Question

If the product of zeros of the polynomial $f(x)=ax^{3}-6x^{2}+11x-6$ is 4,then a=
(a) $\frac{3}{2}$
(b) $-\frac{3}{2}$
(c) $\frac{2}{3}$
(d) $-\frac{2}{3}$

Answer 1

Solution :  

The correct option is (a ) : 3/2

Let α,β,γ are the three Zeroes of the cubic  polynomial.

Given : f(x) = ax³ - 6x² + 11x - 6 and product of zeroes (αβγ) = 4

On comparing with ax³ + bx² + cx + d ,

a = a , b = -6 , c = 11 , d = -6

product of zeroes of cubic polynomial = - constant term /coefficient of x³

αβγ = −d/a

4 = -(-6)/a

4 = 6/a  

4a = 6

a= 6/4

a = 3/2

Hence, the value of ‘a’ is 3/2

HOPE THIS ANSWER WILL HELP YOU..

Answer 2

HEY MATE!!!

HERE IS YOUR ANSWER!!!

GIVEN THAT F (x)=ax^3-6x^2+11x-6 and product of zeroes (aby)=4

ON COMPARING WITH ax^3+bx^2+cx+d,

a=a,b=-6,c=11,d=-6

PRODUCT OF ZEROES OF CUBIC POLYNOMIAL =CONSTANT TERM /COEFFICIENT OF X^3

aby =-d/a

4=-(-6)/a

4=6/a

a=6/4

a=3/2

HENCE,A VALUE IS 3/2

HOPE IT HELPS YOU.....

PLZ MARK BRAINLIEST♥♥♥