Answer:
Factor over complex numbers of [tex]2 x^{4}+36 x^{2}+162[/tex] is [tex]2(x+3 i)^{2}(x-3 i)^{2}[/tex]
Solution:
From question given that
[tex]2 x^{4}+36 x^{2}+162[/tex] → (1)
On substituting [tex]x^{2}=y[/tex] in equation (1),
[tex]=2 y^{2}+36 y+162[/tex]
Taking 2 as a common in above expression,
[tex]=2\left(y^{2}+18 y+81\right)[/tex]
Rewrite the above expression,
[tex]=2\left[y^{2}+2(y)(9)+9^{2}\right][/tex]
[tex]=2(y+9)^{2}[/tex] [tex]\left[\text { Using }(a+b)^{2}=a^{2}+2 a b+b^{2}\right][/tex]
[tex]=2\left(x^{2}+9\right)^{2}[/tex] [tex]\left[\text { since } y=x^{2}\right][/tex]
[tex]\left[\text { Using } a^{2}+b^{2}=(a+i b)(a-i b), \text { where } i=\sqrt{-1}\right][/tex]
[tex]=2[(x+3 i)(x-3 i)]^{2}[/tex]
[tex]=2(x+3 i)^{2}(x-3 i)^{2}[/tex]
Hence Factor over complex numbers of [tex]2 x^{4}+36 x^{2}+162[/tex] is [tex]2(x+3 i)^{2}(x-3 i)^{2}[/tex]
