Question

What are all the zeros of the equation x^4 -6x^2-7x-6=0

Answer 1

So, the roots of the given equation are:

`x=3,x=-2,x=(-1)^{(2)/(3)},x=\sqrt[3]{-1}`

Explanation:

We have been given an equation:

`x^4-6x^2-7x-6=0`

The equation we have has degree four that means roots of the equation will be four.

The given equation can be rewritten as:

`(x-3)(x+2)(x^2+x+1)=0`

We will equate the above factors to zero we get:

(x-3)=0

x=3

(x+2)=0

x=-2

`x^2+x+1` (1)

we will solve equation (1) by discriminant method:

`D=b^2-4ac`

Here, a=1,b=1,c=1

`D=(1)^2-4(1)(1)`

`\Rightarrow D=-3`

`x=(-b\pm√(D))/(2a)`

`\Rightarrow x=(-1\pm√(-3))/(2)`

`x=(-1)^((2)/(3))`

`x=\sqrt[3]{-1}`

So, the roots of the given equation are:

`x=3,x=-2,x=(-1)^((2)/(3)),x=\sqrt[3]{-1}`

Answer 2

-2,3 and \frac{-1±i\sqrt{3} }{2}

Explanation:

Given is an equation of power 4 in polynomial as

`x^4 -6x^2-7x-6=0`

We have to find the zeroes of the funciton

By rational roots theorem since constant term = -6 and Leading coefficient =1

possible rational factors can be

±1,±2,±3,±6

By trial and error we try one by one.

`f(3) =3^4 -6(3)^2-7(3)-6=0`

So x=3 is zero and x-3 is factor

Try with -2

`f(-2) =(-2)^4 -6(-2)^2-7(-2)-6\\=16-24+14-6=0`

x=-2 is the zero

Now divide the given polynomial by (x+2)(x-3) to get quadratic form

Given polynomial =
`(x+2)(x-3)(x^2+x+1)`

Find the roots of quadratic by formula

x=
`x=(-1±√(1-4) )/(2) \\=(-1±i√(3) )/(2)`

So roots are -2,3 and \frac{-1±i\sqrt{3} }{2}