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:

$\displaystyle

\[\begin{array}{l}

(x{}^6 - 2x{}^3 + 1)\\

\\

(x - 1){}^2 = (x{}^2 - 2x + 1)\\

\\

?

\end{array}\]

$

Sam

Re: Factoring Perfect Squares

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

Do you mean:

$\displaystyle \left \{ \begin{array}{rcl} x^6-2x^3+1 \\ (x-1)^2=x^2-2x+1 \end{array}\right.$

?

If you have difficulties to factore $\displaystyle x^6-2x^3+1$ let $\displaystyle x^3=t$ and so you can write it as:

$\displaystyle t^2-2t+1$

Re: Factoring Perfect Squares

Equivalently, $\displaystyle x^6- 2x^2+ 1= (x^3)^2- 2(x^3)+ 1$

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

Re: Factoring Perfect Squares

Quote:

Originally Posted by

**ArcherSam** 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

$\displaystyle x^6 - 2x^3 + 1 = (x^3 - 1)^2$

now ... what is the factoring pattern for the binomial form $\displaystyle a^3 - b^3$ ?

Re: Factoring Perfect Squares

Quote:

Originally Posted by

**ArcherSam** 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 $\displaystyle \frac{{x^3 - 1}}{{x + 1}}$.

Note that I removed the \[ & \], along with any linefeeds.

1 Attachment(s)

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:

Attachment 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:

$\displaystyle

\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}

$

Help please!

Sam

Re: Factoring Perfect Squares

Quote:

Originally Posted by

**ArcherSam** 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:

Attachment 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:

$\displaystyle

\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]\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.

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

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!

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