Answer:
-9, 3 and 6
Step-by-step explanation:
Given cubic polynomial equation:
[tex]x^3-63x+162=0[/tex]
- Let α = first zero
- Let β = second zero
- Let 2β = third zero (since one zero is double another zero).
[tex]\textsf{The sum of the zeros of a cubic polynomial }ax^3+bx^2+cx+d \textsf{ is }-\dfrac{b}{a}.[/tex]
[tex]\implies \alpha + \beta + 2 \beta = -\dfrac{0}{1}[/tex]
[tex]\implies \alpha + 3 \beta = 0[/tex]
[tex]\implies \beta=- \dfrac{\alpha}{3}[/tex]
[tex]\textsf{The product of the zeros of a cubic polynomial }ax^3+bx^2+cx+d \textsf{ is }-\dfrac{d}{a}.[/tex]
[tex]\implies \alpha \cdot \beta \cdot 2\beta=-\dfrac{162}{1}[/tex]
[tex]\implies 2\alpha\beta^2=-162[/tex]
Substitute the found expression for β into the product equation:
[tex]\implies 2\alpha\beta^2=-162[/tex]
[tex]\implies 2\alpha\left(- \dfrac{\alpha}{3}\right)^2=-162[/tex]
[tex]\implies 2\alpha\left(\dfrac{\alpha^2}{9}\right)=-162[/tex]
[tex]\implies \dfrac{\alpha^3}{9}\right)=-81[/tex]
[tex]\implies \alpha^3=-729[/tex]
[tex]\implies \alpha=\sqrt[3]{-729}[/tex]
[tex]\implies \alpha=-9[/tex]
Substitute the found value of α into the expression for β :
[tex]\implies \beta=- \dfrac{-9}{3}=3[/tex]
Therefore, the zeros are:
[tex]\implies \alpha=-9[/tex]
[tex]\implies \beta=3[/tex]
[tex]\implies 2\beta=6[/tex]
