+0  
 
0
241
1
avatar

p(x)=(x ​2 ​​ +3x+2)(x+1

Guest Apr 14, 2017
 #1
avatar+7348 
0

p(x)=(x ​2 ​​ +3x+2)(x+1

 

Do you need the zeros? Or the first derivative?

 

\(p(x)=(x ^2 ​​ +3x+2)(x+1)=0\\x+1=0\\\color{blue}x_1=-1\)

 

\(x^2+3x+2=0\\x=-\frac{p}{2}\pm(\sqrt{(\frac{p}{2})^2-q}=-1.5\pm(\sqrt{1.5^2-2}\)

\(x=-1.5\pm0.5\)

\(x_2=-1\)

\(x_3=-2\)

 

x1 and x2 = -1 is a double zero in a minimum of the function.

 

\(p(x)=(x ^2 ​​ +3x+2)(x+1)=x^3+3x^2+2x+x^2+3x+2\)

 

\(p(x)=x^3+4x^2+5x+2\)

 

\(p'(x)=3x^2+8x+5\)

 

\(3x^2+8x+5=0\)

 

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

 

\(x = {-8 \pm \sqrt{64-60} \over 6}\) 

 

\( x_{max}=-\frac{5}{3}\\ \\x_{min}=-1\)

 

laugh  !

asinus  Apr 14, 2017

22 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.