In this page 5th degree polynomial question4 we are going to see how to solve the polynomial which is having degree 5.
Question 4 :
Solve x⁵  5 x³ + 5 x²  1
Solution :
The degree of this equation is 5. Therefore we can say there will be 5 roots for this equation.
This is a reciprocal equation of odd degree with unlike terms. So 1 is one of the root of this equation.
The other roots are given by
x⁴ + x³  4 x² + x + 1= 0
Dividing the entire equation by x²
x⁴/x² + x³/x²  4 x²/x² + x/x² + 1/x² = 0
x² + x  4 + 1/x + 1/x² = 0
(x² + 1/x²)  4 + (1/x) + x = 0
(x² + 1/x²) + (x + 1/x)  4 = 0  (1)
Let x + 1/x = y
To find the value of x² + 1/x² from this we have to take squares on both sides
(x + 1/x)² = y²
x² + 1/x² + 2 x (1/x) = y²
x² + 1/x² + 2 = y²
x² + 1/x² = y²  2
So we have to plug y²  2 instead of x² + 1/x²
Let us plug this value in the first equation
(y²  2) + y  4 = 0
y²  2 + y  4 = 0
y² + y  6 = 0
y² + 3 y  2 y  6 = 0
y (y + 3)  2 (y + 3) = 0
(y + 3) (y  2) = 0
y + 3 = 0
y = 3
y = 3
y  2 = 0
y = 2
y = 2
x + 1/x = y
(x² + 1)/x = 3
(x² + 1) = 3 x
x² + 1 + 3x = 0
x² + 3x + 1 = 0 5th degree polynomial question4
x = (3 ± √5)/2
x + 1/x = y
(x² + 1)/x = 2
(x² + 1) = 2x
x²  2 x + 1 = 0
x²  1 x  1 x + 1 = 0
x (x  1)  1(x  1) = 0
(x  1) (x  1) = 0
(x  1) = 0 (x  1) = 0
x = 1 x = 1
x = 1
Therefore the 5 roots are x = 1,1,1,(3 ± √5)/2
5th degree polynomial question1
This is the example problem in the topic solving polynomial of degree5. You can try the following sample test to understand this topic much better.
Questions 
Solution 
Question 1 : Solve 6 x⁵  x⁴  43 x³ + 43 x² + x  6 

Question 2 : Solve 8 x⁵  22 x⁴  55 x³ + 55 x² + 22 x  8 

Question 3 : Solve x⁵  5 x⁴ + 9 x³  9 x² + 5 x  1 

Question 5 : Solve 6 x⁵ + 11 x⁴  33 x³  33 x² + 11 x + 6 
