# Given x^3-4x^2+2x+1=0. There are 2 possible positive roots 1 possible neg. root and 1 possible rational root a. Using synthetic substitution, which of the possible rational roots is actually...

Given x^3-4x^2+2x+1=0.

There are 2 possible positive roots 1 possible neg. root and 1 possible rational root

a. Using synthetic substitution, which of the possible rational roots is actually a root of the equation? explain

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x^3-4x^2+2x+1=0

Let us substitute with x=1

==> 1-4+2+1 =0

Then x1=1 is on of the roots for the equation, then (x-1) is a factor;

==> f(x) = (x-1)Q(x)

To determine Q(x) we will divide f(x) by (x-1)

==> f(x) = (x-1)(x^2-3x-1)

Now let us calculate the roots:

x1=1 (rational root)

x2= [3+ sqrt(9-5)]/2= (3+sqrt5)/2 (positive irrational)

x3= (3-sqrt5)/2 (positive irrational)

f(x) = x^3-4x^2+2x+1 = 0. To find the roots.

Solution:

Obviously the rational root is 1 , as f(1) = 1-4+2+1= 0.

So x^3-4x^2+2x+1 = (x-1)Q(x). We shall find Q(x) by division:

x-1)x^3-4x^2+2x+1(x^2-3x-1 is Q(x) by division.

x^3-x^2

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

-3x^2+ 2x

-3x^2+ 3x

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

-x+1

-x+1

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

0

So Q(x) = x^2-3x-1.

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

So the other two roots are determined from the quadratic x^2-3x-1 = 0.

So x1 = {-(-3)+sqrt((-3)^2-4*1)} = (3+sqrt5)/2 is the positive irrational root.

x2 = (3-sqrt5)/2 is another positive irrational root.

x= 1 is the positive rational root.