Question

If 3 is one of the zeros ofx3+ 3x2+kx-24

Answer 1

Answer:

k=$\frac{-30}{3}$

Step-by-step explanation:

(x+3)=0

x=3

so

p(x)=$x^{3}+ 3x^{2} +kx- 24$

p(3)= $3^{3} +3(3)^{2} + k(3)-24$

     = 27+27+3k-24

     = 30+3k

   k= $\frac{-30}{3}$

Answer 2

substitute 3 in the value x

0 = (3)³ + 3(3)² + 3k - 24

0 = 27 + 27 + 3k - 24

0 = 3k = -30

k = $\large\boxed{-10}$