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Question

2x⁵ +11x⁴+10x³-20x²-12x+9=0

Answer 1

2x⁵ + 11x⁴ + 10x³ - 20x² - 12x + 9 = 0

p: ±1, ±3, ±9

q: ±1, ±2

p/q: ±1, ±3, ±9, ±1/2, ±3/2, ±9/2

 

-3] 2   11   10   -20   -12   9

           -6   -15   15     15  -9

      2     5    -5    -5       3   0

 

2x⁴ + 5x³ - 5x²  -5x + 3 = 0

p: ±1, ±3

q: ±1, ±2

p/q: ±1, ±1/2, ±3, ±3/2

 

-1] 2   5   -5   -5   3

          -2   -3    8  -3 

      2   3   -8    3    0

 

2x³ + 3x - 8x + 3 = 0

p: ±1, ±3

q: ±1, ±2

p/q: ±1, ±3, ±1/2, ±3/2

 

1]   2   3   -8   3

            2    5  -3

       2   5   -3   0

 

2x² + 5x - 3 = 0

(2x - 1)(x + 3) = 0

2x - 1 = 0   or  x + 3 = 0

2x = 1              x = -3

x = 1/2

 

x = -3, multiplicity of 2, x = -1, x = 1, x = 1/2

 

 

 

 

Answer 2

Rational root theorem:

 

P = { factors of 9} = { +-1, +-3, +- 9}

 

Q = {factors of 2} = {+-1, +-2}

 

{ possible rational roots } = P/Q = { +-1, +-3, +-9, +-1/2, +-3/2, +-9/2}

 

each one must be tested, one at a time. Plug each one into the polynomial

to see which one, if any, makes the polynomial zero.

 

It turns out that -1 works.

 

2(-1)^5 + 11(-1)^4 + 10(-1)^3 - 20(-1)^2 - 12(-1) + 9 =

2(-1) + 11(1) + 10(-1) - 20(1) - 12(-1) + 9 =   <--- (-1) to even power is 1 while (-1) to odd power is -1

 

-2 + 11 - 10 - 20 + 12 + 9 =  9 - 10 - 20 + 12 + 9 = -1 - 20 + 12 + 9 = -21 + 12 +9 = -9 + 9 = 0

 

So -1 is a solution. Remember that if x=-1 is the solution then x+1 is a factor.

 

Synthetic division:

 

-1 |    2   11   10   -20   -12    9

               -2   -9   -1     21    -9

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

          2   9     1   -21    9     0

 

So the equation is now ( x+1)(2x^4 + 9x^3 + x^2 - 21x + 9) = 0

 

 

Now we have to repeat the entire procedure for 2x^4 + 9x^3 + x^2 - 21x + 9.

Notice that the constant term is 9 and the leading coefficient is 2, which  is the

same as the original polynomial. So by coincidence and luck, it is the same P/Q.

 

P/Q = { +-1, +-3, +-9, +-1/2, +-3/2, +-9/2}

 

This time, x=1 works: 2(1)^4 + 9(1)^3 + (1)^2 - 21(1) + 9 = 2(1) + 9(1) + 1 - 21(1) + 9 =

                                                                                          = 2 + 9 + 1 - 21 + 9 =

                                                                                          = 11 + 1 - 21 + 9 

                                                                                          = 12 - 21 + 9

                                                                                           = -9 + 9 = 0

 

So x=1 is another solution. 

Synthetic division:

 

1 |   2      9      1     -21      9

               2     11      12    -9

 --------------------------------

       2    11     12     -9      0

 

So now the equation is (x+1)(x-1)(2x^3 + 11x^2 + 12x - 9) = 0

 

We have to repeat the procedure just one more time on 2x^3 + 11x^2 + 12x - 9.

Notice that again by coincidence and luck, the constant term is -9 and the leading coefficient is 2.

However, the -9 does NOT change the set P/Q becuase BOTH positive and negative factors are included.

So we can use the same P/Q

P/Q = { +-1, +-3, +-9, +-1/2, +-3/2, +-9/2} 

 

This time it is -3 that works!!!

2(-3)^3 +11(-3)^2+12(-3)-9 =

2(-27) + 11(9) + 12(-3) - 9 =

-54 + 99 + -36 - 9 =

 45 + -36 - 9 =

  9 - 9 = 0

 

Synthethic division:

 

-3 |   2      11       12   -9

                -6      -15    9

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

         2      5        -3    0

 

So the equation is now :  (x+1)(x-1)(x+3)(2x^2 + 5x - 3) = 0

 

2x^2 + 5x - 3 = (  2x    -  1 )(  x   +    3   )      <--- factors

 

The complete factorization is ( x + 1)(x - 1 )( x + 3) ( 2x - 1)( x + 3) = 0

 

The solutions are x = -1, x = 1, x = -3, x = -3, x = 1/2

The solution x=-3 has multiplicity of 2

But sets are typically written without duplciates

The solution set without duplicates are x = { -1, 1, -3, 1/2 } 

The solution set with multiplicity is x={-1,1,-3,-3,1/2}