Quart equation

[ahmedulkavi]

What are the roots [Four of them] of this quart equation:
x^4+2x^3+4x^2+8x+16=0
[Please show the process.]

[nayel]

Hint:

DIvide by $16=2^4$ to get
\[\left(\frac x2\right)^4+\left(\frac x2\right)^3+\left(\frac x2\right)^2+\left(\frac x2\right)+1=0.\]
Now write $a$ for $x/2$ and the equation is
\[a^4+a^3+a^2+a+1=0.\]

If you still can't solve it, here are some more steps:

Divide by $a^2$ and write it in the following form
\[\left(a^2+\frac{1}{a^2}\right)+\left(a+\frac 1a\right)+1=0.\]

If you still can't solve it, more steps:

Write $a^2+1/a^2$ as $(a+1/a)^2-2$, and then our equation becomes
\[\left(a+\frac 1a\right)^2+\left(a+\frac 1a\right)-1=0.\]

If you still can't do it, more steps:

Write $b=a+1/a$. Then
\[\begin{align*}b^2+b-1&=0\\
b^2+2\cdot b\cdot\frac 12+\left(\frac 12\right)^2-\left(\frac 12\right)^2-1&=0\\
\left(b+\frac 12\right)^2&=\frac 54\\
b&=\frac{-1\pm\sqrt 5}{2}\end{align*}\]
So
\[x+\frac 1x=\frac{-1\pm\sqrt 5}{2}.\]
Now find the value of $x$ in the same way as above.