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p(x)=(x ​2 ​​ +3x+2)(x+1

Guest Apr 14, 2017
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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

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