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

**Example #1**

Factor x^{4} - 23x^{2} - 50 by using a quadratic pattern

**Step 1**

Write x^{4} - 23x^{2} - 50 in the pattern of a quadratic expression so you can factor it like one by making a temporary substitution of variables.

Let y = x^{2} and substitute y for x^{2}

x^{4} - 23x^{2} - 50 = (x^{2})^{2} - 23(x^{2}) - 50

x^{4} - 23x^{2} - 50 = (y)^{2} - 23(y) - 50

**Step 2**

Factor y^{2} - 23y - 50

y^{2} - 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.

y^{2} - 23y - 50 = (y - 25)(y + 2)

**Step 3**

Substitute back to the original variable

(y - 25)(y + 2) = (x^{2} - 25)(x^{2} + 2)

Factor completely

(y - 25)(y + 2) = (x^{2} - 25)(x^{2} + 2) = (x - 5)(x + 5)(x^{2} + 2)

**Step 4**

Set the expression (x - 5)(x + 5)(x^{2} + 2) equal to zero.

(x - 5)(x + 5)(x^{2} + 2) = 0

Solve the following 3 equations

(x - 5) = 0, (x + 5) = 0, and (x^{2} + 2) = 0

x - 5 = 0

x - 5 + 5 = 0 + 5

**x = 5**

x + 5 = 0

x + 5 - 5 = 0 - 5

**x = -5**

x^{2} + 2 = 0

x^{2} + 2 - 2 = 0 - 2

x^{2} = -2

x^{2} = 2(-1)

x^{2} = 2i^{2} (since i^{2} = -1)

x = ±√(2i^{2})

**x = ±i√(2)**

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

**Example #2**

Factor x^{4} - 5x^{2} + 4 by using a quadratic pattern

**Step 1**

Write x^{4} - 5x^{2} + 4 in the pattern of a quadratic expression so you can factor it like one by making a temporary substitution of variables.

Let y = x^{2} and substitute y for x^{2}

x^{4} - 5x^{2} + 4 = (x^{2})^{2} - 5(x^{2}) + 4

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

**Step 2**

Factor y^{2} - 5y + 4

y^{2} - 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.

y^{2} - 5y + 4 = (y - 4)(y - 1)

**Step 3**

Substitute back to the original variable

(y - 4)(y - 1) = (x^{2} - 4)(x^{2} - 1)

Factor completely

(y - 4)(y - 1) = (x^{2} - 4)(x^{2} - 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