Question

Q.22 Find the zeroes of the polynomial , If it is given that the product of its two zeroes is 12.​

Answer 1

Step-by-step explanation:

ANSWER

x

3

−5x

2

−2x+24=0

(x+2)

x+2

x

3

−5x

2

−2x+24

(x+2)(x

2

−7x+12)

(x+2)(x

2

−4x−3x−12)

(x+2)(x(x−4)−3(x−4))

(x+2)(x−4)(x−3)

x=−2,3,4

Given, products of 2 zeros is 12⇒(3×4)

Therefore the required solution is 3 and 4

Answer 2

Correct question

Step-by-step explanation:

Question- Find the zeroes of the polynomial x

^3

−5x^2

−2x+24, it is given that the product of its two zeroes is 12.

Solve- x^3

−5x^2

−2x+24=0

(x+2) x^3 - 5x^2 - 2x + 24 / (x+2)

(x+2)(x^2

−7x+12)

​(x+2)(x^2

−4x−3x−12)

(x+2){x(x−4)−3(x−4)}

(x+2)(x−4)(x−3)

x=−2,3,4

Given, products of 2 zeros is 12⇒(3×4)

Therefore the required solution is 3 and 4.