# Factoring Perfect Squares

• Sep 1st 2011, 08:02 AM
ArcherSam
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:

$
$\begin{array}{l} (x{}^6 - 2x{}^3 + 1)\\ \\ (x - 1){}^2 = (x{}^2 - 2x + 1)\\ \\ ? \end{array}$
$

Sam
• Sep 1st 2011, 08:28 AM
Siron
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:
$t^2-2t+1$
• Sep 1st 2011, 10:08 AM
HallsofIvy
Re: Factoring Perfect Squares
Equivalently, $x^6- 2x^2+ 1= (x^3)^2- 2(x^3)+ 1$
• Sep 1st 2011, 01:41 PM
ArcherSam
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
• Sep 1st 2011, 03:23 PM
skeeter
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

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

now ... what is the factoring pattern for the binomial form $a^3 - b^3$ ?
• Sep 1st 2011, 03:43 PM
Plato
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 $$\frac{{x^3 - 1}}{{x + 1}}$$ gives $\frac{{x^3 - 1}}{{x + 1}}$.

Note that I removed the $&$, along with any linefeeds.
• Sep 1st 2011, 04:55 PM
ArcherSam
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:

$
\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
• Sep 1st 2011, 05:07 PM
Plato
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:

$
\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.
$\begin{array}{l}a{}^3 - b{}^3 = (x - a)(x{}^2 + ax + a{}^2)\\x{}^6 - 2x{}^3 + 1 = (x{}^3 - 1){}^2\\(x - 1){}^2(x{}^2 + ( - 1)x + ( - 1){}^2){}^2\\(x - 1){}^2(x{}^2 - x + 1){}^2\end{array}$
• Sep 2nd 2011, 04:03 AM
ArcherSam
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!

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