SOLUTION: Hi, can someone please help me find all the zeros in order using synthetic division and quadratic formula for equation: P(x)= 2x^9 - x^8 - 16x^7 + 17x^6 - 6x^5 - 5x^4 + 68x^3 - 77

Algebra ->  Test -> SOLUTION: Hi, can someone please help me find all the zeros in order using synthetic division and quadratic formula for equation: P(x)= 2x^9 - x^8 - 16x^7 + 17x^6 - 6x^5 - 5x^4 + 68x^3 - 77      Log On


   

Question 438947: Hi, can someone please help me find all the zeros in order using synthetic division and quadratic formula for equation:
P(x)= 2x^9 - x^8 - 16x^7 + 17x^6 - 6x^5 - 5x^4 + 68x^3 - 77x^2 + 18
Thankyou!...
Found 2 solutions by Edwin McCravy, richard1234:
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Hi, can someone please help me find all the zeros in order using synthetic division and quadratic formula for equation:

P(x)= 2x⁹ - x⁸ - 16x⁷ + 17x⁶ - 6x⁵ - 5x⁴+ 68x³ - 77x² + 18

Try 1 as a zero

     1|2   -1   -16   17   -6   -5   68   -77    0   18
      |     2     1  -15    2   -4   -9    59  -18  -18 
       2    1   -15    2   -4   -9   59   -18  -18    0

That factors P(x) as

P(x) = (x - 1)(2x⁸ + x⁷ - 15x⁶ + 2x⁵ - 4x⁴- 9x³ - 59x² - 18x - 18)

Try 1 as a zero again

     1|2    1   -15    2   -4   -9   59   -18  -18
      |     2     3  -12  -10  -14  -23    36   18  
       2    3   -12  -10  -14  -23   36    18    0

That factors P(x) as

P(x) = (x - 1)(x - 1)(2x⁷ + 3x⁶ - 12x⁵ - 10x⁴- 14x³ - 23x² + 36x + 18)

Try 1 as a zero again

     1|2    3   -12  -10  -14  -23   36   18 
      |     2     5   -7  -17  -31  -54  -18 
       2    5    -7  -17  -31  -54  -18    0

That factors P(x) as

P(x) = (x - 1)(x - 1)(x - 1)(2x⁶ + 5x⁵ - 7x⁴- 17x³ - 31x² - 54x - 18)

Try -3 as a zero 

    -3|2    5   -7  -17  -31  -54  -18 
      |    -6    3   12   15   48   18 
       2   -1   -4   -5  -16   -6    0

That factors P(x) as

P(x) = (x - 1)(x - 1)(x - 1)(x + 3)(2x⁵ - x⁴- 4x³ - 5x² - 16x - 6)

Try -3/2 as a zero 

  -3/2|2   -1   -4  -5  -16  -6   
      |    -3    6  -3   12   6
       2   -4    2  -8   -4   0

That factors P(x) as

P(x) = (x - 1)(x - 1)(x - 1)(x + 3)(x + 3/2)(2x⁴- 4x³ + 2x² - 8x - 4)

We can factor 2 out of that last parentheses:

P(x) = (x - 1)(x - 1)(x - 1)(x + 3)(x + 3/2)(2)(x⁴- 2x³ + x² - 4x - 2)

P(x) = 2(x - 1)(x - 1)(x - 1)(x + 3)(x + 3/2)(x⁴- 2x³ + x² - 4x - 2)


Now we must factor the last polynomial:

x⁴- 2x³ + x² - 4x - 2

Rearrange as

x⁴+ x² - 2 - 2x³ - 4x

Factor -2x out of the last two terms:

(x⁴+ x² - 2) - 2x(x² + 2)

Factor the first parentheses:

(x² + 2)(x² - 1) - 2x(x² + 2)

Factor out (x² + 2)

(x² + 2)[(x² - 1) - 2x]

(x² + 2)(x² - 1 - 2x)

(x² + 2)(x² - 2x - 1)

Now we have factored P(x) as

P(x) = 2(x - 1)(x - 1)(x - 1)(x + 3)(x + 3/2)(x² + 2)(x² - 2x - 1)

P(x) = 2(x - 1)³(x + 3)(x + 3/2)(x² + 2)(x² - 2x - 1)

We set x² + 2 = 0
           x² = -2 _
            x = ±i√2

We set x² - 2x - 1 = 0

    
   
x+=+%282+%2B-+sqrt%284%2B4%29%29%2F2+

x+=+%282+%2B-+sqrt%288%29%29%2F2

x+=+%282+%2B-+sqrt%284%2A2%29%29%2F2+

x+=+%282+%2B-+2sqrt%282%29%29%2F2+

x+=+%282%281+%2B-+sqrt%282%29%29%29%2F2+

x+=+%28cross%282%29%281+%2B-+sqrt%282%29%29%29%2Fcross%282%29+
         _
x = 1 ± √2

So, 1 is a zero of multiplicity 3,

3/2 is a zero
  _         _
i√2  and -i√2  are zeros
     _           _
1 + √2  and 1 - √2  are zeros.

That's really a humdinger of a problem!

Edwin

Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
In all aspects, ninth degree polynomials are quite ugly. The fact is, when you or I see the problem, we have absolutely no idea what any potential zeros could be. You could try the rational root test to knock off a few zeros, but it would take a very long time. In fact, the Abel-Ruffini theorem states that there is no general formula for a polynomial of degree 5 or higher.

Here is what WolframAlpha has to say:
http://www.wolframalpha.com/input/?i=2x^9+-+x^8+-+16x^7+%2B+17x^6+-+6x^5+-+5x^4+%2B+68x^3+-+77x^2+%2B+18+