# HOW TO FIND COMPLEX ROOTS OF A 4TH DEGREE POLYNOMIAL

Complex roots of an equation will occur in conjugate pair. That is, if (a + ib) is a root, then (a - ib) is the other root of the same equation.

Example 1 :

Solve the following equation, if (2 + i√3) is a root.

x4 - 4x2 + 8x + 35 = 0

Solution :

Since (2 + i√3) is a complex root, (2 - i√3) must be the other root.

x = 2 + i√3   or   x - (2 + i√3) = 0

x = 2 - i√3   or   x - (2 - i√3) = 0

Quadratic polynomial with the roots (2 + i√3) and (2 - i√3) :

= x2 - (sum of the roots)x + product of the roots

= x2 - [(2 + i√3) + (1 - i2√3)]x + (2 + i√3)(2 - i√3)

= x2 - [2 + i√3 + 2 - i√3]x + 22 - (√3i)2

= x2 - 4x + 4 - (√3)2i2

= x2 - 4x + 4 - 3(-1)

= x2 - 4x + 4 + 3

= x2 - 4x + 7

(x2 - 4x + 7) is a factor of (x4 - 4x2 + 8x + 35).

Let (x2 + ax + b) be the other factor.

Then,

(x2 + ax + b)(x2 - 4x + 7) = x4 - 4x2 + 8x + 35

Comparing the constant terms,

7b = 35

b = 5

Comparing the coefficient of x terms,

7a - 4b = 8

Substitute b = 5.

7a - 4(5) = 8

7a - 20 = 8

7a = 28

a = 4

x2 + ax + b ----> x2 + 4x + 5

Solving the equation x2 + 4x + 5 = 0, we get

x = -2 + i or -2 - i

Therefore, the roots of the given equation are

2 + i√3, 2 - i√3, -2 + i and -2 - i

Example 2 :

Solve the following equation, if (3 + i) is a root.

x4 - 8x3 + 24x2 - 32x + 20 = 0

Solution :

Since (3 + i) is a complex root, (3 - i) must be the other root.

x = 3 + i   or   x - (3 + i) = 0

x = 3 - i   or   x - (3 - i) = 0

Quadratic polynomial with the roots (3 + i) and (3 - i) :

= x2 - (sum of the roots)x + product of the roots

= x2 - [(3 + i) + (3 - i)]x + (3 + i)(3 - i)

= x2 - [3 + i + 3 - i]x + 32 - i2

= x2 - 6x + 9 - (-1)

= x2 - 6x + 9 + 1

= x2 - 6x + 10

(x2 - 6x + 10) is a factor of (x4 - 8x3 + 24x2 - 32x + 20).

Let (x2 + ax + b) be the other factor.

Then,

(x2 + ax + b)(x2 - 6x + 10) = x4 - 8x3 + 24x2 - 32x + 20

Comparing the constant terms,

10b = 20

b = 2

Comparing the coefficient of x terms,

10a - 6b = -32

Substitute b = 2.

10a - 6(2) = -32

10a - 12 = -32

10a = -20

a = -2

x2 + ax + b ----> x2 - 2x + 2

Solving the equation x2 - 2x + 2 = 0, we get

x = 1 + i or 1 - i

Therefore, the roots of the given equation are

3 + i, 3 - i, 1 + i and 1 - i

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