How may I align the following set of equations as represented in the image?

Question

If IEEEeqnarray environment is not best-suited for this particular alignment, what could be the best environment to be used?

\documentclass[11pt,a4paper]{article}
\usepackage{blindtext}
\usepackage{mathtools}
\usepackage{IEEEtrantools}
\begin{document}
\begin{IEEEeqnarray*}{rCl}
(x-a)(x-2a)(x-3a)(x-4a)=x^{4}-P_{1}x^{3}+P_{2}x^{2}-P_{3}x+P_{4}.\\
\shortintertext{Here}
P_{1}&=&a+2a+3a+4a=10a,\\
P_{2}&=&1\times 2a^{2}+1\times 3a^{2}+1\times 4a^{2}+2\times 4a^{2}+3\times 4a^{2}=35a^{2},\\
P_{3}&=&2\times 3\times 4a^{3}+1\times 3\times 4a^{2}+1\times 2\times 4a^{3}+1\times 2\times 3a^{3}=50a^{3},\\
P_{4}&=&1\times 2\times 3\times 4a^{4}=24a^{4}.
\shortintertext{so that}
(x-a)(x-2a)(x-3a)(x-4a)=x^{4}-10ax^{3}+35a^{2}x^{2}-50a^{3}x+24a^{4}.
\end{IEEEeqnarray*}
\end{document}

Answer 1

The presentation of your alignment is not really that accurate. So, I assume the following will be sufficient:

\documentclass{article}
\usepackage{array}
\begin{document}
\[
  (x - a)(x - 2a)(x - 3a)(x - 4a) = x^4 - P_1 x^3 + P_2 x^2 - P_3 x + P_4.
\]
Here
\[
  \renewcommand{\arraystretch}{1.3}
  \begin{array}{r@{}>{{}}l@{}r@{}>{{}}l}
    P_1 &= a+2a+3a+4a &&= 10a, \\
    P_2 & \multicolumn{3}{@{}l}{{}= 1 \times 2a^2 + 1 \times 3a^2 + 1 \times 4a^2 + 2 \times 4a^2 + 3 \times 4a^2} \\
      &&&= 35a^2,\\
    P_3 & \multicolumn{3}{@{}l}{{}= 2 \times 3 \times 4a^3 + 1 \times 3 \times 4a^2 + 1 \times 2\times 4a^3 + 1 \times 2 \times 3a^3} \\
      &&&= 50a^3,\\
    P_4 &= 1 \times 2 \times 3 \times 4a^4 &&= 24a^4, \\
  \end{array}
\]
so that
\[
  (x - a)(x - 2a)(x - 3a)(x - 4a) = x^4 - 10ax^3 + 35a^2 x^{2} - 50a^3 x + 24a^4.
\]
\end{document}

Multiple alignments that's not strictly adhered to (some lines use the alignment points and some don't) is difficult to do with standard align and friends. Using an array may circumvent this difficulty with the aid of \multicolumn.

The use of the array package above is not really needed, but I've used it anyway.

Answer 2

Why not this simple way?

\documentclass{article}

\usepackage{mathtools}

\begin{document}

\noindent We have
\begin{align*}
\MoveEqLeft (x - a)(x - 2a)(x - 3a)(x - 4a)
  = x^{4} - P_{1}x^{3} + P_{2}x^{2} - P_{3}x + P_{4}\\
\intertext{where}
  P_{1} &= a + 2a + 3a + 4a = 10a,\\
  P_{2} &= 1 \cdot 2a^{2} + 1 \cdot 3a^{2} + 1 \cdot 4a^{2} + 2 \cdot 4a^{2} + 3 \cdot 4a^{2} = 35a^{2},\\
  P_{3} &= 2 \cdot 3 \cdot 4a^{3} + 1 \cdot 3 \cdot 4a^{2} + 1 \cdot 2 \cdot 4a^{3} + 1 \cdot 2 \cdot 3a^{3}= 50a^{3},\\
  P_{4} &= 1 \cdot 2 \cdot 3 \cdot 4a^{4} = 24a^{4},
\intertext{so that}
\MoveEqLeft (x - a)(x - 2a)(x - 3a)(x - 4a)
  = x^{4} - 10ax^{3} + 35a^{2}x^{2} - 50a^{3}x + 24a^{4}.
\end{align*}

\end{document} 

Answer 3

Here is how I would do it:

\documentclass{article}

\usepackage{mathtools}

\begin{document}

\noindent We have
\begin{equation*}
  (x - a)(x - 2a)(x - 3a)(x - 4a)
  = x^{4} - P_{1}x^{3} + P_{2}x^{2} - P_{3}x + P_{4}
\end{equation*}
where
\begin{align*}
  P_{1} &= a + 2a + 3a + 4a\\
        &= 10a,\\
  P_{2} &= 1 \cdot 2a^{2} + 1 \cdot 3a^{2} + 1 \cdot 4a^{2} + 2 \cdot 4a^{2} + 3 \cdot 4a^{2}\\
        &= 35a^{2},\\
  P_{3} &= 2 \cdot 3 \cdot 4a^{3} + 1 \cdot 3 \cdot 4a^{2} + 1 \cdot 2 \cdot 4a^{3} + 1 \cdot 2 \cdot 3a^{3}\\
        &= 50a^{3},\\
  P_{4} &= 1 \cdot 2 \cdot 3 \cdot 4a^{4}\\
        &= 24a^{4},
\end{align*}
so that
\begin{equation*}
  (x - a)(x - 2a)(x - 3a)(x - 4a)
  = x^{4} - 10ax^{3} + 35a^{2}x^{2} - 50a^{3}x + 24a^{4}.
\end{equation*}

\end{document}