Answer
The solutions are $x = 8, -1$.
Work Step by Step
We are asked to solve this quadratoc equation. We can try to factor it, but first, we want to rewrite it in standard form $(ax^2 + bx + c = 0)$:
$x^2 - 7x - 8 = 0$
To factor a quadratic polynomial in the standard form $x^2 + bx + c = 0$, we look at factors of $c$ such that, when added together, equal $b$.
For the equation $x^2 - 7x - 8 = 0$, $c=-8$ so look for factors of $-8$ that when added together will equal $b$ or $-7$. Here are the possibilities:
$-8=(-8)(1)$
$-8+1=-7$
$-8=(8)(-1)$
$8+(-1)=7$
$-8=(-4)(2)$
$-4+2= -2$
$-8=(4)(-2)$
$4+(-2)= 2$
The first pair, $-8$ and $1$, gives the correct sum.
This means that the factored form of the trinomial is $(x - 8)(x + 1)$'
Thus, the equation above is equivalent to $(x-8)(x+1)=0$.
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:
First factor:
$x - 8 = 0$
$x = 8$
Second factor:
$x + 1 = 0$
$x = -1$
The solutions are $x = 8, -1$.
