Show Polynomial Is Prime

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

  1. nycmathguy

    nycmathguy

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    College Algebra
    Chapter 1/Section 5

    IMG_20220717_124156.jpg
     
    nycmathguy, Jul 17, 2022
    #1
  2. nycmathguy

    MathLover1

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    it is prime because cannot be factored
     
    MathLover1, Jul 17, 2022
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  3. nycmathguy

    nycmathguy

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    Ok but can you show this to be true algebraically?
     
    nycmathguy, Jul 17, 2022
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  4. nycmathguy

    MathLover1

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    upload_2022-7-17_16-4-0.gif

    = upload_2022-7-17_16-7-13.gif

    = upload_2022-7-17_16-8-13.gif
     

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    MathLover1, Jul 17, 2022
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  5. nycmathguy

    nycmathguy

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    nycmathguy, Jul 17, 2022
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  6. nycmathguy

    MathLover1

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    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
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  7. nycmathguy

    MathLover1

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    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
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  8. nycmathguy

    nycmathguy

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    This is what I'm talking about.
     
    nycmathguy, Jul 18, 2022
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  9. nycmathguy

    nycmathguy

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    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
    #9
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