Question

the roots of x^3-3x^2-16c+48=0 , given that the sum of two roots is zero is

Answer 1

x³ - 3 x² -16 c + 48 = 0

For the cubic polynomial

A x³ + B x² + C x + D = 0

Vieta's formulas:

x₁ + x₂ + x₃ = - B / A

x₁ ∙ x₂ + x₁ ∙ x₃ + x₂ ∙ x₃ = C / A

x₁ ∙ x₂ ∙ x₃ = - D / A

Where x₁, x₂ and x₃ are the roots of this polynomial.

In this cace:

A = 1 , B = - 3 , C = 0 , D = -16 c + 48

so:

x₁ + x₂ + x₃ = - B / A = - ( - 3 ) / 1 = 3 / 1 = 3

x₁ ∙ x₂ + x₁ ∙ x₃ + x₂ ∙ x₃ = C / A = 0 / 1 = 0

x₁ ∙ x₂ ∙ x₃ = - D / A = - (-16 c + 48 ) / 1 = ( 16 c - 48 ) / 1 = 16 c - 48

When the sum of two roots is zero then:

x₁ = - x₂

Now:

x₁ + x₂ + x₃ =

- x₂ + x₂ + x₃ = 3

x₃ = 3


x₁ ∙ x₂ + x₁ ∙ x₃ + x₂ ∙ x₃ = 0

( - x₂ ) ∙ x₂ + ( - x₂ ) ∙ 3 + x₂ ∙ 3 = 0

- x₂² - 3 x₂ + 3 x₂ = 0

- x₂² = 0 Multiply both sides by -1

x₂² = 0

x₂ = 0


x₁ = - x₂

x₁ = 0


Also by Vieta's formulas:

A x³ + B x² + C x + D = ( x - x₁ ) ( x - x₂ ) ( x - x₃ )


x³ - 3 x² -16 c + 48 = ( x - x₁ ) ( x - x₂ ) ( x - x₃ ) =

( x - 0 ) ( x - 0 ) ( x - 3 ) = x ∙ x ∙ ( x - 3 ) = x² ∙ ( x - 3 ) = x³ - 3 x²


x³ - 3 x² -16 c + 48 = x³ - 3 x² Subtract x³ - 3 x² to both sides

( x³ - 3 x² ) -16 c + 48 - ( x³ - 3 x² ) = ( x³ - 3 x² ) - ( x³ - 3 x² )

-16 c + 48 = 0 Subtract 48 to both sides

-16 c + 48 - 48 = - 48

-16 c = - 48 Divide both sides by - 16

c = - 48 / - 16

c = 3


The solutions:

c = 3

x₁ = 0

x₂ = 0

x₃ = 3

Answer 2

One correction.


A x³ + B x² + C x + D = A ( x - x₁ ) ( x - x₂ ) ( x - x₃ )

Answer 3

x^3-3x^2-16c+48
= x^2(x-3)-16(c-3)
If we let c=x, then we have
(x^2-16)(x-3)
= (x-4)(x+4)(x-3)
the roots are -4,4,3 so our cubic is

x^3-3x^2-16x+48=0

or, as above, let c=3, and we have

x^2(x-3) = 0
and the roots are 0,0,3