5th degree polynomial question3





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


Solution

Question 2 :

Solve 8 x⁵ - 22 x⁴ - 55 x³ + 55 x² + 22 x - 8


Solution

Question 4 :

Solve  x⁵ - 5 x³ + 5 x² - 1


Solution

Question 5 :

Solve  6 x⁵ + 11 x⁴ - 33 x³ - 33 x² + 11 x + 6


Solution






5th Degree Polynomial Question3 to Algebra