we have the equation
[tex]4x^4+6x^3-6x^2-4x=0[/tex][tex]x(4x^3+6x^2-6x^{}-4)=0[/tex][tex](4x^3+6x^2-6x^{}-4)=0[/tex]Note that
For x=1
[tex]\begin{gathered} (4(1)^3+6(1)^2-6(1)^{}-4)=0 \\ 0=0 \end{gathered}[/tex]x=1 is a root of the given equation
so
Divide
(4x^3+6x^2-6x-4) : (x-1)
4x^2+10x+4
-4x^3+4x^2
---------------------------
10x^2-6x-4
-10x^2+10x
-----------------------
4x-4
-4x+4
-----------
0
therefore
(4x^3+6x^2-6x-4)=(x-1)(4x^2+10x+4)
Solve the quadratic equation
(4x^2+10x+4)=0
a=4
b=10
c=4
substitute in the formula
[tex]x=\frac{-10\pm\sqrt[]{10^2-4(4)(4)}}{2(4)}[/tex][tex]x=\frac{-10\pm\sqrt[]{36}}{8}[/tex]the values of x are
x=-1/2 and x=-2
therefore
(4x^2+10x+4)=4(x+1/2)(x+2)=(4x+2)(x+2)
and the answer is
