Question

if two zeros of the polynomial x4-6x3-26x2+138x-35are 2+-root 3,find other zeros...
please help fast ..

Answer 1

Given,


Quatric equation = x⁴ - 6x³ - 26 x² + 138 x - 35.

Here coefficient of x⁴ ( a ) = 1 

Coefficient of x³ ( b ) = -6

Coefficient of x² ( c ) = -26

Coefficient of x ( d ) = 138

Constant term ( e ) = -35.

Two zeroes are = 2 + √3 ( α ) and 2 - √3 ( β ).

Let other 2 zeroes are Γ and Θ.

Sum of zeroes = - b/a

⇒ α + β + Г + Θ =  - ( -6 ) / 1 = 6

⇒ 2 + √3 + 2 - √3 + Γ + Θ = 6

⇒ 4 + Г + Θ = 6

⇒ Γ + Θ = 6 - 4 = 2. ---- ( 1 )

Now ,

Product of zeroes = e / a

⇒ α × β × Г × Θ = -35 / 1

⇒ ( 2 + √3 ) ( 2 - √3 ) × Г x Θ = -35

⇒ { ( 2 )² - ( √3 )² } × Γ × Θ = -35

⇒ ( 4 - 3 ) × Γ × Θ = -35

⇒ Γ × Θ = -35  ------ ( 2 )

Now , 

 ( Γ - Θ )² = ( Γ + Θ )² - 4 × Γ × Θ

By substituting the values of ( 1 ) and ( 2 ) ,

⇒ ( Γ - Θ )² = ( 2 )² - 4 × ( - 35 )

⇒ (Γ - Θ )² = 4 + 140

⇒ ( Γ - Θ )² = 144

⇒ ( Γ - Θ ) = √144

⇒ ( Γ - Θ ) = 12  --------- ( 3 ).

By adding ( 1 ) and ( 3 ) ,

⇒ Γ + Θ + Γ - Θ = 12 + 2

⇒ 2 Γ = 14

⇒ Γ = 14 ÷ 2

∴ Γ = 7.

By substituting the value of Γ in ( 3 ),

⇒ Γ - Θ = 12

⇒ 7 - Θ = 12

⇒ Θ = 7 - 12

 ∴  Θ = -5.


Therefore , other 2 zeroes are ( -5 ) and ( 7 ).

Answer 2

hey friend...


hope it will help you.