Learn application problems involving factoring using substitution and combining multiple formulas.
Factoring Using Substitution
Example 1
Factor (x＋1)²＋3(x＋1)＋2.
Solution:
Let A＝x＋1: A²＋3A＋2
Product is 2・sum is 3 → (A＋1)(A＋2)
Substitute back: (x＋2)(x＋3)
(x＋2)(x＋3)
Factor (x＋1)²＋3(x＋1)＋2.
Solution:
Let A＝x＋1: A²＋3A＋2
Product is 2・sum is 3 → (A＋1)(A＋2)
Substitute back: (x＋2)(x＋3)
(x＋2)(x＋3)
Combining Common Factors and the Cross-Multiplication Method
Example 2
Factor 2x²－8.
Solution:
Factor out the common factor 2: 2(x²－4)
Difference of squares: 2(x＋2)(x－2)
2(x＋2)(x－2)
Factor 2x²－8.
Solution:
Factor out the common factor 2: 2(x²－4)
Difference of squares: 2(x＋2)(x－2)
2(x＋2)(x－2)
Applications of Factoring (Equations)
Example 3
Solve x²－7x＋12＝0.
Solution:
Factoring: (x－3)(x－4)＝0
x＝3 or x＝4
Solve x²－7x＋12＝0.
Solution:
Factoring: (x－3)(x－4)＝0
x＝3 or x＝4
Key Points
- Simplify complex expressions using “substitution (A=...)”
- First look for common factors, then consider formulas
- Develop the habit of expanding after factoring to check your answer
Practice Problems
- Factor 3x²－12.
- Factor (x－2)²－5(x－2)＋6.
Answers and Explanations
- Answer:3(x＋2)(x－2)
Explanation: Factor out the common factor 3: 3(x²－4). Next, use the difference of squares: x²－4＝x²－2²＝(x＋2)(x－2). Therefore, 3x²－12＝3(x＋2)(x－2). - Answer:(x－4)(x－1)
Explanation: Let A＝x－2: A²－5A＋6. The two numbers whose product is 6 and whose sum is －5 are －2 and －3. Thus, (A－2)(A－3). Substituting back: (x－2－2)(x－2－3)＝(x－4)(x－5). Check: Let A=x-2 and expand (A-2)(A-3)=(x-4)(x-5): x²-9x+20. (x-2)²-5(x-2)+6=x²-4x+4-5x+10+6=x²-9x+20 ✓