Find The Rational Zero Test && Then The Linear Factor Of h(x)=x^3-3x^2+4x-2Please Do The Step By Step. Like The Rational Zero && If Use Synthetic...

f(x) =
x^3-3x^2x+4x-2

Rational zero test is a way of finding the
zeros of a polynomial by looking at the factors of the  constant term divided by the
coefficient of the laeding term of the polymials.; and we can try around + or - factors
also.

In this case , constant term = -2 and leading term is
x^3 and coefficient of x^3 is 1. So, constant term / coeff of x^3 = -2. The factors of
-2 are : -2 , -1 ,2 and 2. Try  which of them make f(x) = 0, by
substitution.

The sum of the coefficients of powers of x is
zero. So f(1) = 1^3-3*1^2+4*1-2 = 0. Thus by rational zero test 1 is zero the polynomial
f(x).

Therefore x-1 is a factor by remainder therem. So we
divide f(x) by (x-1).

x-1)x^3-3x^2+4x-2 (
x^2

     
x^3-x^2

-----------------

       x-1)-2x^2
+4x (  x^2 -2x

             -2x^2
+2x

-------------------------------           

                   x-1) 2x-2
(
x^2-2x(x^2-2x+2

                          2x-2

---------------------------------------

                             
0

So  f(x) /x-1 =
x^2-2x+2.

Therefore  f(x) = x^3-3x^2+4x-2 = (
x-1)(x^2-2x+2).

Bur x^2-2x+2  has no real factors ( no 
real zeros of the polynomial f(x) ),  as the discriminant  of x^2-2x+2 is (-2)^2 - 4*1*2
= 4-8 =  -4  < 0.

So the only linear factor of
x^3-3x^2+4x+2 is (x-1).

Hope this
helps.