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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