Question

(04.05 LC)

Answer 1

correct answer is option d.

Step-by-step explanation:

Here, we have The following expression to check Which of the following is a polynomial with roots: −square root of 3 , square root of 3, and 2 :

a. x3 + 3x2 − 5x − 15

For $-\sqrt{3}$ ,  $x^{3} + 3x^{2} -5x -15$  ⇒ $(-\sqrt{3}) ^{3} + 3(-\sqrt{3} )^{2} -5(-\sqrt{3} ) -15$ $\neq 0$ . So, this isn't the one. We move on further!

b. x3 + 2x2 − 3x − 6

For $-\sqrt{3}$ ,  $x^{3} + 2x^{2} -3x -6$  ⇒ $(-\sqrt{3}) ^{3} + 2(-\sqrt{3} )^{2} -3(-\sqrt{3} ) -6$  = 0 .

For $\sqrt{3}$ ,  $x^{3} + 2x^{2} -3x -6$  ⇒ $(\sqrt{3}) ^{3} + 2(\sqrt{3} )^{2} -3(\sqrt{3} ) -6$ = 0.

For 2 ,  $x^{3} + 2x^{2} -3x -6$  ⇒ $(2) ^{3} + 2(2)^{2} -3(2 ) -6$$\neq 0$ . So, this isn't the one. We move on further!

c. x3 − 3x2 − 5x + 15

For $-\sqrt{3}$ ,  $x^{3}- 3x^{2} -5x -15$  ⇒ $(-\sqrt{3}) ^{3}- 3(-\sqrt{3} )^{2} -5(-\sqrt{3} ) -15$ $\neq 0$ . So, this isn't the one. We move on further!

d. x3 − 2x2 − 3x + 6

For $-\sqrt{3}$ ,  $x^{3} - 2x^{2} -3x +6$  ⇒ $(-\sqrt{3}) ^{3} - 2(-\sqrt{3} )^{2} -3(-\sqrt{3} ) +6$  = 0 .

For $\sqrt{3}$ ,  $x^{3} - 2x^{2} -3x +6$  ⇒ $(\sqrt{3}) ^{3} -2(\sqrt{3} )^{2} -3(\sqrt{3} ) +6$ = 0.

For 2 ,  $x^{3} - 2x^{2} -3x +6$  ⇒ $(2) ^{3} - 2(2)^{2} -3(2 ) +6$ = 0.

Therefore, correct answer is option d.