Solve the polynomial equation by finding all roots

\(\mbox{There is no constant term so clearly }x=0 \mbox{ is one of the roots.}\\ \mbox{Dividing through by }x-0=x \mbox{ we get} \\ x^3-5x^2 +3x+1 = 0 \\ \mbox{Because the constant term is 1 if anything divides this polynomial it will be }(x-1) \mbox{ or }(x+1) \\ \mbox{Checking }(x-1)\mbox{ we plug 1 in and see if it works}\\ 1-5+3+1 = 0 \mbox{ so }1 \mbox{ is indeed a root. Dividing by }(x-1) \mbox{we get}\\ x^2-4 x-1\\ \)

\(\mbox{This can't be factored over rational numbers so we'll have to use the quadratic formula.}\\ r=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a} \\ a=1, b=-4, c=-1 \\ r_1=\dfrac{4+\sqrt{16+4}}{2}= (2+\sqrt{5}) \\ r_2 =\dfrac{4-\sqrt{16+4}}{2}= (2-\sqrt{5})\)

\(\mbox{So gathering all this up we have roots of } \\ x \in \{0,1,2+\sqrt{5},2-\sqrt{5}\}\)

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x^4-5x^3+3x^2+x=0 

 

Solve for x over the real numbers:
x^4-5 x^3+3 x^2+x = 0

The left hand side factors into a product with three terms:
x (x-1) (x^2-4 x-1) = 0

Split into three equations:
x-1 = 0 or x = 0 or x^2-4 x-1 = 0

Add 1 to both sides:
x = 1 or x = 0 or x^2-4 x-1 = 0

Add 1 to both sides:
x = 1 or x = 0 or x^2-4 x = 1

Add 4 to both sides:
x = 1 or x = 0 or x^2-4 x+4 = 5

Write the left hand side as a square:
x = 1 or x = 0 or (x-2)^2 = 5

Take the square root of both sides:
x = 1 or x = 0 or x-2 = sqrt(5) or x-2 = -sqrt(5)

Add 2 to both sides:
x = 1 or x = 0 or x = 2+sqrt(5) or x-2 = -sqrt(5)

Add 2 to both sides:
Answer: | 
| x = 1 or x = 0 or x = 2+sqrt(5) or x = 2-sqrt(5)