Answer:
-2,3 and \frac{-1±i\sqrt{3} }{2}
Step-by-step explanation:
Given is an equation of power 4 in polynomial as
[tex]x^4 -6x^2-7x-6=0[/tex]
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.
[tex]f(3) =3^4 -6(3)^2-7(3)-6=0[/tex]
So x=3 is zero and x-3 is factor
Try with -2
[tex]f(-2) =(-2)^4 -6(-2)^2-7(-2)-6\\=16-24+14-6=0[/tex]
x=-2 is the zero
Now divide the given polynomial by (x+2)(x-3) to get quadratic form
Given polynomial = [tex](x+2)(x-3)(x^2+x+1)[/tex]
Find the roots of quadratic by formula
x=[tex]x=\frac{-1±\sqrt{1-4} }{2} \\=\frac{-1±i\sqrt{3} }{2}[/tex]
So roots are -2,3 and \frac{-1±i\sqrt{3} }{2}
Answer:
So, the roots of the given equation are:
[tex]x=3,x=-2,x=(-1)^{\frac{2}{3}},x=\sqrt[3]{-1}[/tex]
Step-by-step explanation:
We have been given an equation:
[tex]x^4-6x^2-7x-6=0[/tex]
The equation we have has degree four that means roots of the equation will be four.
The given equation can be rewritten as:
[tex](x-3)(x+2)(x^2+x+1)=0[/tex]
We will equate the above factors to zero we get:
(x-3)=0
x=3
(x+2)=0
x=-2
[tex]x^2+x+1[/tex] (1)
we will solve equation (1) by discriminant method:
[tex]D=b^2-4ac[/tex]
Here, a=1,b=1,c=1
[tex]D=(1)^2-4(1)(1)[/tex]
[tex]\Rightarrow D=-3[/tex]
[tex]x=\frac{-b\pm\sqrt{D}}{2a}[/tex]
[tex]\Rightarrow x=\frac{-1\pm\sqrt{-3}}{2}[/tex]
[tex]x=(-1)^(\frac{2}{3})[/tex]
[tex]x=\sqrt[3]{-1}[/tex]
So, the roots of the given equation are:
[tex]x=3,x=-2,x=(-1)^(\frac{2}{3}),x=\sqrt[3]{-1}[/tex]
