Question

find all Integral zeroes of x^3 - 3x^2 - 2x +6​

Answer 1

first, put value of x as 1 then, equation will be 1-3-2+6=2

now, put the value of x as -1 then, eq^n will be -1-3+2+6= 4

put value of x as 2 then, equation will be 8-12-4+6=-2

put value of x as -2 then, equation will be -8-4+4+6=-2

put value of x as 3 then, equation will be 27-27-6+6=0

therefore the value of x=3

similary,other two values are √2 and -√2

Answer 2

Answer:

Zeroes = 3, -√2,√2

Step-by-step explanation:

p(x) = x^3 - 3x^2 - 2x + 6

By splitting the middle term and taking common term,we get

x^2 (x-3) -2(x-3)

(x^2 - 2)(x-3)

For the values of x,

CASE I _

(x^2 - 2) = 0

x^2 = 2

x = √2

x = +√2,-√2

CASE II_

(x-3) = 0

x = 3

Hence all the zeroes of the given polynomial are :

3 , +√2 , -√2