HOW TO FIND THE ROOTS OF A POLYNOMIAL OF DEGREE 4

Example 1 :

Solve the equation

x4+4x3+5x2+2x-2  =  0

which one root is -1 + i

Solution :

Let 

f(x)  =  x4+4x3+5x2+2x-2

Since one of the root is complex number, the other root may be its conjugate.

So,  α = -1 + i  β = - 1 - i

By using these two roots we can find a quadratic equation which is the part of the original equation. First, let us find the quadratic equation.

General form of any quadratic equation:

 x²-(α+β)x+αβ = 0

α  =  -1+i  β  =  -1-i

Sum of roots (α+β)  =  -1+i-1-i

= - 2

Product of roots (αβ)  =  (-1+i) (-1-i)

=  (-1)2 - i2

=  2

Therefore the required quadratic equation 

x2-(-2)x+2  =  0

x²+2x+2  =  0

This is the part of the equation of polynomial of degree 4.

By solving this quadratic equation

x2+2x-1  =  0

we get other two roots.

a = 1, b = 2, c = -1

x  =  [- b ± √(b2-4 a c)]/2 a

x  =  [-2 ± √(22-4 (1)(-1))]/2 (1)

x  =  [- 2 ± √8]/2

x  =  [- 2 ± 2 √2]/2

x  =  -1±√2

Therefore the four roots are

-1 + i ,- 1 - i , - 1 + √2 , - 1 - √2.

Example 2 :

Solve the equation

6x4 − 35x3 + 62x2 − 35x + 6 = 0

Solution :

By suing synthetic division repeatedly, we may solve this problem.

2 and 3 are the solutions of the given polynomial. To know the other solutions, let us solve the quadratic equation.

6x2 - 5x + 1  =  0

6x2 - 2x - 3x + 1  =  0

2x (3x - 1) - 1(3x - 1)  =  0

(2x - 1) (3x - 1)  =  0

2x - 1  =  0 and 3x - 1  =  0

x  =  1/2  and x  =  1/3

Hence the solutions are 2, 3, 1/2 and 1/3

Example 3 :

Solve

x4 + 3x3 - 3x - 1  =  0

Solution :

1 and -1 are the solutions of the given polynomial, to find other two solutions, let us solve the quadratic equation

x2 + 3x + 1  =  0

x = -b ± √(b2 - 4ac) / 2a

x = -3 ± √(9 - 4) / 2(1)

x = (-3 ± √5)/2

So, roots are -1, 1, (-3 ± √5)/2.

Example 4 :

Solve the equation 6x4 − 5x3 − 38x2 − 5x + 6 = 0 if it is known that 1/3 is a solution.

Solution :

1/3 and 3 are solutions of the given polynomial, to find other two solutions, let us solve the quadratic equation.

6x2 + 15x + 6  =  0

6x2 + 12x + 3x + 6  =  0

6x(x + 2) + 3(x + 2)  =  0

(6x + 3)(x + 2)  =  0

x  =  -1/2 and x = -2

Hence the solutions are -1/2, -2, 3 and 1/3.

Example 5 :

Solve the equation x4 −14x2 + 45 = 0

Solution :

Let x2  =  t

t2 − 14t + 45 = 0

t2 − 9t - 5t + 45 = 0

t(t - 9) - 5(t - 9)  =  0

(t - 5) (t - 9)  =  0

t  =  5 and t  =  9

x2  =  5, x2  =  9

x  =  ±√5, x = ±3

Hence the solutions are √5, -√5, 3 and -3.

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