Roots of Polynomial ofdegree4





In this page roots of polynomial ofdegree4 we are going to see how to find relationship between roots and coefficients of polynomial with degree 4.

Let a x⁴ + b x³ + c x² + d x + e be the polynomial of degree 4 whose roots are α,β,γ and δ

Formula:

α + β + γ + δ = - b (co-efficient of x³)

α β + β γ +  γ δ + δ α = c (co-efficient of x²)

α β γ + β γ δ + γ δ α + δ α β = - d (co-efficient of x)

α β γ δ = e

Example 1:

Solve the equation x⁴ + 2x³ - 25 x² - 26 x + 120 = 0 given that the product of two roots is 8.

Solution :

Let α,β,γ,δ be the four roots.

Product of two roots = 8

                       α β = 8             

x⁴ + 2x³ - 25 x² - 26 x + 120 = (x - α)(x - β)(x - γ)(x - δ)

                                        = [x² - (α+β)x + αβ][x² - (γ+δ)x + γδ]

=x - x³[(γ+δ)+(α+β)]+[γδβ++β)(γ+δ)]x²-[αβ+δ)+γδ(α+β)]xβγδ

equating the coefficients of x³,x²,x and constant

-[(γ+δ)+(α+β)] = 2

α+β+γ+δ = -2       --------------- (1)

γδβ++β)(γ+δ) = -25  -----------(2)

-[αβ+δ) + γδ(α+β)] = -26 ------------(3)

αβγδ = 120 ---------(4)

8γδ = 120

  γδ = 120/8

   γδ = 15

substitute α β = 8 and γ δ = 15 in the third equation

       8+δ) + 15(α+β) = 26 ----------(5)

Multiplying the first equation by 8

8(α+β)+8(γ+δ) = -16 -------------(6)

(5) - (6) => 15(α+β) + 8+δ) = 26

                 8(α+β) + 8(γ+δ) = -16

                  (-)         (-)         (+)   

                ____________________

                   7(α+β) = 42

                     (α+β) = 42/7

                     (α+β) = 6

substitute (α+β) = 6 in the fifth equation

               15(6) + 8+δ) = 26

                 90 + 8+δ) = 26

                       8+δ) = 26 - 90

                       8+δ) = -64

                        +δ) = -64/8

                        +δ) = -8

substitute αβ = 8,(α+β) = 6,+δ) = -8 in the second equation

γδβ++β)(γ+δ) = -25 

γδ+8+(6)(-8) = -25 

γδ+8-48 = -25 

γδ-40 = -25

γδ = -25 + 40

γδ = 15

Now let us take the two quadratic equations

(i) (x² - (α+β)x + αβ)

(ii) (x² - (γ+δ)x + γδ)

by solving (x² - 6x + 8) we will get values of the roots α and β

x² - 6x + 8 = 0

(x-4)(x-2) = 0 

x = 4 and x = 2

by solving (x² - 8x + 15) we will get values of the roots γ and δ

x² - 8x + 15 = 0

(x-3)(x-5) = 0

x = 3 and x = 5

Therefore the four roots are 2,3,4,5

roots of polynomial ofdegree4 roots of polynomial ofdegree4




Roots of Polynomial ofdegree4 to Algebra