Is it possible to factor a quadratic equation when $a$, $b$, and $c$ are all equal?

Question

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?

Answer 1

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.

Answer 2

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, $$\Bigl(2x-(-1+\sqrt{-3})\Bigr)\Bigl(2x-(-1-\sqrt{-3})\Bigr)$$

Answer 3

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 $$4x^2+4x+4=4\left(x-\frac{-1+i\sqrt3}{2}\right)\left(x-\frac{-1-i\sqrt3}{2}\right)$$ $$=4\left(x+\frac{1-i\sqrt3}{2}\right)\left(x+\frac{1+i\sqrt3}{2}\right)$$ $$=\left(2x+1-i\sqrt3\right)\left(2x+1+i\sqrt3\right)$$

Answer 4

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.

Answer 5

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.