## HOW TO SOLVE POLYNOMIAL EQUATION OF DEGREE 5

To solve a polynomial equation of degree 5, we have to factor the given polynomial as much as possible. After having factored, we can equate factors to zero and solve for the variable.

Example 1 :

Solve :

6x5 - x4 - 43x3 + 43x2 + x - 6 = 0

Solution :

By trial and error, we can check the values 1 or -1 or 2 or -2...... as a zero for the above equation using synthetic division.

When we check with the value 1, we get remainder zero. So, x = 1 is one of the zeros.

Resulting equation :

6x4 + 5x3 - 38x2 + 5x + 6 = 0

Dividing both sides by x2.

6x4/x+ 5x3/x2 - 38x2/x2 + 5x/x2 + 6/x2 = 0

6x2 + 5x - 38 + 5/x + 6/x2 = 0

6(x2 + 1/x2) + 5 (x + 1/x) - 38 = 0 ----(1)

Let y = x + 1/x.

y2 = (x + 1/x)2

y= x2 + 2(x)(1/x) + (1/x)2

y= x2 + 2 + 1/x2

y- 2 = x2 + 1/x2

(1)----> 6(y2 - 2) + 5y - 38 = 0

6y2 - 12 + 5y - 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 = 0y = -10/3 2y - 5 = 0y = 5/2

When y  =  -10/3,

(x2 + 1)/x = -10/3

3(x2 + 1) = -10x

3x2 + 3 = -10x

3x2 + 10x + 3 = 0

(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

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

x = 1/2 and 2

Therefore the five zeros are 1, -1/3, 3, 1/2 and 2.

Example 2 :

Solve :

8x5 - 22x4 - 55x3 + 55x2 + 22x - 8 = 0

Solution :

When we check with the value 1, we get remainder zero. So, x = 1 is one of the zeros.

Resulting equation :

8x4 - 14x3 - 69x2 - 14x + 8 = 0

Dividing both sides by x2.

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

8x- 14x - 69 - 14/x + 8/x2 = 0

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

Let y = x + 1/x.

y2 = (x + 1/x)2

y= x2 + 2(x)(1/x) + (1/x)2

y= x2 + 2 + 1/x2

y- 2 = x2 + 1/x2

(1)----> 8(y2 - 2) - 14y - 69 = 0

8y2 - 16 - 14y - 69 = 0

8y2 - 14y - 85 = 0

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

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

When y = -5/2,

x + 1/x = y

(x2 + 1)/x = -5/2

2(x2 + 1) = -5x

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

x = -1/2 and -2

When y = 17/4,

x + 1/x = y

(x2 + 1)/x = 17/4

4(x2 + 1) = 17x

4x2 + 4 = 17x

4x2 - 17x + 4 = 0

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

x = 1/4 and 4

Therefore the five zeros are  1, -1/2, -2, 1/4 and 4.

Example 3 :

Solve :

6x5 + 11x4 - 33x3 - 33x2 + 11x + 6 = 0

Solution :

When we check with the value -1, we get remainder zero. So, x = -1 is one of the zeros.

Resulting equation :

6x4 + 5x3 - 38x2 + 5x + 6 = 0

Dividing both sides by x2.

6x4/x+ 5x3/x2 - 38x2/x2 + 5x/x2 + 6/x2 = 0

6x+ 5x - 38 + 5/x + 6/x2 = 0

6(x2 + 1/x2) + 5(x + 1/x) - 38 = 0 ----(1)

Let y = x + 1/x.

y2 = (x + 1/x)2

y= x2 + 2(x)(1/x) + (1/x)2

y= x2 + 2 + 1/x2

y- 2 = x2 + 1/x2

(1)----> 6(y2 - 2) + 5y - 38 = 0

6y2 - 12 + 5y - 38 = 0

6y2 + 5y - 50 = 0

6y2 - 15y + 20y - 50 = 0

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

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

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

When y = -10/3,

x + 1/x = y

(x2 + 1)/x = -10/3

3(x2 + 1) = -10x

3x2 + 3 = -10x

3x2 + 10x + 3 = 0

3x2 + 9x + x + 3 = 0

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

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

x = -3 and -1/3

When y = 5/2,

x + 1/x = y

(x2 + 1)/x = 5/2

2(x2 + 1) = 5x

2x2 + 2 - 5x = 0

2x2 - 5x + 2 = 0

2x2 - 4x - x + 2 = 0

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

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

x = 1/2 and 2

Therefore the five zeros are -1, -3, -1/3, 1/2 and 2.

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