Question

If one of the zeroes of cubic polynomial x^3+ax^2+bx+c is 2 then the product of other two zeroes is:

a)c/2
b)-2/c
c)-c/2
d)2/c

Answer 1

The answer is d) 2/c.

Explanation :

If one of the zeroes of cubic polynomial x^3+ax^2+bx+c is 2, then the product of the other two zeroes can be found by dividing the polynomial by (x-2), which gives:

x^3+ax^2+bx+c = (x-2)(x^2+(a+2)x+(c-2a))

We can solve the quadratic equation x^2+(a+2)x+(c-2a) to find the other two zeroes. By Vieta's formulas, the product of the roots of a quadratic equation ax^2+bx+c=0 is c/a. Therefore, the product of the other two zeroes is:

pq = (c-2a)/1 * 1/(a+2) = (c-2a)/(a+2)

Substituting the value of c/2a from the cubic polynomial gives:

pq = (c-2a)/(a+2) = (2c-4a)/(2a+4) = -2(c/2a) + 2

Therefore, the answer is d) 2/c.