Question

One of the zeros of the equation x³ − 63 + 162 = 0 is double another zero. Find all the three zeros.​

Answer 1

-9, 3 and 6

Explanation:

Given cubic polynomial equation:

`x^3-63x+162=0`

• Let α = first zero
• Let β = second zero
• Let 2β = third zero (since one zero is double another zero).

`\textsf{The sum of the zeros of a cubic polynomial }ax^3+bx^2+cx+d \textsf{ is }-(b)/(a).`

`\implies \alpha + \beta + 2 \beta = -(0)/(1)`

`\implies \alpha + 3 \beta = 0`

`\implies \beta=- (\alpha)/(3)`

`\textsf{The product of the zeros of a cubic polynomial }ax^3+bx^2+cx+d \textsf{ is }-(d)/(a).`

`\implies \alpha \cdot \beta \cdot 2\beta=-(162)/(1)`

`\implies 2\alpha\beta^2=-162`

Substitute the found expression for β into the product equation:

`\implies 2\alpha\beta^2=-162`

`\implies 2\alpha\left(- (\alpha)/(3)\right)^2=-162`

`\implies 2\alpha\left((\alpha^2)/(9)\right)=-162`

`\implies (\alpha^3)/(9)\right)=-81`

`\implies \alpha^3=-729`

`\implies \alpha=\sqrt[3]{-729}`

`\implies \alpha=-9`

Substitute the found value of α into the expression for β :

`\implies \beta=- (-9)/(3)=3`

Therefore, the zeros are:

`\implies \alpha=-9`

`\implies \beta=3`

`\implies 2\beta=6`