Question

factorize X^3 - 3x - 1 + 3/x - 1/x^3

Answer 1

Let f(x) = x^3 + 1/x^3 - 2 - 3x + 1/x

Put x =1, We get f(1) = -2

Now put x =2, We get f(2) = 5/8.

Now f(1) < 0 i.e a negative quantity

and f(2) > 0 i.e a positive quantity.

It clearly means that between 1 and 2, there is a value which makes the function as Zero.

This value is nothing but the root of the equation lying between 1 & 2. — ——-(1)

Now follow simple bisection method, which is as follows-

If x1 and x2 are 2 assumed values for a function where f(x1)<0 and f(x2)>0,

then the next assumption will be lying between the two values x1 and x2.

This value will be given by x3 = (x1 + x2)/2. ————————————(2)

In our case, x1 =1, x2 = 2, so becomes x3= (1+2)/2 = 1.5

Put the value of x3 in the function and obtain f(x3).

Here f(x3) = -0.45.

Now f(x2) >0 and f(x3) < 0. ——————————-(3)

Which means our root lies between 1.5 and 2, following eqn(1).

Repeating same procedure as Eqn(2),

x4 = (x3+x2)/2 = (1.5+2)/2 = 1.75 ——————-(4)

f(x4) = f(1.75) = -1.15. which is again a negative quantity.

Repeat the process until same values of x is obtained. When even after iteration, same value is obtained, that value of x becomes root of the equation. U can obtain it within 7 iterations normally.

Consider that root as dominant root of the equation. (Or First root).

Suppose we obtain the value as p.

Now divide the polynomial by this root i.e divide by (x-p).

Perform long division method.

[x^3 + 1/x^3 - 2 - 3x + 1/x] ÷ (x-p) ———— (5)

U can obtain other factors as well. My assumption is that the value of p would be around 1.9.

The benefit of using this method is that it will provide u with the real root. By long division method u can obtain other factors also, containing real/complex roots.