1. ## 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

2. ## 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:
$\displaystyle t^2-2t+1$

3. ## Re: Factoring Perfect Squares

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

4. ## 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

5. ## Re: Factoring Perfect Squares

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 $\displaystyle a^3 - b^3$ ?

6. ## Re: Factoring Perfect Squares

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

Note that I removed the $&$, along with any linefeeds.

7. ## 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:

@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:

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

Sam

8. ## Re: Factoring Perfect Squares

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:

@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:

$\displaystyle \begin{array}{l} a{}^3 - b{}^3 = (x - a)(x{}^2 + ax + a{}^2)\\ x{}^6 - 2x{}^3 + 1 = (x{}^3 - 1){}^2\\ (x - 1){}^2(x{}^2 + ( - 1)x + ( - 1){}^2){}^2\\ (x - 1){}^2(x{}^2 - x + 1){}^2 \end{array}$
$$\begin{array}{l}a{}^3 - b{}^3 = (x - a)(x{}^2 + ax + a{}^2)\\x{}^6 - 2x{}^3 + 1 = (x{}^3 - 1){}^2\\(x - 1){}^2(x{}^2 + ( - 1)x + ( - 1){}^2){}^2\\(x - 1){}^2(x{}^2 - x + 1){}^2\end{array}$$
I told you REMOVE ALL LINEFEEDS.
$\displaystyle \begin{array}{l}a{}^3 - b{}^3 = (x - a)(x{}^2 + ax + a{}^2)\\x{}^6 - 2x{}^3 + 1 = (x{}^3 - 1){}^2\\(x - 1){}^2(x{}^2 + ( - 1)x + ( - 1){}^2){}^2\\(x - 1){}^2(x{}^2 - x + 1){}^2\end{array}$

9. ## 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!

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