If f(x)=x^3-9x^2+11x+21f(x)=x 3 −9x 2 +11x+21 and x-3x−3 is a factor of f(x)f(x), then find all of the zeros of f(x)f(x) algebraically.

x = -1, 3, and 7

Step-by-step explanation:

Using grouping:

f(x) = x³ − 9x² + 11x + 21

f(x) = x³ − 9x² + 18x − 7x + 21

f(x) = x (x² − 9x + 18) − 7 (x − 3)

f(x) = x (x − 6) (x − 3) − 7 (x − 3)

f(x) = (x² − 6x) (x − 3) − 7 (x − 3)

f(x) = (x² − 6x − 7) (x − 3)

f(x) = (x + 1) (x − 7) (x − 3)

To use long division, see the picture.

The zeros are x = -1, 3, and 7.

step-by-step explanation:

because 1/3 is a fraction. we can tell because there is a divisor symbol between the 1 and the 3. if we were to add 2 fractions together, we would add the numerators, which is the number on top, but not the denominators, which is the number on the bottom

1/3+1/3=2/3

hope this !

you need to give us more information since we can't answer