1

I want to make these lines below 'b(x)', but I don't know how to do it. Can someone help me, please?enter image description here

4

Again with pst-node and align:

\documentclass{article}
\usepackage{mathtools}
\usepackage{pst-node} 
\usepackage{auto-pst-pdf} 

\begin{document}

  \begin{align*} P_2(x) & = ax^2 + \rnode[b]{B}{bx\vphantom{j}} + c \\[1ex]
  &\qquad b_{1}\rnode[t]{B1}{(^{\vphantom{1}}x})\hspace{1.2em} \rnode[t]{B2}{b_{2}^{\vphantom{2}} }(x)
  \psset{linewidth=0.56pt, linejoin=1}
  \psline(B1)(B)(B2)
  \end{align*}

\end{document} 

enter image description here

| improve this answer | |
3

Try the following it will not give you the lines but it will position the b_1(x) and b_c(x).

\documentclass{article}
\usepackage{amsmath} 
\begin{document}
\begin{equation}
P_2(x)=ax^2+\underset{b_1(x) b_2(x)}{\mathrm{bx}}+c
\end{equation}
\end{document}
| improve this answer | |
3

With TikZ:

\documentclass{article}
\usepackage{tikz}
\usepackage{amsmath}
\newcommand{\tm}[1]{\tikz[overlay, anchor=base] \node (#1) {};}
\tikzstyle{every picture}+=[remember picture]
\begin{document}
\begin{tikzpicture}

   $ \tm{P}P_2(x) =
    \tm{a} ax^2+
    \tm{b} b
    \tm{bx}x+
    \tm{c}c$

 \draw (bx.-90)--++(-30:0.5)node[below]{$b_2(x)$};
 \draw (bx.-90)--++(-150:0.5)node[below]{$b_1(x)$};

 \end{tikzpicture}
\end{document}

enter image description here

| improve this answer | |
2

Also with istgame package it is possible to obtain a nice solution.

enter image description here

\documentclass[a4paper,12pt]{article}
\usepackage{istgame}
\begin{document}
\begin{istgame}
\xtdistance{7mm}{10mm}
\istroot(0)[null node]{$\mkern-80mu P(x)=ax^2+bx+c$}
\istb{}[al]{b_1(x)}
\istb{}[ar]{b_2(x)}
\endist
\end{istgame}
\end{document}
| improve this answer | |

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