Certainly! Let's solve the given equation step-by-step:
Given the equation:
[tex]\[
(x+1)(x-2)=0
\][/tex]
To solve this equation, we apply the property that if the product of two factors is zero, then at least one of the factors must be zero. This means we can set each factor equal to zero and solve for [tex]\( x \)[/tex].
1. First, set the first factor equal to zero:
[tex]\[
x + 1 = 0
\][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[
x = -1
\][/tex]
2. Next, set the second factor equal to zero:
[tex]\[
x - 2 = 0
\][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[
x = 2
\][/tex]
Thus, the solutions to the equation [tex]\((x+1)(x-2)=0\)[/tex] are [tex]\( x = -1 \)[/tex] and [tex]\( x = 2 \)[/tex].
Among these solutions, the one with the lowest value is:
[tex]\[
x = -1
\][/tex]
So, the solution with the lowest value is:
[tex]\[
\boxed{-1}
\][/tex]
