Question

One root of f(x)=x^3+10x^2-25-250 is x = –10. What are all the roots of the function? Use the Remainder Theorem. x = –25 or x = 10 x = –25, x = 1, or x = 10 x = –10 or x = 5 x = –10, x = –5, or x = 5

Answer 1

Answer:D

Step-by-step explanation:

Answer 2

Answer:

All zeros are

x=-10 , x=-5 , x=5

Step-by-step explanation:

we are given function as

$f(x)=x^3+10x^2-25x-250$

we are given one of zero is x=-10

we have to use Remainder theorem

we can find all possible factor of 250

$250=-5\times 5\times -10$

so, we will check zeros at x=-5 and x=5

At x=-5:

we can plug x=-5

$f(-5)=(-5)^3+10(-5)^2-25(-5)-250$

$f(-5)=-125+250-\left(-125\right)-250$

$f(-5)=0$

At x=5:

we can plug x=5

$f(5)=(5)^3+10(5)^2-25(5)-250$

$f(5)=125+250-\left(125\right)-250$

$f(5)=0$

So, other zeros are

x=-5 and x=5

All zeros are

x=-10 , x=-5 , x=5