Answer
$x = 4, 3$
Work Step by Step
To factor a quadratic polynomial equation in the form $ax^2 + bx + c = 0$, we look at factors of the product of $a$ and $c$ such that, when added together, equal $b$.
For the equation $x^2 - 7x + 12 = 0$, $(a)(c)$ is $(1)(12)$, or $12$, but when added together will equal $b$ or $-7$. Both factors need to be negative in this case. This is because a negative number multiplied with a negative number will equal a positive number; however, when a negative number is added to a negative number, the result will also be a negative number.
We came up with the following possibilities:
$(a)(c)$ = $(-12)(-1)$
$b = -13$
$(a)(c)$ = $(-4)(-3)$
$b = -7$
$(a)(c)$ = $(-6)(-2)$
$b = -8$
The second pair, $-4$ and $-3$, will work. Let us factor the polynomial incorporating these factors:
$(x - 4)(x - 3) = 0$
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:
The first factor:
$x - 4 = 0$
Add $4$ to each side to solve for $x$:
$x = 4$
The second factor:
$x - 3 = 0$
Add $3$ to each side to solve for $x$:
$x = 3$
The solution is $x = 4, 3$.
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. Let's plug in $x = 4$ first:
$(4)^2 - 7(4) + 12 = 0$
Multiply first, according to order of operations:
$16 - 28 + 12 = 0$
Now add and subtract from left to right:
$-12 + 12 = 0$
Subtract once again:
$0 = 0$
The left and right sides are equal; therefore, this solution is correct.
Let's check $x = 3$:
$(3)^2 - 7(3) + 12 = 0$
Multiply first, according to order of operations:
$9 - 21 + 12 = 0$
Now add and subtract from left to right:
$-12 + 12 = 0$
Add to simplify:
$0 = 0$
The left and right sides are equal; therefore, this solution is also correct.
