Answer:
1,-3,-5
Step-by-step explanation:
Given:
f(x)=x^3+7x^2+7x-15
Finding all the possible rational zeros of f(x)
p= ±1,±3,±5,±15 (factors of coefficient of last term)
q=±1(factors of coefficient of leading term)
p/q=±1,±3,±5,±15
Now finding the rational zeros using rational root theorem
f(p/q)
f(1)=1+7+7-15
=0
f(-1)= -1 +7-7-15
= -16
f(3)=27+7(9)+21-15
=96
f(-3)= (-3)^3+7(-3)^2+7(-3)-15
= 0
f(5)=5^3+7(5)^2+7(5)-15
=320
f(-5)=(-5)^3+7(-5)^2+7(-5)-15
=0
f(15)=(15)^3+7(15)^2+7(15)-15
=5040
f(-15)=(-15)^3+7(-15)^2+7(-15)-15
=-1920
Hence the rational roots are 1,-3,-5 !
The answer to the question is use photomath
Answer:
that's wrong
Step-by-step explanation:
for n = 1
n^(2n)=1²=1
and 2 doesnt devide 1
It’s the 3 and for number 2 it’s 1
Answer:
what's the question ndndn
