Special Products And Factorization

(2x + 3)²

Step 1: Check if this is a perfect square trinomial of form a² + 2ab + b². Step 2: Here, 4x² = (2x)², 9 = 3², and 12x = 2(2x)(3). Step 3: Since all conditions match a² + 2ab + b² where a = 2x and b = 3, we get (2x + 3)². Step 4: Verify: (2x + 3)² = 4x² + 12x + 9 ✓

x² + 8x + 15

Step 1: Use the formula (x + a)(x + b) = x² + (a + b)x + ab. Step 2: Here, a = 5 and b = 3. Step 3: Calculate: x² + (5 + 3)x + (5 × 3) = x² + 8x + 15. Step 4: The answer is x² + 8x + 15. The coefficient of x is the sum of the constants, and the constant term is their product.

(x - 3)(x - 4)

Step 1: We need two numbers that multiply to +12 and add to -7. Step 2: List factor pairs of 12: (1,12), (2,6), (3,4). Step 3: Check which pair adds to -7: -3 + (-4) = -7 ✓. Step 4: Therefore: x² - 7x + 12 = (x - 3)(x - 4). Verify by expanding: (x - 3)(x - 4) = x² - 4x - 3x + 12 = x² - 7x + 12 ✓

2x² + 7x + 3

Step 1: Use FOIL method: First, Outer, Inner, Last terms. Step 2: First: 2x × x = 2x²; Outer: 2x × 3 = 6x; Inner: 1 × x = x; Last: 1 × 3 = 3. Step 3: Combine: 2x² + 6x + x + 3 = 2x² + 7x + 3. Step 4: The answer is 2x² + 7x + 3. Always combine like terms carefully.