Question

please help in solving this question

Answer 1

Consider the cubic equation,
x³ + ax² + bx + c.
Given, one of the root to be = -1
As, the total number of roots of this equation will be 3.
Assume, roots be as m, n and -1.

For cubic equation,
Product of roots = -c [As per the equation ]
m * n * -1 = -c
So, product of other two zeros or root = c.

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Answer 2

Given Equation is x^3 + ax^2 + bx + c .

Let the zeroes of the polynomial be p,q,r

Given that one of the zeroes of the polynomial is -1.

= > p + q + r = -b/a

 -1 + q + r= -(a)/1

-1 + q + r = -a 

 q + r= 1 - a ------- (1)

= > pq + qr + rp = c/a

     (-1)q + qr + r(-1) = b/1

     -q + qr - r = b 

     qr = q + r + b   --------------------- (2)

= > pqr = -d/a

   (-1)qr = -c 

   c = qr   ------ (3)

Substitute (1) in (2), we get

= > qr = 1 - a + b.

Therefore the product of other two zeroes is 1 - a + b.

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