Answer
$x = -3, 2$
Work Step by Step
Let's rewrite the equation so that all terms are on the left side of the equation:
$$x^2 + x - 6 = 0$$
To factor a quadratic polynomial in the form $x^2 + bx + c = 0$, we look at factors of $c$ such that, when added together is equal to $b$.
For the equation $x^2 + x - 6 = 0$, $c=-6$, so look for factors of $-6$ that when added together will equal $b$ or $1$. One of the factors needs to be positive and the other one negative, but the positive factor should be the one with the greater absolute value. This is because a negative number multiplied with a positive number will equal a negative number; however, when a negative number is added to a positive number, the result can be either negative or positive, depending on which number has the greater absolute value.
Below are the possibilities:
$-6=(6)(-1)$
$6+(-1) = 5$
$-6=(3)(-2)$
$3+(-2) = 1$
The second pair, $3$ and $-2$, is the one we are looking for.
Thus, the factored form of the trinomial is $(x + 3)(x - 2)$ and the equation above is equivalent to
$$(x+3)(x-2)=0$$
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 then solve each one.
First factor:
$x + 3 = 0$
$x = -3$
Second factor:
$x - 2 = 0$
$x = 2$
The solutions are $-3$ and $2$.
To check if our solutions are correct, we plug our solutions back into the equation to see if the left and right sides equal one another.
When $x = -3$:
$x^2=-x+6\\
(-3)^2=-(-3)+6\\
9=3+6\\
9=9$
When $x = 2$:
$x=-x+6\\
2^2=-2+6\\
4=4$
Thus, both solutions are correct.
The solutions are $-3$ and $2$.
