Determine the rational zeros for the function f(x)=x^3+7x^2+7x-15

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 !

Actual rational zeros of f(x) are x=-5, x=-3, and x=1.

Step-by-step explanation:

Given function is .

constant term = p = -15

coefficient of leading term q= 1

then possible rational roots of given function f(x) are the divisors of

Now we can plug those possible roots into given function to see which one of them gives output 0.

test for x=1

Hence x=1 is the actual rational zero.

Similarly testing other roots, we get final answer as:

Actual rational zeros of f(x) are x=-5, x=-3, and x=1.

care to share were the trapezoid shown is ?

the answer is 150.6

step-by-step explanation:

you multiply .02 by 30 and get .6.

and then you multiply 5.00 by 30 and get 150.

hope this !