SOLUTION: Find the roots of the polynomial equation. x^4 - 5x^3 + 11x^2 - 25x + 30 = 0; 2,3

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Find the roots of the polynomial equation. x^4 - 5x^3 + 11x^2 - 25x + 30 = 0; 2,3      Log On


   



Question 196689: Find the roots of the polynomial equation.
x^4 - 5x^3 + 11x^2 - 25x + 30 = 0; 2,3

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
Find the roots of the polynomial equation.
x%5E4+-+5x%5E3+%2B+11x%5E2+-+25x+%2B+30+=+0; 2,3

Since you are given that 2 is a root, we
do this synthetic division

2 | 1   -5   11  -25  30
  |      2   -6   10 -30 
   ---------------------
    1   -3    5  -15   0

So now we have factored the original polynomial as

%28x-2%29%28x%5E3-3x%5E2%2B5x-15%29=0

Now we factor the cubic polynomial in 
the 2nd parentheses, using the fact that 3 is
a root of the original, so it must be a 
root of the cubic polynomial in the 2nd
parentheses:

3 | 1 -3  5 -15
  |    3  0  15 
   ---------------
    1  0  5   0

So now we have factored the original polynomial as

%28x-2%29%28x-3%29%28x%5E2%2B0x%2B5%29=0

or

%28x-2%29%28x-3%29%28x%5E2%2B5%29=0

Setting the first two parentheses = 0 gives
the given two roots, 2 and 3.  To find the
other roots, we set the third factor = 0

x%5E2%2B5=0

Add -5 to both sides

x%5E2=-5

Take the square roots of both sides:

sqrt%28x%5E2%29=%22+%22%2B-sqrt%28-5%29

x+=+%22+%22%2B-sqrt%285%29i

So the roots are 2, 3, sqrt%285%29i, and -sqrt%285%29i

Edwin