Solving polynomial of degree5 :

Here we are going to see how to solve the polynomial which is having degree 5.

**Example 1 :**

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

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**

Hence the 5 roots are **x = -1/3,-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.

(1) Solve 6 x⁵ - x⁴ - 43 x³ + 43 x² + x - 6 Solution

(2) Solve 8 x⁵ - 22 x⁴ - 55 x³ + 55 x² + 22 x - 8

(3) Solve x⁵ - 5 x⁴ + 9 x³ - 9 x² + 5 x - 1

(4) Solve x⁵ - 5 x³ + 5 x² - 1

(5) Solve 6 x⁵ + 11 x⁴ - 33 x³ - 33 x² + 11 x + 6

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