FACTORING HIGHER DEGREE POLYNOMIALS WITH SYNTHETIC DIVISION

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

x5 - 5x4 + 9x3 - 9x2 + 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

x4 - 4 x3 + 5 x2 - 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 ± √b2 - 4ac)/2a

By comparing the quadratic equation with the general form of a quadratic equation ax2 + bx + c  =  0, we get

a  =  1, b  =  -1 and c  =  1

b2 - 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 ± √b2 - 4ac)/2a

By comparing the quadratic equation with the general form of a quadratic equation ax2 + bx + c  =  0, we get

a  =  1, b  =  -3 and c  =  1

b2 - 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 

x5 - 5x3 + 5x2 - 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 ± √b2 - 4ac)/2a

By comparing the quadratic equation with the general form of a quadratic equation ax2 + bx + c  =  0, we get

a  =  1, b  =  3 and c  =  1

b2 - 4ac  =  32 - 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.

Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Angular Speed and Linear Speed

    Dec 07, 22 05:15 AM

    Angular Speed and Linear Speed - Concepts - Formulas - Examples

    Read More

  2. Linear Speed Formula

    Dec 07, 22 05:13 AM

    Linear Speed Formula and Examples

    Read More

  3. Angular Speed and Linear Speed Worksheet

    Dec 07, 22 05:08 AM

    Angular Speed and Linear Speed Worksheet

    Read More