How many distinct, real number roots does p pp have ?

Question

P(x)=x⁵+5x³+17x²+85

Answer 1

P(x)=x⁵+5x³+17x²+85
 
Factor this by grouping to get a better idea
P(x) = (x⁵ + 5x³) + (17x² + 85)
P(x) = x³(x² + 5) + 17(x² + 5)
       = (x³ + 17)(x² + 5)

 

Looking at x³ + 17

Make this the sum of two cubes

x³ + (³√17)³

 

There is a formula for the sum of two cubes:

a³ + b³ = (a+b)(a² - ab + b²)

with a=x, b = ³√17

 

x³ + 17 = (x + ³√17)(x² - x(³√17) + (³√17)²)

 

P(x) = (x²+5)(x+³√17)(x² - x(³√17) + 17^2/3)

 

One real root will be from x = -³√17

 

For x² + 5 = 0
     x² = -5
     x = ±√(-5)
     x = ± i√5

Here we have two complex roots

 

For x² - (³√17)x + 17^2/3 = 0

Look at the quadratic equation with a=1, b= -³√17, c = 17^2/3

 

x = [³√17 ± √(17^2/3 -4(17^2/3))]/2

The radicand will be negative and the two root answers will be complex

 

So there are 5 roots as expected but 4 are complex and one is real

x = -³√17

Answer 2

P(x) has only one real root, -³√17.

The other four roots are complex.