Respuesta :

absor201

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 !

lublana

Answer:

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

Step-by-step explanation:

Given function is [tex]f(x)=x^3+7x^2+7x-15[/tex].

constant term = p = -15

coefficient of leading term q= 1

then possible rational roots of given function f(x) are the divisors of [tex]\pm \frac{p}{q}=\pm 1, \quad \pm 3, \quad \pm 5, \quad \pm 15[/tex]

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

test for x=1

[tex]f(x)=x^3+7x^2+7x-15[/tex]

[tex]f(1)=1^3+7(1)^2+7(1)-15[/tex]

[tex]f(1)=1+7+7-15[/tex]

[tex]f(1)=0[/tex]

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.

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