Alright, let's solve each of these quadratic equations by factoring step-by-step.
### 1. [tex]\( x^2 + 11x = 0 \)[/tex]
First, we factor out the common term [tex]\( x \)[/tex]:
[tex]\[ x(x + 11) = 0 \][/tex]
Set each factor equal to zero:
[tex]\[ x = 0 \][/tex]
[tex]\[ x + 11 = 0 \][/tex]
Solving the second equation:
[tex]\[ x = -11 \][/tex]
Therefore, the solutions are:
[tex]\[ x = 0, -11 \][/tex]
### 2. [tex]\( m^2 + 8m + 7 = 0 \)[/tex]
To factor this quadratic, we look for two numbers that multiply to 7 and add up to 8. Those numbers are 7 and 1. We can rewrite and factor as:
[tex]\[ (m + 7)(m + 1) = 0 \][/tex]
Set each factor equal to zero:
[tex]\[ m + 7 = 0 \][/tex]
[tex]\[ m + 1 = 0 \][/tex]
Solving these:
[tex]\[ m = -7 \][/tex]
[tex]\[ m = -1 \][/tex]
Therefore, the solutions are:
[tex]\[ m = -7, -1 \][/tex]
### 3. [tex]\( x^2 + 5x - 14 = 0 \)[/tex]
We need two numbers that multiply to [tex]\(-14\)[/tex] and add up to 5. The numbers [tex]\(-2\)[/tex] and 7 meet these criteria. We can factor the quadratic as:
[tex]\[ (x + 7)(x - 2) = 0 \][/tex]
Set each factor equal to zero:
[tex]\[ x + 7 = 0 \][/tex]
[tex]\[ x - 2 = 0 \][/tex]
Solving these:
[tex]\[ x = -7 \][/tex]
[tex]\[ x = 2 \][/tex]
Therefore, the solutions are:
[tex]\[ x = -7, 2 \][/tex]
### 4. [tex]\( 2y^2 + 8y - 10 = 0 \)[/tex]
First, we factor out the common term (if any), but since there is none, we proceed with factoring. We need numbers multiplying to [tex]\(2 \cdot (-10) = -20\)[/tex] and adding up to 8. Those numbers are 10 and [tex]\(-2\)[/tex]. Thus, we rewrite and factor as:
[tex]\[ 2y^2 + 10y - 2y - 10 = 0 \][/tex]
[tex]\[ 2y(y + 5) - 2(y + 5) = 0 \][/tex]
[tex]\[ (2y - 2)(y + 5) = 0 \][/tex]
Set each factor equal to zero:
[tex]\[ 2y - 2 = 0 \][/tex]
[tex]\[ y + 5 = 0 \][/tex]
Solving these:
[tex]\[ y = 1 \][/tex]
[tex]\[ y = -5 \][/tex]
Therefore, the solutions are:
[tex]\[ y = -5, 1 \][/tex]
### 5. [tex]\( x^2 - 4x + 4 = 0 \)[/tex]
We need numbers that multiply to 4 and add up to [tex]\(-4\)[/tex]. The numbers are [tex]\(-2\)[/tex] and [tex]\(-2\)[/tex]. This is a perfect square quadratic:
[tex]\[ (x - 2)^2 = 0 \][/tex]
Since it is a perfect square, setting the factor to zero gives:
[tex]\[ x - 2 = 0 \][/tex]
So we solve:
[tex]\[ x = 2 \][/tex]
Therefore, the only solution is:
[tex]\[ x = 2 \][/tex]
### Summary of Solutions:
1. [tex]\( x^2 + 11x = 0 \)[/tex] : [tex]\( x = 0, -11 \)[/tex]
2. [tex]\( m^2 + 8m + 7 = 0 \)[/tex] : [tex]\( m = -7, -1 \)[/tex]
3. [tex]\( x^2 + 5x - 14 = 0 \)[/tex] : [tex]\( x = -7, 2 \)[/tex]
4. [tex]\( 2y^2 + 8y - 10 = 0 \)[/tex] : [tex]\( y = -5, 1 \)[/tex]
5. [tex]\( x^2 - 4x + 4 = 0 \)[/tex] : [tex]\( x = 2 \)[/tex]
