Question

The zeros of the polynomial 6 y ( y + 2 )( y – 3 )

0 , 2 , -3

0 , - 2 , 3

- 2 , 3

2 , -3​

Answer 1

EXPLANATION.

Zeroes of the polynomial.

⇒ 6y(y + 2)(y - 3).

As we know that,

We can write equation as,

⇒ 6y(y² - 3y + 2y - 6).

⇒ 6y(y² - y - 6).

⇒ 6y³ - 6y² - 36y.

Take common 6y from the equation, we get.

⇒ 6y(y² - y - 6).

If we factorizes the equation, we get.

⇒ y² - y - 6.

Factorizes the equation into middle term splits, we get.

⇒ y² - 3y + 2y - 6.

⇒ y(y - 3) + 2(y - 3).

⇒ (y + 2)(y - 3).

Now, we can write equation as,

⇒ 6y(y + 2)(y - 3).

This equations is equal to 0.

⇒ 6y(y + 2)(y - 3) = 0.

⇒ 6y = 0.

⇒ y = 0. - - - - - (1).

⇒ (y + 2) = 0.

⇒ y = - 2. - - - - - (2).

⇒ (y - 3) = 0.

⇒ y = 3. - - - - - (3).

Zeroes of the polynomial are = 0, - 2, 3.

Option [B] is correct answer.

                                                                                                                       

MORE INFORMATION.

Conjugate roots.

(1) = If D < 0.

One roots = α + iβ.

Other roots = α - iβ.

(2) = If D > 0.

One roots = α + √β.

Other roots = α - √β.

Answer 2

Note -Here I solve this problem using two methods

•First method

Given polynomial

$6 y ( y + 2 )( y – 3 )$

Zeroes of polynomial

$6y +(y + 2)(y - 3) = 0$

$i.e \: \: y = \:0 \: or \: \: y +2 = 0 \\ \\ or \: \: y -3 =0$

$y = 0 \: \: or \: y = - 2 \: \: \\ \\ y = 3$

• The zeroes of the polynomial is

$0, - 2,3$

_______________________

•Second method

f (y) =6y +(y + 2)(y - 3)

f (y) =0

$6y +(y + 2)(y - 3) = 0$

• 6 y =0 ,y = 0

• y+2 =0,y = -2

• y-3 =0,y=3

So answer will be 0,-2 and 3

Refer the attachment

• For detail information