In this page 5th degree polynomial question5 we are going to see how to solve the polynomial which is having degree 5.
Question 5 :
Solve 6 x⁵ + 11 x⁴  33 x³  33 x² + 11 x + 6
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 like terms. So 1 is one of the root of this equation.
The other roots are given by
6x⁴ + 5 x³  38 x² + 5 x + 6 = 0
Dividing the entire equation by x²
6x⁴/x² + 5 x³/x²  38 x²/x² + 5 x/x² + 6/x² = 0
6 x² + 5 x  38 + (5/x) + (6/x²) = 0
6 x² + (6/x²) + 5 x + (5/x)  38 = 0
6 (x² + 1/x²) + 5 (x + 1/x)  38 = 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
6 (y²  2) + 5 y  38 = 0
6 y²  12 + 5 y  38 = 0
6 y² + 5 y  12  38 = 0
6 y² + 5 y  50 = 0
6 y² + 20 y  15 y  50 = 0
2 y (3y + 10)  5 (3y + 10) = 0
(2y  5) (3y + 10) = 0
2y  5 = 0
2y = 5
y = 5/2
y = 5/2
3y + 10 = 0
3y = 10
y = 10/3
y = 10/3
x + 1/x = y
(x² + 1)/x = 5/2
2(x² + 1) = 5 x
2x² + 2  5x = 0
2x²  5x + 2 = 0
2x²  4 x  1 x + 2 = 0
2x(x  2)  1(x  2) = 0
(2x  1) (x  2) = 0
2 x  1 = 0 x  2 = 0
2x = 1 x = 2
x = 1/2
x + 1/x = y
(x² + 1)/x = 10/3
3(x² + 1) = 10 x
3x² + 3 + 10x = 0
3x² + 10x + 3 = 0
3x² + 1 x + 9 x + 3 = 0
x(3x + 1) + 3(3x + 1) = 0
(3x + 1) (x + 3) = 0
3 x + 1 = 0 x + 3 = 0
3x = 1 x = 3
x = 1/3
x = 1/3
5th degree polynomial question5
Therefore the 5 roots are x = 3,1/3,2,1/2,1
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 4 : Solve x⁵  5 x³ + 5 x²  1 
