#### Solution:

Let,

F(x)=x^3-2x-5=0

Now,

Put x=0:- 0^(3)-2*0-5= -5 (-ve)

Put x=1:- 1^(3)-2*1-5= -6 (-ve)

Put x=2:- 2^(3)-2*2-5= -1 (-ve)

Put x=3:- 3^(3)-2*3-5= 16 (+v)

Therefore the root lie between 2 and 3:

**1 ^{st} stage:-**

Hence,

x0= 2+3/2= 5/2

x0=2.5

now,

f(x0)=2.5^(3)-2*2.5-5

f(x0)=5.625

So, the roots lie between 2 and x0(which is 2.5):

**2 ^{nd} stage:-**

Hence,

x1= 2+2.5/2=4.5/2

x1=2.25

now,

f(x1)=2.25^(3)-2*2.25-5

f(x1)=1.89

So, the roots lie between 2 and x1(which is 2.25):

**3 ^{rd} stage:-**

Hence,

x2= 2+2.25/2= 4.25/2

x2= 2.125

now,

f(x2)= 2.125^(3) – 2*2.125 – 5=

f(x2)=0.346

So, the roots lie between 2 and x2(which is 2.125):

4^{th} stage:-

Hence,

x3= 2+2.125/2=2.0625

x3=2.0625

now,

f(x3)=2.0625^(3) – 2*2.0625 – 5=

f(x3)= -0.351

here, the roots lie between x2 and x3:

**5 ^{th} stage:-**

Hence,

x4= 2.125+2.0625/2

x4= 2.093

now,

f(x4)= 2.093^(3) – 2*2.093 – 5=

f(x4)= -0.017