**Factoring Higher Degree Polynomials with Synthetic Division :**

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

x^{5} - 5x^{4} + 9x^{3} - 9x^{2} + 5x - 1

**Solution :**

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

The other roots are given by

x^{4} - 4 x^{3} + 5 x^{2} - 4x + 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) - 4y + 5 = 0

y² - 2 - 4y + 5 = 0

1y² - 4y - 2 + 5 = 0

1y² - 4y + 3 = 0

(y - 1) (y - 3) = 0

Solving for y, we get

y - 1 = 0 and y - 3 = 0

y = 1 and y = 3

When y = 1

x + 1/x = y

(x² + 1)/x = 1

(x² + 1) = 1 x

x² - 1x + 1 = 0

Solving the quadratic equation using the formula

x = (-b ± √b^{2} - 4ac)/2a

By comparing the quadratic equation with the general form of a quadratic equation ax^{2} + bx + c = 0, we get

a = 1, b = -1 and c = 1

b^{2} - 4ac = (-1)^{2} - 4(1)(1)

= -3

So,

x = (1 ± √-3)/2

x = (1 ± √3i)/2

When y = 3

x + 1/x = y

(x² + 1)/x = 3

(x² + 1) = 3x

x² - 3x + 1 = 0

Solving the quadratic equation using the formula

x = (-b ± √b^{2} - 4ac)/2a

By comparing the quadratic equation with the general form of a quadratic equation ax^{2} + bx + c = 0, we get

a = 1, b = -3 and c = 1

b^{2} - 4ac = (-3)^{2} - 4(1)(1)

= 5

So,

x = (1 ± √5)/2

Hence the values of x are (1 ± √-3)/2, (1 ± √5)/2 and 1.

**Example 2 :**

Solve

x^{5} - 5x^{3} + 5x^{2} - 1

Dividing the entire equation by x²

x⁴/x² + x³/x² - 4 x²/x² + x/x² + 1/x² = 0

x² + x - 4 + 1/x + 1/x² = 0

(x² + 1/x²) - 4 + (1/x) + x = 0

(x² + 1/x²) + (x + 1/x) - 4 = 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

(y² - 2) + y - 4 = 0

y² - 2 + y - 4 = 0

y² + y - 6 = 0

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

Solving for y, we get

y + 3 = 0 and y - 2 = 0

y = -3 and y = 2

When y = -3

x + 1/x = y

(x² + 1)/x = -3

(x² + 1) = -3 x

x² + 1 + 3x = 0

x² + 3x + 1 = 0

Solving the quadratic equation using the formula

x = (-b ± √b^{2} - 4ac)/2a

By comparing the quadratic equation with the general form of a quadratic equation ax^{2} + bx + c = 0, we get

a = 1, b = 3 and c = 1

b^{2} - 4ac = 3^{2} - 4(1)(1)

= 5

x = (-1 ± √5)/2

When y = 2

x + 1/x = y

(x² + 1)/x = 2

(x² + 1) = 2x

x² - 2x + 1 = 0

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

Solving for x, we get

x = 1

Therefore the 5 roots are x = 1, 1, 1, (-3 ± √5)/2.

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