KUTA SOFTWARE- INFINITE ALGEBRA 1

Question

SOLVING QUADRATIC EQUATIONS BY FACTORING
SOLVE EACH EQUATION BY FACTORING.
1) (K+ 1) (K - 5) = 0
2) (a + 1 (a + 2) = 0
3) (4k + 5) ( k+ 1) = 0
4) (2m + 3) ( 4m + 3)= 0
5) x 2 - 11 x + 19 + -5
6) n 2 + 7n + 15 = 5

Answer 1

To solve each equation by factoring, we will set each factor equal to zero and solve for the variable.

1. \( (K+ 1)(K - 5) = 0 \)

   Set each factor to zero:

   • \( K + 1 = 0 \) → \( K = -1 \)
   • \( K - 5 = 0 \) → \( K = 5 \)

   Solutions: \( K = -1 \) and \( K = 5 \)

2. \( (a + 1)(a + 2) = 0 \)

   Set each factor to zero:

   • \( a + 1 = 0 \) → \( a = -1 \)
   • \( a + 2 = 0 \) → \( a = -2 \)

   Solutions: \( a = -1 \) and \( a = -2 \)

3. \( (4k + 5)(k + 1) = 0 \)

   Set each factor to zero:

   • \( 4k + 5 = 0 \) → \( 4k = -5 \) → \( k = -\frac{5}{4} \)
   • \( k + 1 = 0 \) → \( k = -1 \)

   Solutions: \( k = -\frac{5}{4} \) and \( k = -1 \)

4. \( (2m + 3)(4m + 3) = 0 \)

   Set each factor to zero:

   • \( 2m + 3 = 0 \) → \( 2m = -3 \) → \( m = -\frac{3}{2} \)
   • \( 4m + 3 = 0 \) → \( 4m = -3 \) → \( m = -\frac{3}{4} \)

   Solutions: \( m = -\frac{3}{2} \) and \( m = -\frac{3}{4} \)

5. \( x^2 - 11x + 19 = -5 \)

   First, move -5 to the other side: \[ x^2 - 11x + 19 + 5 = 0 \] \[ x^2 - 11x + 24 = 0 \]

   Now, factor: \[ (x - 3)(x - 8) = 0 \]

   Set each factor to zero:

   • \( x - 3 = 0 \) → \( x = 3 \)
   • \( x - 8 = 0 \) → \( x = 8 \)

   Solutions: \( x = 3 \) and \( x = 8 \)

6. \( n^2 + 7n + 15 = 5 \)

   First, move 5 to the other side: \[ n^2 + 7n + 15 - 5 = 0 \] \[ n^2 + 7n + 10 = 0 \]

   Factor: \[ (n + 2)(n + 5) = 0 \]

   Set each factor to zero:

   • \( n + 2 = 0 \) → \( n = -2 \)
   • \( n + 5 = 0 \) → \( n = -5 \)

   Solutions: \( n = -2 \) and \( n = -5 \)

Final Summary of Solutions:

1. \( K = -1, 5 \)
2. \( a = -1, -2 \)
3. \( k = -\frac{5}{4}, -1 \)
4. \( m = -\frac{3}{2}, -\frac{3}{4} \)
5. \( x = 3, 8 \)
6. \( n = -2, -5 \)