One more question, I think! Find all the real zeros of the function?

Question

f(x)= x^3 - 5x^2 + 5x -1

Answer 1

Rational roots will be all the factors of the last coefficient (-1) divided by all the factors of the coefficient on x^3 (1).

That leads to the following set of possible roots:
{-1, 1}

Trying x = 1, we find that results in f(x) = 0. So 1 is one root. Now you can factor out (x - 1) using synthetic division:

(x - 1)(x² - 4x + 1)

You can use the quadratic formula on the second part and you'll get the other two roots.

a = 1, b = -4, c = 1

x = ( 4 ± sqrt( 16 - 4 ) ) / 2
x = ( 4 ± sqrt( 12 ) ) / 2
x = ( 4 ± 2 sqrt( 3 ) ) / 2
x = 2 ± sqrt( 3 )

The roots are:
x = 1
x = 2 + √3
x = 2 - √3

Answer 2

Since 1 is a zero of this, we may factor as follows:

f(x) = (x-1)(x^2-5x+1) = 0

So, x=1 is one zero.

The other two may be obtained by the quadratic formula: 3.732050808 and .2679491924

Answer 3

(x-1)(x^2-4x+1)=0
x^2-4x+1
x=4+-sqrt12/2=2+-sqrt3
the zeros are 1, 2+-sqrt3