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!]
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Answers & Comments
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!]