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Math Help - Factoring Perfect Squares

  1. #1
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    Unhappy Factoring Perfect Squares

    Hey!

    I need help factoring this polynomial. I can see the perfect square but I am unable to get the correct answer when factoring this polynomial:

    <br />
\[\begin{array}{l}<br />
(x{}^6 - 2x{}^3 + 1)\\<br />
\\<br />
(x - 1){}^2 = (x{}^2 - 2x + 1)\\<br />
\\<br />
?<br />
\end{array}\]<br />
    Sam
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Factoring Perfect Squares

    You have to use [ tex ] ... [ / tex] in stead of [ math ] ... [ / math] but it's still disordered.

    Do you mean:
    \left \{ \begin{array}{rcl} x^6-2x^3+1 \\ (x-1)^2=x^2-2x+1 \end{array}\right.
    ?

    If you have difficulties to factore x^6-2x^3+1 let x^3=t and so you can write it as:
    t^2-2t+1
    Last edited by Siron; September 1st 2011 at 07:44 AM.
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  3. #3
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    Re: Factoring Perfect Squares

    Equivalently, x^6- 2x^2+ 1= (x^3)^2- 2(x^3)+ 1
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  4. #4
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    Re: Factoring Perfect Squares

    Wow, thanks for replying so fast.

    I exported the markup code from MathType, but I kept getting errors! I'm not too familiar with Tex and LaTex .

    What I meant was I could not get this equation:
    x^6-2x^3+1

    to!:
    (x-1)^2(x^2+x+1)

    I also don't understand how the answer multiplies:
    (x-1)^2(x^2+x+1)

    to get!:
    x^6-2x^3+1

    I was able to identify the perfect square:
    (x-1)^2

    But I did not know how to extract this perfect square without unsatisfying the equation!

    Listing the steps from:
    x^6-2x^3+1

    to,:
    (x-1)^2(x^2+x+1)

    would be greatly appreciated. Also could someone explain how extracting a perfect square does not unsatisfy the equation.

    Sam
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  5. #5
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    Re: Factoring Perfect Squares

    Quote Originally Posted by ArcherSam View Post
    Wow, thanks for replying so fast.

    I exported the markup code from MathType, but I kept getting errors! I'm not too familiar with Tex and LaTex .

    What I meant was I could not get this equation:
    x^6-2x^3+1

    to!:
    (x-1)^2(x^2+x+1)

    I also don't understand how the answer multiplies:
    (x-1)^2(x^2+x+1)

    to get!:
    x^6-2x^3+1

    I was able to identify the perfect square:
    (x-1)^2

    But I did not know how to extract this perfect square without unsatisfying the equation!

    Listing the steps from:
    x^6-2x^3+1

    to,:
    (x-1)^2(x^2+x+1)^2 ... correction

    would be greatly appreciated. Also could someone explain how extracting a perfect square does not unsatisfy the equation.

    Sam
    x^6 - 2x^3 + 1 = (x^3 - 1)^2

    now ... what is the factoring pattern for the binomial form a^3 - b^3 ?
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  6. #6
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    Re: Factoring Perfect Squares

    Quote Originally Posted by ArcherSam View Post
    I exported the markup code from MathType, but I kept getting errors! I'm not too familiar with Tex and LaTex
    I use MathType all the time.
    For example:\[
    \frac{{x^3 - 1}}
    {{x + 1}}
    \]
    was generated be MathType.

    CLEAN it up [tex]\frac{{x^3 - 1}}{{x + 1}}[/tex] gives \frac{{x^3  - 1}}{{x + 1}}.

    Note that I removed the \[ & \], along with any linefeeds.
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  7. #7
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    Wink Re: Factoring Perfect Squares

    Thanks to all.

    @skeeter

    Thanks for giving insight to the simplicity of the equation. I am known to over-think everything. Probably from a lack in confidence.

    @All

    From prior post I was able to deduce the solutions:

    Factoring Perfect Squares-solution.gif

    @Plato
    This is what MathType generated when I copy and pasted the equation above(start and end tags contained math and /math):

    []\begin{array}{l}
    a{}^3 - b{}^3 = (x - a)(x{}^2 + ax + a{}^2)\\
    x{}^6 - 2x{}^3 + 1 = (x{}^3 - 1){}^2\\
    (x - 1){}^2(x{}^2 + ( - 1)x + ( - 1){}^2){}^2\\
    (x - 1){}^2(x{}^2 - x + 1){}^2
    \end{array}[]

    Then I edited the markup code and got:

    <br />
\begin{array}{l}<br />
a{}^3 - b{}^3 = (x - a)(x{}^2 + ax + a{}^2)\\<br />
x{}^6 - 2x{}^3 + 1 = (x{}^3 - 1){}^2\\<br />
(x - 1){}^2(x{}^2 + ( - 1)x + ( - 1){}^2){}^2\\<br />
(x - 1){}^2(x{}^2 - x + 1){}^2<br />
\end{array}<br />

    Help please!

    Sam
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  8. #8
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    Re: Factoring Perfect Squares

    Quote Originally Posted by ArcherSam View Post
    Thanks to all.

    @skeeter

    Thanks for giving insight to the simplicity of the equation. I am known to over-think everything. Probably from a lack in confidence.

    @All

    From prior post I was able to deduce the solutions:

    Click image for larger version. 

Name:	Solution.gif 
Views:	18 
Size:	4.0 KB 
ID:	22165

    @Plato
    This is what MathType generated when I copy and pasted the equation above(start and end tags contained math and /math):

    []\begin{array}{l}
    a{}^3 - b{}^3 = (x - a)(x{}^2 + ax + a{}^2)\\
    x{}^6 - 2x{}^3 + 1 = (x{}^3 - 1){}^2\\
    (x - 1){}^2(x{}^2 + ( - 1)x + ( - 1){}^2){}^2\\
    (x - 1){}^2(x{}^2 - x + 1){}^2
    \end{array}[]

    Then I edited the markup code and got:

    <br />
\begin{array}{l}<br />
a{}^3 - b{}^3 = (x - a)(x{}^2 + ax + a{}^2)\\<br />
x{}^6 - 2x{}^3 + 1 = (x{}^3 - 1){}^2\\<br />
(x - 1){}^2(x{}^2 + ( - 1)x + ( - 1){}^2){}^2\\<br />
(x - 1){}^2(x{}^2 - x + 1){}^2<br />
\end{array}<br />
    [tex]\begin{array}{l}a{}^3 - b{}^3 = (x - a)(x{}^2 + ax + a{}^2)\\x{}^6 - 2x{}^3 + 1 = (x{}^3 - 1){}^2\\(x - 1){}^2(x{}^2 + ( - 1)x + ( - 1){}^2){}^2\\(x - 1){}^2(x{}^2 - x + 1){}^2\end{array}[/tex]
    I told you REMOVE ALL LINEFEEDS.
    \begin{array}{l}a{}^3 - b{}^3 = (x - a)(x{}^2 + ax + a{}^2)\\x{}^6 - 2x{}^3 + 1 = (x{}^3 - 1){}^2\\(x - 1){}^2(x{}^2 + ( - 1)x + ( - 1){}^2){}^2\\(x - 1){}^2(x{}^2 - x + 1){}^2\end{array}
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  9. #9
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    Re: Factoring Perfect Squares

    @Plato

    Thanks! I was unsure what you meant by linefeeds. I thought some code inside the tags was preventing the desired output, but you meant the formatting of the code. I used textfixer text tools to get rid of the linefeeds. I also download AutoUnbreak for future use. If you know of any offline software for text cleanup, please let me know!

    \begin{array}{l} a{}^3 - b{}^3 = (x - a)(x{}^2 + ax + a{}^2)\\ x{}^6 - 2x{}^3 + 1 = (x{}^3 - 1){}^2\\ (x - 1){}^2(x{}^2 + ( - 1)x + ( - 1){}^2){}^2\\ (x - 1){}^2(x{}^2 - x + 1){}^2 \end{array}
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