Question

Find zeroes of the polynomial x³–2 .

This is very important Question for my maths test. So, Anyone plz answer me this question...plz​

Answer 1

Simple, since x³–2's zero must be found:

             p(x) = 0

           x³–2 =  0

               x³ = 2

              x  = $\sqrt[3]{2\\}$

Answer 2

Answer:

${x}^{3} - (a + b + c) {x}^{2} + (ab + bc + ac)x - (abc) = 0$

where a,b,c are zeroes of polynomial

from above

a+b+c = 0-----1

ab+bc+ac = 0------2

abc = 2-----3

a+b = -c----- from 1

ab+c(a+b) = 0-----from 2

$ab - {c}^{2} = 0$

c^2 = ab------4

from 3 and 4

c^3 = 2

$c = \sqrt[3]{2}$

a+b = -c

ab=c^2

we can find a, b