Solution 3.4:1e - Förberedande kurs i matematik 2

-                      Welcome to the course
                       
                        About the course 
                        Examinations
                      
                    
                      1. Derivatives
                       
                        1.1 Introduction 
                        1.2 Rules of differentiation 
                        1.3 Max and min problems
                      
                    
                      2. Integrals
                       
                        2.1 Introduction 
                        2.2 Substitution 
                        2.3 Integration by parts
                      
                    
                      3. Complex numbers
                       
                        3.1 Calculations 
                        3.2 Polar form 
                        3.3 Powers and roots 
                        3.4 Complex polynomials
                      
                    
                    
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+			Solution 3.4:1e
			
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- {{NAVCONTENT_START}} + Imagine for a moment taking away all the terms in the numerator apart from <math>x^3</math>. If we are to  make <math>x^3</math> divisible by the denominator 
- <center> [[Image:3_4_1e.gif]] </center> + <math>x^2+3x+1</math>, we need to add and subtract <math>3x^2+x</math> in order to obtain the expression <math>x^3+3x^2+x=x(x^2+3x+1)</math>,
- {{NAVCONTENT_STOP}} +  
  + {{Displayed math||<math>\begin{align}
  + \frac{x^3+2x^2+1}{x^2+3x+1}
  + &= \frac{x^3\bbox[#FFEEAA;,1.5pt]{{}+3x^2+x-3x^2-x}+2x^2+1}{x^2+3x+1}\\[5pt]
  + &= \frac{x^3+3x^2+x}{x^2+3x+1} + \frac{-3x^2-x+2x^2+1}{x^2+3x+1}\\[5pt]
  + &= \frac{x(x^2+3x+1)}{x^2+3x+1} + \frac{-x^2-x+1}{x^2+3x+1}\\[5pt]
  + &= x+\frac{-x^2-x+1}{x^2+3x+1}\,\textrm{.} 
  + \end{align}</math>}}
  +  
  + Now, we carry out the same procedure with the new quotient. To the term <math>-x^2</math>, we add and subtract <math>-3x-1</math> and obtain
  +  
  + {{Displayed math||<math>\begin{align}
  + x + \frac{-x^2-x+1}{x^2+3x+1}
  + &= x + \frac{-x^2\bbox[#FFEEAA;,1.5pt]{{}-3x-1+3x+1}-x+1}{x^2+3x+1}\\[5pt]
  + &= x + \frac{-x^2-3x-1}{x^2+3x+1} + \frac{3x+1-x+1}{x^2+3x+1}\\[5pt]
  + &= x - 1 + \frac{2x+2}{x^2+3x+1}\,\textrm{.} 
  + \end{align}</math>}}
Current revision
Imagine for a moment taking away all the terms in the numerator apart from \displaystyle x^3. If we are to  make \displaystyle x^3 divisible by the denominator 
\displaystyle x^2+3x+1, we need to add and subtract \displaystyle 3x^2+x in order to obtain the expression \displaystyle x^3+3x^2+x=x(x^2+3x+1),





\displaystyle \begin{align}
\frac{x^3+2x^2+1}{x^2+3x+1}
&= \frac{x^3\bbox[#FFEEAA;,1.5pt]{{}+3x^2+x-3x^2-x}+2x^2+1}{x^2+3x+1}\\[5pt]
&= \frac{x^3+3x^2+x}{x^2+3x+1} + \frac{-3x^2-x+2x^2+1}{x^2+3x+1}\\[5pt]
&= \frac{x(x^2+3x+1)}{x^2+3x+1} + \frac{-x^2-x+1}{x^2+3x+1}\\[5pt]
&= x+\frac{-x^2-x+1}{x^2+3x+1}\,\textrm{.} 
\end{align}



Now, we carry out the same procedure with the new quotient. To the term \displaystyle -x^2, we add and subtract \displaystyle -3x-1 and obtain





\displaystyle \begin{align}
x + \frac{-x^2-x+1}{x^2+3x+1}
&= x + \frac{-x^2\bbox[#FFEEAA;,1.5pt]{{}-3x-1+3x+1}-x+1}{x^2+3x+1}\\[5pt]
&= x + \frac{-x^2-3x-1}{x^2+3x+1} + \frac{3x+1-x+1}{x^2+3x+1}\\[5pt]
&= x - 1 + \frac{2x+2}{x^2+3x+1}\,\textrm{.} 
\end{align}




Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse2-ImperialCollege/index.php/Solution_3.4:1e"
@@ Line 1: / Line 1: @@
-{{NAVCONTENT_START}}
+Imagine for a moment taking away all the terms in the numerator apart from <math>x^3</math>. If we are to  make <math>x^3</math> divisible by the denominator
-<center> [[Image:3_4_1e.gif]] </center>
+<math>x^2+3x+1</math>, we need to add and subtract <math>3x^2+x</math> in order to obtain the expression <math>x^3+3x^2+x=x(x^2+3x+1)</math>,
-{{NAVCONTENT_STOP}}
+{{Displayed math||<math>\begin{align}
+\frac{x^3+2x^2+1}{x^2+3x+1}
+&= \frac{x^3\bbox[#FFEEAA;,1.5pt]{{}+3x^2+x-3x^2-x}+2x^2+1}{x^2+3x+1}\\[5pt]
+&= \frac{x^3+3x^2+x}{x^2+3x+1} + \frac{-3x^2-x+2x^2+1}{x^2+3x+1}\\[5pt]
+&= \frac{x(x^2+3x+1)}{x^2+3x+1} + \frac{-x^2-x+1}{x^2+3x+1}\\[5pt]
+&= x+\frac{-x^2-x+1}{x^2+3x+1}\,\textrm{.}
+\end{align}</math>}}
+Now, we carry out the same procedure with the new quotient. To the term <math>-x^2</math>, we add and subtract <math>-3x-1</math> and obtain
+{{Displayed math||<math>\begin{align}
+x + \frac{-x^2-x+1}{x^2+3x+1}
+&= x + \frac{-x^2\bbox[#FFEEAA;,1.5pt]{{}-3x-1+3x+1}-x+1}{x^2+3x+1}\\[5pt]
+&= x + \frac{-x^2-3x-1}{x^2+3x+1} + \frac{3x+1-x+1}{x^2+3x+1}\\[5pt]
+&= x - 1 + \frac{2x+2}{x^2+3x+1}\,\textrm{.}
+\end{align}</math>}}
 \displaystyle \begin{align}
\frac{x^3+2x^2+1}{x^2+3x+1}
&= \frac{x^3\bbox[#FFEEAA;,1.5pt]{{}+3x^2+x-3x^2-x}+2x^2+1}{x^2+3x+1}\\[5pt]
&= \frac{x^3+3x^2+x}{x^2+3x+1} + \frac{-3x^2-x+2x^2+1}{x^2+3x+1}\\[5pt]
&= \frac{x(x^2+3x+1)}{x^2+3x+1} + \frac{-x^2-x+1}{x^2+3x+1}\\[5pt]
&= x+\frac{-x^2-x+1}{x^2+3x+1}\,\textrm{.} 
\end{align}
 \displaystyle \begin{align}
x + \frac{-x^2-x+1}{x^2+3x+1}
&= x + \frac{-x^2\bbox[#FFEEAA;,1.5pt]{{}-3x-1+3x+1}-x+1}{x^2+3x+1}\\[5pt]
&= x + \frac{-x^2-3x-1}{x^2+3x+1} + \frac{3x+1-x+1}{x^2+3x+1}\\[5pt]
&= x - 1 + \frac{2x+2}{x^2+3x+1}\,\textrm{.} 
\end{align}