To solve the quadratic equation [tex]\((x+4)(x+1)=0\)[/tex], we need to find the values of [tex]\(x\)[/tex] that make this equation true.
Start by setting each factor equal to zero:
1. [tex]\(x + 4 = 0\)[/tex]
2. [tex]\(x + 1 = 0\)[/tex]
Solve each equation for [tex]\(x\)[/tex]:
1. [tex]\(x + 4 = 0\)[/tex]
[tex]\[
x = -4
\][/tex]
2. [tex]\(x + 1 = 0\)[/tex]
[tex]\[
x = -1
\][/tex]
Thus, the solutions to the equation [tex]\((x+4)(x+1)=0\)[/tex] are [tex]\(x = -4\)[/tex] and [tex]\(x = -1\)[/tex].
The correct answer is:
[tex]\[ x = -4 \text{ or } x = -1 \][/tex]
