Question

Determine the roots of p(x)=3(x+1)^4-5(x+1)^2+1
"Show your work"

Answer 1

`p(x)=3(x+1)^4-5(x+1)^2+1=3\left[(x+1)^2\right]^2-5(x+1)^2+1\\\\\text{subtitute:}\ (x+1)^2=t\geq0`


`p(t)=3t^2-5t+1\\\\p(t)=0\iff3t^2-5t+1=0\\\\a=3;\ b=-5;\ c=1\\\\\Delta=b^2-4ac\to\Delta=(-5)^2-4\cdot3\cdot1=25-12=13 > 0\\\\\sqrt\Delta=√(13)\\\\x_1=(-b-\sqrt\Delta)/(2a);\ x_2=(-b+\sqrt\Delta)/(2a)\\\\x_1=(-(-5)-√(13))/(2\cdot3)=(5-√(13))/(6) > 0\\\\x_2=(-(-5)+√(13))/(2\cdot3)=(5+√(13))/(6) > 0`


`(x+1)^2=(5-√(13))/(6)\ \vee\ (x+1)^2=(5+√(13))/(6)\\\\x+1=\pm\sqrt{(5-√(13))/(6)}\ \vee\ x+1=\pm\sqrt{(5+√(13))/(6)}`


`Answer:\\\\\boxed{x\in\left\{-\sqrt{(5-√(13))/(6)}-1;\ -\sqrt{(5+√(13))/(6)}-1;\ \sqrt{(5-√(13))/(6)}-1;\ \sqrt{(5+√(13))/(6)}-1\right\}}`