Factoring X4+1: Step-by-Step Guide

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The discussion revolves around factoring the polynomial x^4 + 1 into two multiplied polynomials, with the initial assumption that the factors take the form (ax^2 + bx + c)(ax^2 + bx + c). The user explores various values for a, b, and c, ultimately finding that b equals the square root of 2 after simplifying the equation. A suggestion is made to approach the problem using the difference of squares method, leading to a more systematic solution by completing the square. The conversation highlights the challenge of factoring polynomials without real roots and the potential for complex roots. Overall, the thread emphasizes the importance of exploring different factoring techniques for polynomials.
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Homework Statement



I am trying to factor x4+1 into two multiplied polynomials

Homework Equations




My teacher gave us this hint that its factored form is (ax2+bx+c)(ax2+bx+c)


The Attempt at a Solution



First i assumed that a and c were equal to 1 so that when x2 is multiplied by the other x2 is gives me x4 and 1 times 1 gives me 1. I knew that b had to be a constant so I multiplied...

(however i didnt know if both b's were the same so i split them into a and b. I also knew one of the constants must be negative so that variables cancel out.)

(x2+ax+1)(x2-bx+1)= x4+1

and I get

x4+ax3-bx3+2x2-abx2+ax-bx+1= x4+1 canceling terms I get

ax3-bx3+2x2-abx2+ax-bx=0

I noticed that to cancel out ax3-bx3 and ax-bx , a and b must be equal to each other. This means 2x2-abx2 = 0

2=ab(but they are the same) 2=b2

b=\sqrt{}2

So i checked my answer and it works out, but I am wondering if there is a more systematic approach to solve this so that I don't have to assume as much as I did.
 
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x4+1 has no real roots, so at most you can factor it into complex roots

using i2=-1

x4-i2=(x2)2-(i)2
 
Welcome to PF!

Hi flyers! Welcome to PF! :smile:

(have a square-root: √ :wink:)
flyers said:
… So i checked my answer and it works out, but I am wondering if there is a more systematic approach to solve this so that I don't have to assume as much as I did.

You could have looked for a way to write it as the difference of two squares …

so complete the square …

x4 + 1 = (x2 + 1)2 - 2x2 :wink:
 
More generally …

x4 + 2(a-b)x2 + a2

= (x2 + a)2 - 2bx2

= (x2 + (√2b)x + a)(x2 - (√2b)x + a) :smile:​
 

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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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