm{4}+\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} \\ $$$$\left({x}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} −\left(\mathrm{2}{x}\right)^{\mathrm{2}} =\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right) \\ $$
Commented by mathocean1 last updated on 15/Oct/20
$${thanks} \\ $$$$ \\ $$
Answered by Bird last updated on 16/Oct/20
![inside R[x] x^4 +4 =(x^2 )^2 +2^2 =(x^2 +2)^2 −4x^2 =(x^2 +2−2x)(x^2 +2+2x) =(x^2 −2x+2)(x^2 +2x +2) inside C[x] let solve z^4 +4=0 ⇒ z^(4 ) =−4 let z=r e^(iθ) ⇒r^(4 ) e^(4iθ) =4e^((2k+1)iπ) ⇒ r =^4 (√4) and θ =(((2k+1)π)/4) the roots are z_k =^4 (√4) e^(i(((2k+1)π)/4)) k∈[[0,3]] ⇒ x^4 +4 =Π_(k=0) ^3 (x−z_k ) =(x−z_0 )(x−z_1 )(x−z_2 )(x−z_3 )](https://www.tinkutara.com/question/Q118215.png)
$${inside}\:{R}\left[{x}\right] \\ $$$${x}^{\mathrm{4}} +\mathrm{4}\:=\left({x}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} \:=\left({x}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} \\ $$$$=\left({x}^{\mathrm{2}} +\mathrm{2}−\mathrm{2}{x}\right)\left({x}^{\mathrm{2}} +\mathrm{2}+\mathrm{2}{x}\right) \\ $$$$=\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}\:+\mathrm{2}\right) \\ $$$${inside}\:{C}\left[{x}\right]\:{let}\:{solve}\:{z}^{\mathrm{4}} +\mathrm{4}=\mathrm{0}\:\Rightarrow \\ $$$${z}^{\mathrm{4}\:} =−\mathrm{4}\:\:{let}\:{z}={r}\:{e}^{{i}\theta} \\ $$$$\Rightarrow{r}^{\mathrm{4}\:} {e}^{\mathrm{4}{i}\theta} \:=\mathrm{4}{e}^{\left(\mathrm{2}{k}+\mathrm{1}\right){i}\pi} \:\Rightarrow \\ $$$${r}\:=^{\mathrm{4}} \sqrt{\mathrm{4}}\:{and}\:\theta\:=\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{4}} \\ $$$${the}\:{roots}\:{are}\:{z}_{{k}} =^{\mathrm{4}} \sqrt{\mathrm{4}}\:{e}^{{i}\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{4}}} \\ $$$${k}\in\left[\left[\mathrm{0},\mathrm{3}\right]\right]\:\Rightarrow \\ $$$${x}^{\mathrm{4}} +\mathrm{4}\:=\prod_{{k}=\mathrm{0}} ^{\mathrm{3}} \left({x}−{z}_{{k}} \right) \\ $$$$=\left({x}−{z}_{\mathrm{0}} \right)\left({x}−{z}_{\mathrm{1}} \right)\left({x}−{z}_{\mathrm{2}} \right)\left({x}−{z}_{\mathrm{3}} \right) \\ $$
Commented by MJS_new last updated on 16/Oct/20
$${x}^{\mathrm{4}} +\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow\:{x}=\pm\mathrm{1}\pm\mathrm{i}\:\left(\mathrm{all}\:\mathrm{combinations}\right) \\ $$