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

the answer to this question is in the pic

8 2/3 is the answer

step-by-step explanation: just use a calculator