Answers pleasee this stuff is hard

According to the integral root theorem the only possible real, integral zeros are +/- factors of (10*3 - 8)

... or factors of 992(+/-).  you can test by direct substitution.  No positive values are roots ... try 1, 1332, but as x increases so does y ...  so there are no positive roots.

  Roots, means y = 0 ...  so (x + 10)^3  = 8 ... take cube root ... x + 10 = 2  .. x = -8

x = -8 is a root

Testing other negatives and the y values are not zero but tend toward - infinity.

So the only real zero is x = -8

it helps you

Step-by-step explanation:

this is for y = (x-10)^3 - 8

According to the integral root theorem the only possible real, integral zeros are +/- factors of (10*3 - 8)

... or factors of 992(+/-). you can test by direct substitution. No positive values are roots ... try 1, 1332, but as x increases so does y ... so there are no positive roots.

Roots, means y = 0 ... so (x + 10)^3 = 8 ... take cube root ... x + 10 = 2 .. x = -8

x = -8 is a root

Testing other negatives and the y values are not zero but tend toward - infinity.

So the only real zero is x = -8

Similarly for y = (x + 7)^3 +2 = 0

(x + 7)^3 = -2

x + 7 = cube root (-2)

x = -7 + cube root (-2) <<< the only real root

d i think

step-by-step explanation:

answer: all of the odd numbers were opened

the first person was the one that was opening all of the lockers. since one is odd, this causes all of the odd number lockers to be opened in the end.