Answer:
The correct option is D) (x+2)²(x-3)² = 0
Step-by-step explanation:
We need to find the polynomial equation of least degree has -2, -2, 3, and 3 as four of its roots
we will check each option by equating them to zero
Check part A)
(x + 2)(x - 3) = 0
if x + 2 =0 ⇒ x = -2
if x - 3 =0 ⇒ x = 3
Check part B)
(x - 2)-2(x - 3)³ = 0
if x - 2 =0 ⇒ x = 2
if x - 3 = 0 ⇒ x = 3, 3 ,3 ( since, multiplicity of (x - 3)³ is 3 )
Check part C)
(x-2+2)(x²-3) = 0
if (x-2+2) = 0 ⇒ x = 0
if x²-3 = 0 ⇒ x² = 3 ⇒ x = √3
Check part D)
(x+2)²(x-3)² = 0
if (x+2)² = 0 ⇒ x = -2, -2 ( since, multiplicity of (x + 2)² is 2 )
if (x-3)² = 0 ⇒ x² = 3, 3, 3 ( since, multiplicity of (x - 3)² is 2 )
In this part polynomial equation of least degree has -2, -2, 3, and 3 as four of its roots
Hence, the correct option is D) (x+2)²(x-3)² = 0
