**How to Solve Polynomial Equation of Degree 5 ?**

To solve a polynomial of degree 5, we have to factor the given polynomial as much as possible. After factoring the polynomial of degree 5, we find 5 factors and equating each factor to zero, we can find the all the values of x.

**Example 1 :**

Solve

6x^{5} - x^{4} - 43 x^{3} + 43x^{2} + x - 6

**Solution :**

Since the degree of the polynomial is 5, we have 5 zeroes. To find the zeroes, we use synthetic division.

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

6y² - 12 + 5 y - 38 = 0

6y² + 5y - 12 - 38 = 0

6y² + 5y - 50 = 0

6y² - 15y + 20 y - 50 = 0

3y (2y - 5) + 10 (2y - 5) = 0

(3y + 10) (2y - 5) = 0

3y + 10 = 0 y = -10/3 |
2y - 5 = 0 y = 5/2 |

When y = -10/3

(x² + 1)/x = -10/3

3(x² + 1) = -10x

3x² + 3 = -10 x

3x² + 10 x + 3 = 0

By factoring we get,

(3x + 1) (x + 3) = 0

x = -1/3 and 3

When 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

By factoring we get,

(2x - 1) (x - 2) = 0

x = 1/2 and 2

Therefore the 5 roots are -1/3, 3, 1/2, 2 and 1.

**Example 2 :**

Solve

8x^{5} - 22 x^{4} - 55 x^{3} + 55x^{2} + 22x - 8

**Solution :**

Dividing the entire equation by x²

8 x⁴/x² - 14 x³/x² - 69 x²/x² - 14 x/x² + 8/x² = 0

8 x² - 14 x - 69 - 14 (1/x) + 8(1/x²) = 0

8 (x² + 1/x²) - 14 (x + 1/x) - 69 = 0 ------ (1)

Let x + 1/x = y

To find the value of x² + 1/x², we 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

8 (y² - 2) - 14y - 69 = 0

8y² - 16 - 14y - 69 = 0

8y² - 14y - 16 - 69 = 0

8y² - 14y - 85 = 0

(2y + 5) (4y - 17) = 0

Solving for y, we get

2y + 5 = 0 and 4y - 17 = 0

y = -5/2 and y = 17/4

When 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

Solving for x, we get

2x + 1 = 0 and x + 2 = 0

x = -1/2 and x = -2

When y = 17/4

x + 1/x = y

(x² + 1)/x = 17/4

4(x² + 1) = 17x

4x² + 4 = 17 x

4x² - 17 x + 4 = 0

(4x - 1)(x - 4) = 0

(4x - 1) = 0 (x - 4) = 0

Solving for x, we get

x = 1/4 and x = 4

Therefore the 5 roots are 1/4, 4, 2, 1/2, 1.

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