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 !
Composite is a number that is NOT prime. So we want to find an odd perfect square that IS prime to be a counterexample.
√36 = Not odd
√49 = 7 Factors only 1 * 7 so it is prime
√81 = 9 Factors 1, 3, 9 composite
√225 = 15 Factors 1,3, 5,9,15, 25, 45, 75, 225 composite
Counterexample for 2∧n - 1 Is prime
2^6 - 1 = 64 - 1 = 63 NOT PRIME Factors are: 1, 3, 7, 9,21, 63
2^5 - 1 = 32 - 1 = 31 Prime
2^3 - 1 = 8 - 1 = 7 Prime
2^2 - 1 = 4 - 1 = 3 Prime
Answer:
A, C
Step-by-step explanation:
Between the two answer is 3 hundred
