# Show Polynomial Is Prime

Discussion in 'Algebra' started by nycmathguy, Jul 17, 2022.

1. ### nycmathguy

Joined:
Jun 27, 2021
Messages:
5,386
420
College Algebra
Chapter 1/Section 5

nycmathguy, Jul 17, 2022

2. ### MathLover1

Joined:
Jun 27, 2021
Messages:
2,989
2,883
it is prime because cannot be factored

MathLover1, Jul 17, 2022
nycmathguy likes this.

3. ### nycmathguy

Joined:
Jun 27, 2021
Messages:
5,386
420
Ok but can you show this to be true algebraically?

nycmathguy, Jul 17, 2022
4. ### MathLover1

Joined:
Jun 27, 2021
Messages:
2,989
2,883

=

=

#### Attached Files:

File size:
1.5 KB
Views:
3
File size:
1.5 KB
Views:
3
File size:
1.4 KB
Views:
3
MathLover1, Jul 17, 2022
nycmathguy likes this.
5. ### nycmathguy

Joined:
Jun 27, 2021
Messages:
5,386
420
Looks good but I don't know what you just did. Ha!

nycmathguy, Jul 17, 2022
6. ### MathLover1

Joined:
Jun 27, 2021
Messages:
2,989
2,883
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, Jul 17, 2022
nycmathguy likes this.
7. ### MathLover1

Joined:
Jun 27, 2021
Messages:
2,989
2,883
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

MathLover1, Jul 17, 2022
nycmathguy likes this.
8. ### nycmathguy

Joined:
Jun 27, 2021
Messages:
5,386
420
This is what I'm talking about.

nycmathguy, Jul 18, 2022
9. ### nycmathguy

Joined:
Jun 27, 2021
Messages:
5,386
420
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?

nycmathguy, Jul 18, 2022