Find all the real zeros of the polynomial

Question

Find all the real zeros of the polynomial 

P(x)=x³-2x²-9x+4

Give them as a comma separated list, and give exact answers - no decimals.
The real zeros of P(x) have x=

Answer 1

Dear Dalia,

 

The real roots of P(x) are as follows:

 

x = -(√2 + 1), x = √2 - 1 and x = 4

 

P(x) can be factored into (x - 4)*(x² + 2x - 1).  From this it is easy to see that x = 4 is one of the real zeros.

 

(x² + 2x - 1) cannot be factored any further, but notice that it is just a quadratic equation with a = 1, b = 2, and c = -1.

 

x = [-b ±(b² - 4ac)^(1/2)]/2a

 

x = {-2 ± [(4 - (4)*(1)*(-1)]^^(1/2)}/2

 

x = [-2 ± (2)√2]/2

 

x = -(√2 + 1), x = √2 - 1     

 

Answer 2

P(x) = x^3 - 2x^2 - 9x + 4

2 sign changes => 2 or 0 positive real zeros

P(-x) = -x^3 - 2x^2 + 9x + 4

1 sign change => 1 negative real zero

 

Any rational zero(s) would be in the set {±1, ±2, ±4}.

 

-1 |  1  -2  -9  4

           -1   3   6

       1  -3  -6 | not zero

 

-2 |  1  -2  -9   4

           -2   8   2      

       1  -4  -1 | not zero

 

-4 | 1  -2  -9   4

          -4  24 -60         

      1  -6  15 | not zero

 

4 | 1 -2 -9    4

         4  8   -4   

     1  2 -1 | 0 = P(4)

 

P(x) = (x - 4)(x^2 + 2x - 1)

 

0 = x^2 + 2x - 1

 

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

 

0 = (x + 1)^2 - 2

 

(x + 1)^2 = 2

 

x + 1 = ±√(2)

 

x = -1 ± √(2)

 

The real zeros of P(x) are x = -1 - √(2), -1 + √(2), 4

Answer 3

Using synthetic division first.

x^3-2x^2-9x+4

   

 

 

      1       -2       -9       4

4              4        8      -4

    

      1        2       -1       0

 

We now have (x-4)(x^2+2x-1)

For x^2+2x-1 use the quadratic equation.

 

[-2+sqrt(4+4)]/2=[-2+sqrt(8)]/2

                         =[-2+2sqrt(2)]/2

                         =-1+sqrt(2)

x=4

x=-1+sqrt(2)

x=-1-sqrt(2)

Answer 4

I would solve this by first graphing the equation, which finds one foot, x = 4; then dividing out x-4 from the original equation, arriving to the other factor being x2+2x-1; finally, solving the quadratic, by using the quadratic formula, and finding the other two roots