Question

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

Answer 1

Answer:

-2,3 and \frac{-1±i\sqrt{3} }{2}

Step-by-step 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=$x=\frac{-1±\sqrt{1-4} }{2} \\=\frac{-1±i\sqrt{3} }{2}$

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

Answer 2

Answer:

So, the roots of the given equation are:

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

Step-by-step 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

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

$\Rightarrow D=-3$

$x=\frac{-b\pm\sqrt{D}}{2a}$

$\Rightarrow x=\frac{-1\pm\sqrt{-3}}{2}$

$x=(-1)^(\frac{2}{3})$

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

So, the roots of the given equation are:

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