In this page 5th degree polynomial question3 we are going to see how to solve the polynomial which is having degree 5.
Question 3 :
Solve x⁵  5 x⁴ + 9 x³  9 x² + 5 x  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  4 x³ + 5 x²  4 x + 1 = 0
Dividing the entire equation by x²
x⁴/x²  4 x³/x² + 5 x²/x²  4 x/x² + 1/x² = 0
1 x²  4 x  5  4 (1/x) + (1/x²) = 0
1 (x² + 1/x²)  4 (x + 1/x) + 5 = 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
1 (y²  2)  4 y + 5 = 0
1 y²  2  4 y + 5 = 0
1 y²  4 y  2 + 5 = 0
1 y²  4 y + 3 = 0
1 y²  1 y  3 y + 3 = 0
1 y (y  1)  3 (y  1) = 0
(y  1) (y  3) = 0
y  1 = 0
y = 1
y = 1
y  3 = 0
y = 3
x + 1/x = y
(x² + 1)/x = 1
(x² + 1) = 1 x
x²  1 x + 1 = 0
x = (1± i√3)/2
x + 1/x = 3
(x² + 1)/x = 3
(x² + 1) = 3x
x² + 1 = 3 x
x²  3 x + 1 = 0 5th Degree Polynomial Question3
x = (3 ± √5)/2
Therefore the 5 roots are x = 1,(1± i√3)/2,(3 ± √5)/2
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 4 : Solve x⁵  5 x³ + 5 x²  1 

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