What are the real zeros of x3 + 4x2 − 9x − 36?

a. 1, 2, 3

b. -6, 2, 3

c. –4, –3, 3

d. –1, 2, 18

C

Step-by-step explanation:

x^3+4x^2-9x-36=0

or x²(x+4)-9(x+4)=0

or (x+4)(x²-9)=0

or (x+4)(x+3)(x-3)=0

x=-4,-3 or 3

you probably already have the answer but anyways it's b. 15.0

step-by-step explanation:

so, converting firstly the mixed fractions to improper fractions and then "adding".