Answer :
To find the solutions to the equation [tex]\((x+6)(x+2) = 0\)[/tex], follow these steps:
1. Identify the given equation:
[tex]\[ (x+6)(x+2) = 0 \][/tex]
2. Understand that a product of two factors is zero if either factor is zero.
This means we need to find the values of [tex]\(x\)[/tex] for which each factor is zero.
3. Set each factor equal to zero and solve for [tex]\(x\)[/tex] individually:
- For the first factor, [tex]\(x+6 = 0\)[/tex]:
[tex]\[ x + 6 = 0 \\ x = -6 \][/tex]
- For the second factor, [tex]\(x+2 = 0\)[/tex]:
[tex]\[ x + 2 = 0 \\ x = -2 \][/tex]
4. Check the given values of [tex]\(x\)[/tex] to see if they are solutions:
The potential solutions provided are [tex]\(x = -6\)[/tex], [tex]\(x = -4\)[/tex], [tex]\(x = 4\)[/tex], and [tex]\(x = 12\)[/tex].
- For [tex]\(x = -6\)[/tex]:
[tex]\[ (x + 6)(x + 2) = (-6 + 6)(-6 + 2) = 0 \times (-4) = 0 \][/tex]
[tex]\[ \text{Since } 0 = 0, \text{ this value is a solution.} \][/tex]
- For [tex]\(x = -2\)[/tex]:
[tex]\[ (x + 6)(x + 2) = (-2 + 6)(-2 + 2) = 4 \times 0 = 0 \][/tex]
[tex]\[ \text{Since } 0 = 0, \text{ this value is a solution.} \][/tex]
- For [tex]\(x = 4\)[/tex]:
[tex]\[ (x + 6)(x + 2) = (4 + 6)(4 + 2) = 10 \times 6 = 60 \][/tex]
[tex]\[ \text{Since } 60 \neq 0, \text{ this value is not a solution.} \][/tex]
- For [tex]\(x = 12\)[/tex]:
[tex]\[ (x + 6)(x + 2) = (12 + 6)(12 + 2) = 18 \times 14 = 252 \][/tex]
[tex]\[ \text{Since } 252 \neq 0, \text{ this value is not a solution.} \][/tex]
5. Conclude the correct solutions:
The values of [tex]\(x\)[/tex] that satisfy the equation [tex]\((x+6)(x+2) = 0\)[/tex] are:
[tex]\[ x = -6 \text{ and } x = -2. \][/tex]
Therefore, the solutions to the equation [tex]\((x + 6)(x + 2) = 0\)[/tex] are [tex]\(-6\)[/tex] and [tex]\(-2\)[/tex]. The values [tex]\(x = 4\)[/tex] and [tex]\(x = 12\)[/tex] are not solutions to the equation.
1. Identify the given equation:
[tex]\[ (x+6)(x+2) = 0 \][/tex]
2. Understand that a product of two factors is zero if either factor is zero.
This means we need to find the values of [tex]\(x\)[/tex] for which each factor is zero.
3. Set each factor equal to zero and solve for [tex]\(x\)[/tex] individually:
- For the first factor, [tex]\(x+6 = 0\)[/tex]:
[tex]\[ x + 6 = 0 \\ x = -6 \][/tex]
- For the second factor, [tex]\(x+2 = 0\)[/tex]:
[tex]\[ x + 2 = 0 \\ x = -2 \][/tex]
4. Check the given values of [tex]\(x\)[/tex] to see if they are solutions:
The potential solutions provided are [tex]\(x = -6\)[/tex], [tex]\(x = -4\)[/tex], [tex]\(x = 4\)[/tex], and [tex]\(x = 12\)[/tex].
- For [tex]\(x = -6\)[/tex]:
[tex]\[ (x + 6)(x + 2) = (-6 + 6)(-6 + 2) = 0 \times (-4) = 0 \][/tex]
[tex]\[ \text{Since } 0 = 0, \text{ this value is a solution.} \][/tex]
- For [tex]\(x = -2\)[/tex]:
[tex]\[ (x + 6)(x + 2) = (-2 + 6)(-2 + 2) = 4 \times 0 = 0 \][/tex]
[tex]\[ \text{Since } 0 = 0, \text{ this value is a solution.} \][/tex]
- For [tex]\(x = 4\)[/tex]:
[tex]\[ (x + 6)(x + 2) = (4 + 6)(4 + 2) = 10 \times 6 = 60 \][/tex]
[tex]\[ \text{Since } 60 \neq 0, \text{ this value is not a solution.} \][/tex]
- For [tex]\(x = 12\)[/tex]:
[tex]\[ (x + 6)(x + 2) = (12 + 6)(12 + 2) = 18 \times 14 = 252 \][/tex]
[tex]\[ \text{Since } 252 \neq 0, \text{ this value is not a solution.} \][/tex]
5. Conclude the correct solutions:
The values of [tex]\(x\)[/tex] that satisfy the equation [tex]\((x+6)(x+2) = 0\)[/tex] are:
[tex]\[ x = -6 \text{ and } x = -2. \][/tex]
Therefore, the solutions to the equation [tex]\((x + 6)(x + 2) = 0\)[/tex] are [tex]\(-6\)[/tex] and [tex]\(-2\)[/tex]. The values [tex]\(x = 4\)[/tex] and [tex]\(x = 12\)[/tex] are not solutions to the equation.