solve using synthetic substitution

Question

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

 

 

1.  using synthetic substitution, which of the possible rational roots is actually a root of the equation

 

 

 

2.  find the irrational roots of the equation.  (use the quadratic formula to solve the dpressed equation)

 

 

thanks

 

Answer 1

x³ - 4x² + 2x + 1 =0

1. using synthetic substitution, which of the possible rational roots is actually a root of the equation

2. find the irrational roots of the equation. (use the quadratic formula to solve the depressed equation)

 

The Rational Root Theorem tells us that the rational roots will be factors of ± c/a, where c is the constant term in the equation and a is the leading coefficient (x³ term):

 

c/a --> ± 1/1  --> +1, -1

 

Try synthetic division for +1 and -1:

 

1| 1  -4  2  1             -1| 1  -4   2   1

          1 -3 -1                    -1   5  -7

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

    1  -3 -1 0                  1  -5   7   -6

 

So +1 works but -1 does not (because it has a remainder, -6).  Using the quotient from the synthetic division using +1, our factoring so far is:

 

(x-1)(x²-3x-1) = 0

 

x²-3x-1 (the depressed equation) cannot be factored, so use the quadratic formula to find its zeros:

 

x = (3/2) ± (1/2)*[(-3)²-(4)(1)(-1)]^1/2 = (3±√13)/2

 

The roots are 1, (3+√13)/2, (3-√13)/2

 

The last two roots are irrational roots.  The first two roots are positive, the last root is negative.