Question

if one of the zeroes of the cubic polynomial x cube + ax square + bx + c is -1 then find the product of the other two zeroes

Answer 1

Answer:

$x^2 + (b - c)x + c \ \ \ OR \ \ \ x^2 + (a - 1)x + c$

Step-by-step explanation:

$Very nice question. \\ \\ Let \ p(x) = x^3 + ax^2 + bx + c \\ \\ p(-1) \\ \\ = (-1)^3 + a(-1)^2 + b(-1) + c = 0 \\ \\ = -1 + a(1) + b(-1) + c = 0 \\ \\ = -1 + a - b + c = 0 \ \ \to \ \ (1) \\ \\ a - 1 = b - c \ \ \to \ \ (2)$

$If \ -1\ is a zero of \ p(x),\ then \ x + 1\ is a factor of \ p(x). \\ \\ Let's divide \ p(x)\ by \ x + 1. \\ \\ x^3 + ax^2 + bx + c \\ \\ = x^3 + x^2 + (a - 1)x^2 + (a - 1)x + (b - a + 1)x + (b - a + 1) - (b - a + 1) + c \\ \\ = x^2(x + 1) + x(a - 1)(x + 1) + (b - a + 1)(x + 1) - b + a - 1 + c \\ \\ = x^2(x + 1) + x(a - 1)(x + 1) + (b - a + 1)(x + 1) + (- b + a - 1 + c) \\ \\ = x^2(x + 1) + x(a - 1)(x + 1) + (b - a + 1)(x + 1) + (- 1 + a - b + c)$

$Substituting \ (1) = - 1 + a - b + c = 0, \\ \\ x^2(x + 1) + x(a - 1)(x + 1) + (b - a + 1)(x + 1) + 0 \\ \\ = x^2(x + 1) + x(a - 1)(x + 1) + (b - a + 1)(x + 1) \\ \\ = (x + 1)(x^2 + (a - 1)x + b - a + 1) \\ \\ = (x + 1)(x^2 + (a - 1)x + b - (a - 1)) \\ \\ Substituting \ (2) = a - 1 = b - c, \\ \\ (x + 1)(x^2 + (b - c)x + b - (b - c)) \\ \\ = (x + 1)(x^2 + (b - c)x + b - b + c) \\ \\ = (x + 1)(x^2 + (b - c)x + c)$

$\\ \\ \\ \therefore\ x^2 + (b - c)x + c\ is the product of other two zeroes. \\ \\ x^2 + (a - 1)x + c\ can also be got by substituting (2).$

$\\ \\ \\ Hope this may be helpful. \\ \\ Please mark my answer as the \ \bold{brainliest}\ if this may be helpful. \\ \\ Thank you. Have a nice day. \\ \\ \\ \#adithyasajeevan$

Answer 2

Answer:

the answer is " b - a + 1 "

hope this is helpful.