Answer:
B is true
Step-by-step explanation:
given f(x) = 2x² - x - 1 and g(x) = x - 1
A
If x = - 1 is a root then f(- 1) = 0
f(- 1) = 2(- 1)² - (- 1) - 1 = 2 + 1 - 1 = 2 ≠ 0 ← False
B
If (x - 1) is a factor then x = 1 is a root and f(1) = 0
f(1) = 2(1)² - 1 - 1 = 2 - 1 - 1 = 0
⇒ g(x) = x - 1 is a factor of f(x) ← True
C
If x = 2 is a root then f(2) = 0
f(2) = 2(2)² - 2 - 1 = 8 - 3 = 5 ≠ 0 ← False
D
[tex]\frac{x2x^2-x-1}{x-1}[/tex] = [tex]\frac{(2x+1)(x-1)}{x-1}[/tex]
Cancel the factor (x - 1) on the numerator/ denominator, leaving
[tex]\frac{f(x)}{g(x)}[/tex] = 2x + 1 with remainder 0 ≠ 2 ← False
