What are the real zeros of y=(x+3)^3+10?

Question

What are the real zeros of y=(x+3)^3+10?

Answer 1

y = (x + 3)³+ 10

0 = (x + 3)³ + 10

-10 = (x + 3)³

-³√10 = x + 3

 

Can you continue?

Answer 2

Hi Nora,

 

I provided an answer to this question a couple of days ago. Please check! 

Answer 3

Set y=0 and solve for x.  We can rewrite this function as a sum of cubes to avoid expanding and using synthetic division.

 

0 = (x + 3)³ + [³√(10)]³

 

0 = [(x + 3) + ³√10][(x + 3)² - (x + 3) ³√(10) + ³√100]

 

0 = (x + 3 + ³√10)[x² + 6x + 9 - x ³√(10) - 3 ³√(10) + ³√100]

 

0 = (x + 3 + ³√10)[x² + (6 - ³√10)x + 9 - 3 ³√(10) + ³√100]

 

 

Now just set the two factors equal to zero.

 

x + 3 + ³√10 = 0            and              x² + (6 - ³√10)x + 9 - 3 ³√(10) + ³√100 = 0

 

 

For the quadratic factor, use the quadratic formula in which

 

a = 1

b = 6 - ³√10

c = 9 - 3 ³√(10) + ³√100

 

where:

 

³√10 ≈ 2.15

³√100 ≈ 4.64

 

Answer 4

Hi Nora,

 

This polynomial can not be factored. Trying the rational roots test doesn't work either. 

 

You will have to take advantage of its behaviour on the coordinate plane. You will see that at x=-5, y = 2 and at x=-6, y= -17. From this, we deduce that the root has to be between -6 and -5 and possibly closer to -5. You will need to try something like the bisection method to approximate the root.

Good luck!