In this page 5th degree polynomial question1 we are going to see how to solve the polynomial which is having degree 5.
Question 1 :
Solve 6 x⁵  x⁴  43 x³ + 43 x² + 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 unlike terms. So 1 is one of the root of this equation.
The other roots are given by
6 x⁴ + 5 x³  38 x² + 5 x + 6= 0
Dividing the entire equation by x²
6 x⁴/x² + 5 x³/x²  38 x²/x² + 5 x/x² + 6/x² = 0
6 x² + 5 x  38 + 5 (1/x) + 6(1/x²) = 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  15 y + 20 y  50 = 0
3 y (2y  5) + 10 (2y  5) = 0
(3 y + 10) (2y  5) = 0
3y + 10 = 0
3 y = 10
y = 10/3
2 y  5 = 0
2 y = 5
y = 5/2
x + 1/x = y
(x² + 1)/x = 5/2
2(x² + 1) = 5 x
2x² + 2  5x = 0
2x²  5x + 2 = 0
2x²  4x  1x + 2 = 0
2x (x  2) 1(x  2) = 0
(2x  1) (x  2) = 0
2x  1 = 0 x  2 = 0
2 x = 1 x = 2
x = 1/2
x + 1/x = y
(x² + 1)/x = 10/3
3(x² + 1) = 10x
3x² + 3 = 10 x
3x² + 10 x + 3 = 0
3x² + 9 x + 1 x + 3 = 0
3x (x + 3) + 1(x + 3) = 0
(3x + 1) = 0 (x + 3) = 0
3x = 1 x = 3
x = 1/3
Therefore the 5 roots are x = 1/3,3,2,1/2,1
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.5th degree polynomial question1
Questions 
Solution 
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 

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

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