Solve a higher degree polynomial equation

Learn how to solve a higher degree polynomial equation using a quadratic pattern.

Example #1

Factor x⁴ - 23x² - 50 by using a quadratic pattern

Step 1

Write x⁴ - 23x² - 50 in the pattern of a quadratic expression so you can factor it like one by making a temporary substitution of variables.

Let y = x² and substitute y for x²

x⁴ - 23x² - 50 = (x²)² - 23(x²) - 50

x⁴ - 23x² - 50 = (y)² - 23(y) - 50

Step 2

Factor y² - 23y - 50

y² - 23y - 50 = (y + ___ )(y + ___ )

To fill in the blank above, look for factors of -50 that will add up to -23.

-25 × 2 = -50 and -25 + 2 = -23. 

Fill in the blank in the expression above with -25 and 2.

y² - 23y - 50 = (y - 25)(y + 2)

Step 3

Substitute back to the original variable

(y - 25)(y + 2) = (x² - 25)(x² + 2)

Factor completely

(y - 25)(y + 2) = (x² - 25)(x² + 2) = (x - 5)(x + 5)(x² + 2)

Step 4

Set the expression (x - 5)(x + 5)(x² + 2) equal to zero.

(x - 5)(x + 5)(x² + 2) = 0

Solve the following 3 equations

(x - 5) = 0, (x + 5) = 0, and (x² + 2) = 0

x - 5 = 0

x - 5 + 5 = 0 + 5

x = 5

x + 5 = 0

x + 5 - 5 = 0 - 5

x = -5

x² + 2 = 0

x² + 2 - 2 = 0 - 2

x² = -2

x² = 2(-1)

x² = 2i² (since i² = -1)

x = ±√(2i²)

x = ±i√(2)

The solutions are 5, -5, √(2), and -√(2)

Example #2

Factor x⁴ - 5x² + 4 by using a quadratic pattern

Step 1

Write x⁴ - 5x² + 4 in the pattern of a quadratic expression so you can factor it like one by making a temporary substitution of variables.

Let y = x² and substitute y for x²

x⁴ - 5x² + 4 = (x²)² - 5(x²) + 4

x⁴ - 5x² + 4 = (y)² - 5(y) + 4

Step 2

Factor y² - 5y + 4

y² - 5y + 4 = (y + ___ )(y + ___ )

To fill in the blank above, look for factors of 4 that will add up to -5.

-4 × -1 = 4 and -1 + -4 = -5. 

Fill in the blank in the expression above with -4 and -1.

y² - 5y + 4 = (y - 4)(y - 1)

Step 3

Substitute back to the original variable

(y - 4)(y - 1) = (x² - 4)(x² - 1)

Factor completely

(y - 4)(y - 1) = (x² - 4)(x² - 1) = (x - 2)(x + 2)(x - 1)(x + 1)

Step 4

Set the expression (x - 2)(x + 2)(x - 1)(x + 1) equal to zero.

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

Solve the following 4 equations

(x - 2) = 0, (x + 2) = 0, (x - 1) = 0 and (x + 1) = 0

x - 2 = 0

x - 2 + 2 = 0 + 2

x = 2

x + 2 = 0

x + 2 - 2 = 0 - 2

x = -2

x - 1 = 0

x - 1 + 1 = 0 + 1

x = 1

x + 1 = 0

x + 1 - 1 = 0 - 1

x = -1

The solutions are 2, -2, 1, and -1