Solving Equations by Factorization
We can solve equations like ( x^2 + 5x + 6 = 0 ) by factorization.
Solution:
( x^2 + 5x + 6 = (x + 2)(x + 3) = 0 )
Hence, ( x + 2 = 0 ) or ( x + 3 = 0 )
Therefore, ( x = -2 ) or ( x = -3 ).
Example 1: Solve ( x^2 + 7x + 10 = 0 )
Step 1: Write the equation
x^2 + 7x + 10 = 0
Step 2: Find two numbers whose product is 10 and sum is 7
5 \text{ and } 2
Step 3: Factorize
(x + 5)(x + 2) = 0
Step 4: Set each factor equal to zero
x + 5 = 0 \quad \text{or} \quad x + 2 = 0
Step 5: Solve for (x)
x = -5 \quad \text{or} \quad x = -2
Final Answer:
\boxed{x = -5,; -2}
Example 2: Solve ( x^2 - 9x + 20 = 0 )
Step 1: Write the equation
x^2 - 9x + 20 = 0
Step 2: Find two numbers whose product is 20 and sum is −9
-5 \text{ and } -4
Step 3: Factorize
(x - 5)(x - 4) = 0
Step 4: Set each factor equal to zero
x - 5 = 0 \quad \text{or} \quad x - 4 = 0
Step 5: Solve for (x)
x = 5 \quad \text{or} \quad x = 4
Final Answer:
\boxed{x = 4,; 5}
Example 3: Solve ( 2x^2 + 7x + 3 = 0 )
Step 1: Write the equation
2x^2 + 7x + 3 = 0
Step 2: Multiply the coefficient of (x^2) by the constant term
2 \times 3 = 6
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Step 3: Find two numbers whose product is 6 and sum is 7
6 \text{ and } 1
Step 4: Rewrite the middle term
2x^2 + 6x + x + 3 = 0
Step 5: Group and factor
(2x^2 + 6x) + (x + 3) = 0
2x(x + 3) + 1(x + 3) = 0
Step 6: Factor out the common bracket
(2x + 1)(x + 3) = 0
Step 7: Set each factor equal to zero
2x + 1 = 0 \quad \text{or} \quad x + 3 = 0
Step 8: Solve for (x)
x = -\frac{1}{2} \quad \text{or} \quad x = -3
Final Answer:
\boxed{x = -\frac{1}{2},; -3}
Example 4: Solve ( 3x^2 - 5x - 2 = 0 )
Step 1: Write the equation
3x^2 - 5x - 2 = 0
Step 2: Multiply the coefficient of (x^2) by the constant term
3 \times (-2) = -6
Step 3: Find two numbers whose product is −6 and sum is −5
-6 \text{ and } 1
Step 4: Rewrite the middle term
3x^2 - 6x + x - 2 = 0
Step 5: Group and factor
(3x^2 - 6x) + (x - 2) = 0
3x(x - 2) + 1(x - 2) = 0
Step 6: Factor out the common bracket
(3x + 1)(x - 2) = 0
Step 7: Set each factor equal to zero
3x + 1 = 0 \quad \text{or} \quad x - 2 = 0
Step 8: Solve for (x)
x = -\frac{1}{3} \quad \text{or} \quad x = 2
Final Answer:
\boxed{x = -\frac{1}{3},; 2}