Is it possible to factor a quadratic equation when $a$, $b$, and $c$ are all equal? I have the equation $4x^2+4x+4$ to factor. I know that need to start with
$$(2x \quad )(2x \quad )$$
to make $4^2$, but I can't seem to factor the rest of the way. What should I do?
 A: The polynomial $4x^2+4x+4$ does not factor over the real numbers. However if you apply the quadratic equation you can deduce that its roots are
$$\frac{-4+\sqrt{16-64}}{2\cdot 4}=\frac{-1+\sqrt{-3}}{2}$$
and
$$\frac{-4-\sqrt{16-64}}{2\cdot 4}=\frac{-1-\sqrt{-3}}{2}$$
so that the factorization of $4x^2+4x+4$ over the complex numbers is
$$4\left(x-\frac{-1+\sqrt{-3}}{2}\right)\left(x-\frac{-1-\sqrt{-3}}{2}\right)$$
or rearranging slightly to fit the form you're looking for,
$$\Bigl(2x-(-1+\sqrt{-3})\Bigr)\Bigl(2x-(-1-\sqrt{-3})\Bigr)$$
A: You could factor out the $4$ first:
$$4x^2+4x+4=4(x^2+x+1)$$
And then factor $x^2+x+1$ using the quadratic formula.
A: We have $$4x^2+4x+4=4(x^2+x+1)$$  Let's consider $$f(x)=x^2+x+1$$ $$x=\frac{-1\pm\sqrt{1^2-4(1)(1)}}{2}=\frac{-1\pm i\sqrt3}{2}$$ Thus we have $$4x^2+4x+4=4\left(x-\frac{-1+i\sqrt3}{2}\right)\left(x-\frac{-1-i\sqrt3}{2}\right)$$ $$=4\left(x+\frac{1-i\sqrt3}{2}\right)\left(x+\frac{1+i\sqrt3}{2}\right)$$ $$=\left(2x+1-i\sqrt3\right)\left(2x+1+i\sqrt3\right)$$
A: Yes, as mentioned above you get $4(x^2+x+1)$. However, if you set this to zero and try to solve for $x$ using the quadratic formula, you will get complex numbers for a solution.
A: Yes, you can factor out a = b = c, and solve for the root using the quadratic formula. In this instance, there are two non-real roots, as others have noted. 
