Show Polynomial Is Prime

[nycmathguy]

College Algebra
Chapter 1/Section 5

 

[MathLover1]

it is prime because cannot be factored

 

[nycmathguy]

it is prime because cannot be factored

Ok but can you show this to be true algebraically?

 

[MathLover1]

=

=

 

[nycmathguy]

View attachment 3944

=View attachment 3948

=View attachment 3949

Looks good but I don't know what you just did. Ha!

 

[MathLover1]

x^2+x+1 .... compete square
(x^2+x+b^2)-b^2+1 ...........b=half of the coefficient of term x , so b=1/2

(x^2+x+(1/2)^2)-(1/2)^2+1

(x+1/2)^2-1/4+1

(x+1/2)^2-1/4+4/4
(x+1/2)^2 +3/4 -> square complete
you can also factor out 1/4 from (x+1/2)^2 to get 1/4 (2 x + 1)^2 + 3/4

 

[MathLover1]

in order to complete square you need to group terms containing x variable and add missing part to get form a^2+2ab+b^2=(a+b)^2
then you need subtract same squared number (b^2) as well in order to not to change original expression

 

[nycmathguy]

x^2+x+1 .... compete square
(x^2+x+b^2)-b^2+1 ...........b=half of the coefficient of term x , so b=1/2

(x^2+x+(1/2)^2)-(1/2)^2+1

(x+1/2)^2-1/4+1

(x+1/2)^2-1/4+4/4
(x+1/2)^2 +3/4 -> square complete
you can also factor out 1/4 from (x+1/2)^2 to get 1/4 (2 x + 1)^2 + 3/4

This is what I'm talking about.

 

[nycmathguy]

in order to complete square you need to group terms containing x variable and add missing part to get form a^2+2ab+b^2=(a+b)^2
then you need subtract same squared number (b^2) as well in order to not to change original expression

You said:

". . .you need subtract same squared number (b^2) as well in order to not to change original expression."

What do you mean here?