Question

List all the possible rational zeros f(x)=x^3-3x^2-14x+12 and determine which are zeros

Answer 1

The zeros are x+3 ,
`x-(3+√(5))` and
`x-(3-√(5))`

and is given by

`f(x)=x^3-3x^2-14x+12=(x+3)(x-(3+√(5)))(x-(3-√(5)))`

Explanation:

Given polynomial function is
`f(x)=x^3-3x^2-14x+12`

To find the zeros:

Equate the polynomial function to zero

That is f(x)=0

`f(x)=x^3-3x^2-14x+12=0`

By using synthetic division method we can find zeros

-3_| 1 -3 -14 12

0 -3 18 -12

_______________

1 -6 4 0

Therefore x+3 is a zero

Ie., x=-3

Now the quadratic equation is
`x^2-6x+4=0`

Here a=1 , b=-6 and c=4

`x=(-b\pm√(b^2-4ac))/(2a)`

`x=(-(-6)\pm√((-6)^2-4(1)(4)))/(2(1))`

`x=(6\pm√(36-16))/(2)`

`x=(6\pm√(20))/(2)`

`x=(6\pm2√(5))/(2)`

`x=3\pm√(5)`

Therefore
`x=3+√(5)` and
`x=3-√(5)`

`x-(3+√(5))` and
`x-(3-√(5))`

Therefore the zeros are x+3 ,
`x-(3+√(5))` and
`x-(3-√(5))`

`f(x)=x^3-3x^2-14x+12=(x+3)(x-(3+√(5)))(x-(3-√(5)))`