Solution 2.1:4a

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
m (Lösning 2.1:4a moved to Solution 2.1:4a: Robot: moved page)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
First, we multiply the second bracket by
-
<center> [[Image:2_1_4a.gif]] </center>
+
<math>x</math>
-
{{NAVCONTENT_STOP}}
+
from the first bracket,
 +
 
 +
 
 +
<math>\left( x+2 \right)\left( 3x^{2}-x+5 \right)=x\centerdot 3x^{2}-x\centerdot x+x\centerdot 5+...</math>
 +
 
 +
 
 +
Then, do the same for
 +
<math>2</math>
 +
from the first bracket:
 +
 
 +
 
 +
<math>\left( x+2 \right)\left( 3x^{2}-x+5 \right)=3x^{3}-x^{2}+5x+2\centerdot 3x^{2}-2\centerdot x+2\centerdot 5</math>
 +
 
 +
 
 +
Now, collect together
 +
<math>x^{3}</math>-,
 +
<math>x^{2}</math>-,
 +
<math>x</math>- and the constant terms:
 +
 
 +
 
 +
<math>3x^{3}+\left( -1+6 \right)x^{2}+\left( 5-2 \right)x+10=3x^{3}+5x^{2}+3x+10</math>
 +
 
 +
 
 +
The coefficient in front of
 +
<math>x^{2}</math>
 +
is
 +
<math>5</math>
 +
and the coefficient in front of x is
 +
<math>3</math>.

Revision as of 13:42, 15 September 2008

First, we multiply the second bracket by \displaystyle x from the first bracket,


\displaystyle \left( x+2 \right)\left( 3x^{2}-x+5 \right)=x\centerdot 3x^{2}-x\centerdot x+x\centerdot 5+...


Then, do the same for \displaystyle 2 from the first bracket:


\displaystyle \left( x+2 \right)\left( 3x^{2}-x+5 \right)=3x^{3}-x^{2}+5x+2\centerdot 3x^{2}-2\centerdot x+2\centerdot 5


Now, collect together \displaystyle x^{3}-, \displaystyle x^{2}-, \displaystyle x- and the constant terms:


\displaystyle 3x^{3}+\left( -1+6 \right)x^{2}+\left( 5-2 \right)x+10=3x^{3}+5x^{2}+3x+10


The coefficient in front of \displaystyle x^{2} is \displaystyle 5 and the coefficient in front of x is \displaystyle 3.

Retrieved from "http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-ImperialCollege/index.php/Solution_2.1:4a"