Question

Find zeroes of the polynomial p(x)= x^3+4x^2+x-6 If it is given that the product of its two zeroes is 6

Answer 1

let the zeros be @,# and $

therefore @#$=-d/a

6$=-(-6/1)

$=1

and hence x-1 is a factor

(x-1)(x²-5x+6)
(x-1)(x-2)(x-3)

hence roots are 2,3 and 1

Answer 2

Answer:

$\alpha \beta \gamma = - d \div a$

Step-by-step explanation:

1. -(-6)/1
2. =6
3. product of the zeros of p(X)=x^3+4x^2+x-6=6