Answer
The solution is $x = -\frac{1}{2}$.
Work Step by Step
Write the equation in standard form:
$$4x^2 + 4x + 1 = 0$$
To factor a quadratic polynomial in the form $ax^2 + bx + c$, we look at factors of the product of $a$ and $c$ such that, when added together, equal $b$.
In the trinomial $4x^2 + 4x + 1$, $(a)(c)$ is $(4)(1)$, or $4$, so we have to look for the factors of $4$ that when added together will equal $b$ or $4$. We came up with the possibilities:
$(a)(c)$ = $(4)(1)$
$b = 5$
$(a)(c)$ = $(2)(2)$
$b = 4$
The second pair, $2$ and $2$, will work. Let us factor the polynomial by splitting the middle term using these factors:
$4x^2 + 2x + 2x + 1 = 0$
Group the first two terms together and the last two terms together:
$(4x^2 + 2x) + (2x + 1) = 0$
For the first group of terms, we can factor out a $2x$:
$2x(2x + 1) + (2x + 1) = 0$
We see that $2x + 1$ is a common factor to both groups, so we factor it out to obtain (the $1$ is implied in front of the second group):
$(2x + 1)(2x + 1) = 0$
According to the Zero-Product Property, if the product of two factors $a$ and $b$ equals zero, then either $a$ is zero, $b$ is zero, or both equal zero. Therefore, we can set each factor equal to zero. Since both factors are exactly the same, we only need to set it equal to $0$ once:
$2x + 1 = 0$
$2x = -1$
Divide both sides by $2$ to solve for $x$:
$x = -\frac{1}{2}$
The solution is $x = -\frac{1}{2}$.
