Factorise x^4 - x^2 + 1 fulling by completing the square

Matt23

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Hello, I need assistance with this following question:

Factorise fully x4- x2 + 1 by completing the square
 

tkhunny

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Hello, I need assistance with this following question:

Factorise fully x4- x2 + 1 by completing the square
Could you do it if it were \(\displaystyle z^{2} - z +1\)?
 

stapel

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Hello, I need assistance with this following question:

Factorise fully x4- x2 + 1 by completing the square
What have you tried (including the suggestion provided earlier)? How far have you gotten? Where are you stuck?

Please be complete. Thank you! ;)
 

Jomo

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An interesting problem is to factor x4 + x2 + 1 as it is the difference of two squares.
 

apple2357

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An interesting problem is to factor x4 + x2 + 1 as it is the difference of two squares.
Can that be thought of like this?

x4+2x2+1-x2​ ? and then another step?
 

Jomo

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Can that be thought of like this?

x4+2x2+1-x2​ ? and then another step?
I'm not sure what you're asking since based on what you wrote I suspect that you know how to do this one.
x4+2x2+1-x2 = (x4+2x2+1)-(x2). Now factor each parenthesis and factor the difference of squares
 

sinx

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Hello, I need assistance with this following question:

Factorise fully x4- x2 + 1 by completing the square
if adding x2 to the square is permitted,
x4- x2 + 1= (x2-1)2+x2

 

Jomo

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if adding x2 to the square is permitted,
x4- x2 + 1= (x2-1)2+x2

What you wrote is correct (that is your equal sign is valid). the problem is that your rhs is not factored.

Is it valid to add x2 is not a clear question. You can only add 0 to an expression without changing the value of the expression. You added 0 in the form of +x2 - x2
 
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Dr.Peterson

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Hello, I need assistance with this following question:

Factorise fully x4- x2 + 1 by completing the square
Some people have been talking about solving a different problem (+ instead of -), and/or by a different method than you were told to use. Have you tried doing what it says?

If you complete the square (possibly by first changing it to u2 - u + 1 to make it look more familiar), you will have something in the form (x2 + a)2 + b. This will be a sum of squares (if you are willing to stretch your mind a bit), which can't be factored unless you allow complex numbers. From there, you can continue until you get four linear factors; but, again, you will have complex numbers all over the place.

If the goal is to factor fully over the integers, or even real numbers, then this method will not work, as far as I can see. Since the expression has no real zeros, the best you can expect is to factor it as the product of two quadratic factors.

Can you confirm that you stated the problem correctly?
 

JeffM

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Some people have been talking about solving a different problem (+ instead of -), and/or by a different method than you were told to use. Have you tried doing what it says?

If you complete the square (possibly by first changing it to u2 - u + 1 to make it look more familiar), you will have something in the form (x2 + a)2 + b. This will be a sum of squares (if you are willing to stretch your mind a bit), which can't be factored unless you allow complex numbers. From there, you can continue until you get four linear factors; but, again, you will have complex numbers all over the place.

If the goal is to factor fully over the integers, or even real numbers, then this method will not work, as far as I can see. Since the expression has no real zeros, the best you can expect is to factor it as the product of two quadratic factors.

Can you confirm that you stated the problem correctly?
Yes, exactly. The issue is what is the problem that the OP is supposed to solve. It would be odd to pose a problem to factor an expression (rather than an equation) by completing the square.
 
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