For the quartic equation: \(\displaystyle{x}^{{4}}-{x}^{{3}}+{4}{x}^{{2}}+{3}{x}+{5}={0}\)

delitzo1d4

delitzo1d4

Answered question

2022-04-12

For the quartic equation:
x4x3+4x2+3x+5=0

Answer & Explanation

maggionmoo

maggionmoo

Beginner2022-04-13Added 16 answers

x4x3+4x2+3x+5
=x4x3+4x2+4x+4x+1
=x4x3x+1+4(x2+x+1)
=x3(x1)(x1)+4(x2+x+1)
=(x1)(x31)+4(x2+x+1)
=(x1)(x1)(x2+x+1)+4(x2+x+1)
=(x2+x+1)(x1)2+4)=(x2+x+1)(x22x+5)
As for the roots, I assume you could solve those two quadratic equations and you could find the results on Wolfram|Alpha.

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