Question

Hi cans omeone solve this please !!...

thanks in advance !!​

Answer 1

Answer:

-3

Step-by-step explanation:

Here,

p(x)=x³+5x²-9x-45=(x+3) (x-3) (x-5)

so zeroes of p(x) = -3, 3, 5

Again,

p'(x) =x³+8x²+15x=x(x+3) (x+5)

so zeroes of p'(x) = 0, -3, -5

so, common zero is -3. (only).

[Kindly report my answer if it is misleading.Please!]