- Thread starter Matt23
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Could you do it if it were \(\displaystyle z^{2} - z +1\)?Hello, I need assistance with this following question:

Factorise fully x^{4}- x^{2}+ 1 by completing the square

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What have you tried (including the suggestion provided earlier)? How far have you gotten? Where are you stuck?Hello, I need assistance with this following question:

Factorise fully x^{4}- x^{2}+ 1 by completing the square

Please be complete. Thank you!

Can that be thought of like this?An interesting problem is to factor x^{4}+ x^{2}+ 1 as it is the difference of two squares.

x

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I'm not sure what you're asking since based on what you wrote I suspect that you know how to do this one.Can that be thought of like this?

x^{4}+2x^{2}+1-x^{2} ? and then another step?

x

if adding xHello, I need assistance with this following question:

Factorise fully x^{4}- x^{2}+ 1 by completing the square

x

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What you wrote is correct (that is your equal sign is valid). the problem is that your rhs is not factored.if adding x^{2}to the square is permitted,

x^{4}- x^{2}+ 1= (x^{2}-1)^{2}+x^{2 }

Is it valid to add x

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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?

Factorise fully x^{4}- x^{2}+ 1 bycompleting the square

If you complete the square (possibly by first changing it to u

If the goal is to factor fully

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.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 u^{2}- u + 1 to make it look more familiar), you will have something in the form (x^{2}+ a)^{2}+ b. This will be asumof 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 fullyover the integers, or evenreal 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?