Factorisation of Polynomials — Practice Quiz

(x - 3), (x + 3)

Step 1: Recognize x² - 9 as difference of squares: a² - b² = (a + b)(a - b) Step 2: x² - 9 = x² - 3² = (x + 3)(x - 3) Step 3: Therefore, both (x - 3) and (x + 3) are factors. Verification: f(3) = 3² - 9 = 0 and f(-3) = (-3)² - 9 = 0

True

This statement is exactly the Factor Theorem. It says that for a polynomial f(x), if a is a real number such that f(a) = 0, then (x - a) must be a factor of f(x), and conversely. Since f(a) = 0 means x = a is a root, dividing f(x) by (x - a) always leaves a remainder of zero, confirming (x - a) is a factor. Hence the statement is True.

k = 2

Step 1: For (x - 1) to be a factor, f(1) must equal 0 Step 2: f(1) = (1)³ - k(1)² + 2(1) - 1 Step 3: f(1) = 1 - k + 2 - 1 = 2 - k Step 4: For f(1) = 0: 2 - k = 0 Step 5: Therefore k = 2

x² + x - 2, x³ + 8, x² - 4

Step 1: For (x + 2) to be a factor, f(-2) must equal 0 for each polynomial Step 2: Check x² + x - 2: f(-2) = 4 - 2 - 2 = 0 ✓ Step 3: Check x³ + 8: f(-2) = -8 + 8 = 0 ✓ Step 4: Check x² - 4: f(-2) = 4 - 4 = 0 ✓ Step 5: Check x² + 4: f(-2) = 4 + 4 = 8 ≠ 0 ✗