SOLUTION: f(x)=x^5-6x^4+11x^3-2x^2-12x+8 has (x-2) as a factor determine which one of the following is true about the set of roots for f(x) a) 2 is a single root with 4 more real and

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: f(x)=x^5-6x^4+11x^3-2x^2-12x+8 has (x-2) as a factor determine which one of the following is true about the set of roots for f(x) a) 2 is a single root with 4 more real and      Log On


   

Question 765222: f(x)=x^5-6x^4+11x^3-2x^2-12x+8 has (x-2) as a factor
determine which one of the following is true about the set of roots for f(x)
a) 2 is a single root with 4 more real and 0 complex roots
b) 2 is a single root with 2 more real and 2 complex roots
c) 2 is a single root with no real roots and 4 complex roots
d) 2 is a double root with 3 more real roots and 0 complex roots
e) 2 is a double root with 1 more real root and 2 complex roots
f) 2 is a triple root with no real roots and 2 complex roots
g) 2 is a triple root with 2 more real and 0 complex roots
please help me with this. i still do not understand the concepts of real, complex roots
Found 2 solutions by MathLover1, Edwin McCravy:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

f%28x%29=x%5E5-6x%5E4%2B11x%5E3-2x%5E2-12x%2B8 ....write 11x%5E3 as -x%5E3%2B12x%5E3
and -2x%5E2 as 6x%5E2-8x%5E2
f%28x%29=x%5E5-x%5E3-6x%5E4%2B6x%5E2%2B12x%5E3-12x-8x%5E2%2B8...group
f%28x%29=%28x%5E5-x%5E3%29-%286x%5E4-6x%5E2%29%2B%2812x%5E3-12x%29-%288x%5E2-8%29...factor
f%28x%29=x%5E3%28x%5E2-1%29-6x%5E2%28x%5E2-1%29%2B12x%28x%5E2-1%29-8%28x%5E2-1%29
f%28x%29=%28x%5E3-6x%5E2%2B12x-8%29+%28x%5E2-1%29
f%28x%29=%28x-2%29%5E3+%28x%5E2-1%29
f%28x%29=%28x-2%29%28x-2%29%28x-2%29%28x-1%29%28x%2B1%29

solutions:
if %28x-2%29=0 => x=2...as you can see above, we have three times this solution, means x=2 is a triple root
if %28x-1%29=0 => x=1
if %28x%2B1%29=0 => x=-1
so, we have three real roots and no complex roots
and answer is: g) 2 is a triple root with 2 more real and 0 complex roots

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Complex roots are like 2+3i, -4+7i, -5-7i, 6-3i, i, 3i, etc.
They are roots with i's.  Real root have no i terms.

Since x5-6x4+11x3-2x2-12x+8 has (x-2) as a factor

2|1 -6 11 -2 -12  8
 |   2 -8  6   8 -8
  1 -4  3  4  -4  0

So we have factored f(x) as

f(x) = (x-2)(x4-4x3+3x2+4x-4) = 0

Since 2 is a factor of the last term (in absolute value),
it may be that 2 is a root more than once, so we try 2 again
on x4-4x3+3x2+4x-4

2|1 -4  3  4 -4
 |   2 -4 -2  4
  1 -2 -1  2  0

So we have factored f(x) as

f(x) = (x-2)(x-2)(x3-2x2-x+2) = 0

We can factor x3-2x2-x+2 by grouping:
             x2(x-2)-1(x-2)
             (x-2)(x2-1)
             (x-2)(x-1)(x+1)

So the final factorization is

f(x) = (x-2)(x-2)(x-2)(x-1)(x+1)

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

So 2 is a root of multiplicity 3 (a triple root, 
1 is a single root, and -1 is a single root.

So g is the correct answer.

Edwin