Make a list of all possible rational zeros of f(x)=3x3−11x2−35x+75, and then use this list to find all of the functions’s rational zeros, if any exist.

Answer 1

Factors of 75 are ±1, ±3 and ±5.

Using the factor theorem, determine which the factors of 75 where f(x) = 0

Turns out only f(5) and f(-3) = 0

Then use synthetic division to divide by (x-5) (Could divide by (x+3) also)

3 -11 -35 75 |5

15 20 -75

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3 4 -15 0

So we need to factor 3x²+4x-15=0

By the quadratic equation x = -3, 5/3

So the zeros are 5, -3, 5/3

Answer 2

The rational zeros theorem says that any possible rational zeros are of the form p/q. Where p is a divisor of the constant term (in this case 75) and q is a divisor of the leading coefficient (in this case 3).

So the possible p's are {±1,±3,±5,±15,±25,±75} and possible q's are {±1,±3}.

So the possible rational zeros are all the possible p/q's (get rid of any repeats!)

{ ±1,±3,±5,±15,±25,±75,±1/3,±5/3,±25/3}

Now to see if any of these are rational zeros you need to plug them each into f(x) and see if it equals zero. If it does, then they are a rational zero, if not then they are not.

f(1)=3(1)^3-11(1)^2-35(1)+75 ≠ 0 (so 1 is not a rational zero)

f(-1)=3(-1)^3-11(-1)^2-35(-1)+75 ≠ 0 (so -1 is not a rational zero)

f(3) ≠ 0

f(-3) = 3(-3)^3-11(-3)^2-35(-3)+75 =-81-99+105+75 = -180+180 = 0 (So -3 is a rational zero!)

f(5) = 3(5)^3-11(5)^2-35(5)+75 = 375-275-175+75 = 0 (So 5 is also a rational zero!)

... continue plugging in and the only other point that is a zero you will find is 5/3

f(5/3) = 3(5/3)^3-11(5/3)^2-35(5/3)+75 = 0

So the functions rational zeros are -3,5, and 5/3.