Question

Which of the following are roots of the polynomial function below f(x)=2x^3-x^2-9x+6

Answer 1

The roots of the given polynomial are

`x=2,(-3+√(33))/(4),(-3-√(33))/(4)`

Explanation:

Given : Polynomial
`f(x)=2x^3-x^2-9x+6`

To find : The roots of the given polynomial?

Solution :

Using hit and trial method one of the root of the polynomial is 2 as 2 satisfy the given polynomial equating zero.

Let x=2 substitute in polynomial,

`f(2)=2(2)^3-2^2-9(2)+6`

`f(2)=2(8)-4-18+6`

`f(2)=16-4-18+6`

`f(2)=22-22`

`f(2)=0`

So, One the root is 2.

Now, We divide the
`f(x)=2x^3-x^2-9x+6` by (x-2) to get the remainder in quadratic form by long division method.

The remainder is
`2x^2+3x-3`

Now, we factor the quadratic equation by using quadratic formula,

General form -
`ax^2+bx+c=0`

`D=b^2-4ac`

Solution is
`x=(-b\pm√(D))/(2a)`

Equation is
`2x^2+3x-3`

where, a=2 , b=3, c=-3

`D=b^2-4ac`

`D=(3)^2-4(2)(-3)`

`D=9+24`

`D=33`

Solution is
`x=(-b\pm√(D))/(2a)`

`x=(-(3)\pm√(33))/(2(2))`

`x=(-3\pm√(33))/(4)`

`x=(-3+√(33))/(4),(-3-√(33))/(4)`

Therefore, The roots of the given polynomial are

`x=2,(-3+√(33))/(4),(-3-√(33))/(4)`