Find all the real zeros of the polynomial

Question

Find all the real zeros of the polynomial 

P(x)=x⁴-2x³-6x²-4x-16

its rational zeros are 

x1=

x2=

x3=

x4=

with x1 ≤x2≤x3≤x4

Answer 1

P(x) = x^4 - 2x^3 - 6x^2 - 4x - 16

 

The numerator of any rational zero must be a factor of -16 and the denominator a factor of 1.

 

So if there is a rational zero it will be in this set: ±1, ±2, ±4, ±8, ±16.

 

±1 don't work by inspection.

 

2 | 1  -2  -6  -4  -16

          2   0  -12 -32

     1   0  -6  -16 | not zero

 

-2 | 1  -2  -6  -4  -16
          -2   8  -4    16

-2 | 1  -4   2  -8  |  0 = P(-2)

          -2 12 -28

      1  -6 14 | not zero

 

4 | 1 -4  2  -8
         4  0   8
     1  0  2 | 0 = P(4)

 

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

 

x^2 + 2 = 0

x^2 = -2

x = ±i√(2)

 

So the real roots are x = -2, 4

 

check:

 

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

= (x^2 - 2x - 8)(x^2 + 2)

= (x^2 - 2x - 8)(x^2) + (x^2 - 2x - 8)*2

= x^4 - 2x^3 - 8x^2 + 2x^2 - 4x - 16

= x^4 - 2x^3 - 6x^2 - 4x - 16   √