Which polynomial equation of least degree has -2, -2, 3, and 3 as four of its roots? (x + 2)(x - 3) = 0 (x - 2)-2(x + 3)3 = 0 (x-2 + 2)(x2- 3) = 0 (x + 2)2(x - 3)2 = 0

last option

(x+2)²(x-3)²=0

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

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: