Mathematics, 22.06.2019 00:30 drewje12

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

Answers: 2

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Answers

Answer from: girlwholikesanime

last option

(x+2)²(x-3)²=0

Answer from: Fetty1738

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

Answer from: Quest

d. bc = 12 and ef = 16

step-by-step explanation:

we are given that δabc : δdef such that $\frac{ab}{de}=\frac{bc}{ef}$ .

also, it is given that $\frac{ab}{de}=\frac{3}{4}$

so, from the options, we need that $\frac{bc}{ef}=\frac{3}{4}$ .

a. $\frac{bc}{ef}=\frac{6}{9}=\frac{2}{3}$

b. $\frac{bc}{ef}=\frac{4}{6}=\frac{2}{3}$

c. $\frac{bc}{ef}=\frac{8}{12}=\frac{2}{3}$

d. $\frac{bc}{ef}=\frac{12}{16}=\frac{3}{4}$

as, we can see that only option d gives the same ratio as $\frac{ab}{de}=\frac{3}{4}$ i.e. $\frac{bc}{ef}=\frac{3}{4}$ .

hence, the possible lengths are bc = 12 and ef = 16.

Answer from: Quest

this is a very good question

step-by-step explanation:

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