Solve the following equation with approximation method can any body solve the equation $x^7 + 2x +(61/20)=0    \tag{A}$
here is my method $\cdots x^7 + 2x +3=0$ hence $x=-1$. 
Then I assume  (A) as $x^7+2x+3+q=0$ where $q=1/20$
writing $x=x(0)+qx(1)+((q^2)x(2))\tag{B}$           (I ignore powers of $x$ which are more than $2$)
then plugging (B) in (A) we may get the answer through approximation method There's my problem : I can't plug (B) in (A) and get the answer!!please help me....
 A: You have
$$x_0^7 + 2 x_0+3 = 0 \implies x_0=-1$$
Now you want to solve
$$x^7+2 x+(3+\delta)=0$$
where $\delta = 1/20$.  Assume $x=x_0+\epsilon$, where $\epsilon$ is small compared with $x_0$.  Then
$$(x_0+\epsilon)^7 + 2 (x_0+\epsilon) + (3+\delta)=0 $$
Taylor expand the lead term to get
$$x_0^7 + 7 x_0^6 \epsilon + 2 x_0 + 2 \epsilon + 3 + \delta = O(\epsilon^2)$$
The $O$ term on the right-hand side means that we are ignoring any powers of $\epsilon$ beyond linear.  Now use the original equation and get a simple equation in $\epsilon$:
$$7 x_0^6 \epsilon+ 2 \epsilon+ \delta = 0$$
or
$$\epsilon = -\frac{\delta}{7 x_0^6+2} = -\frac{1}{180}$$
Thus, the approximate solution is $x_0+\epsilon= -181/180 = -1.055\bar{5}$.
WA gives the respective solution as $\approx -1.00548$; not bad but clearly some room for improvement.
A: A straightforward and efficient method is Newton--Raphson: Let $f(x)=x^7+2x+61/20\;(x\in \Bbb R)$. Start with $x_0=-1$, and iterate with $$x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}\quad(n=0,1,\dots),$$where $f'(x)=7x^6+2\;(x\in \Bbb R)$. Two or three iterations should suffice for good accuracy.
A: Since it seems that
a root $x_0 \approx -1$,
let $x = -1+c$ and see what happens
(historically, I think this is akin to Horner's rule).
If $f(x) = x^7+2x+(61/20)$,
and $c$ is small,
$\begin{align}
f(-1+c)
&=(-1+c)^7+2(-1+c)+61/20\\
&\approx (-1)^7 + 7(-1)^6 c -2+2c+61/20\\
&=-3+7c+2c+61/20\\
&=1/20+9c\\
\end{align}
$
or,
to make $f(-1+c)=0$,
$c = -1/180$.
So, a better root is
$1-1/180$.
