Use the rational zeros theorem to write a list of all possible rational zeros of the function. f(x) = 2x3 + 8x2 + 7x - 8

±1/2, ±1, ±2, ±4 and ±8

Step-by-step explanation:

The Rational Zeros Theorem is defined as when a polynomial has all coefficients  integer, then any rational zeroes of the polynomial have to be in the form ±p/q, where q is the coefficient of the highest power of the variable and p is declared  as the constant term.

Furthermore, a rational  "zero" is for a polynomial.  when the polynomial is p(x), a "zero" is a value of x when  p(x) = 0

Secondly, we have to know  what a "rational zero" is.  A "rational zero" is a zero that its number is rational.  Some polynomials  have some rational zeros and some irrational zeros, and some only have zeros that are rational numbers.

By applying this theorem,  all  possible factors of the constant term must be considered . In this example they are 1, 2, 4, and 8.  After that Then you consider all possible factors of the coefficient of the highest power of the variable. we take the x³ term, whose coefficient is 2. the  the possible factors of 2 are 1 and 2.

Therefore, the possible list of rational zeroes are given below

 ±1/1 = ±1

±1/2

±2/1 = ±2

±2/2 = ±1

±4/1 = ±4

±4/2 = ±2

±8/1 = ±8

±8/2 = ±4

By removing the duplicates, we arrive at the following,

±1/2, ±1, ±2, ±4 and ±8

The list is -1,1,-2,2,-4,4,-8,8,-1/2,1/2

Step-by-step explanation:

Possible rational zeros are the constant factors/leading coefficient factors

So factors of -8:  -1,1,-2,2,-4,4,-8,8

So factors of 2:  -1,1,-2,2

Now put every number in the first list over every number in the second list:

The possible rational zeros are:

-1/1=-1

1/1=1

-2/1=-2

2/1=2

-4/1=-4

4/1=4

-8/1=-8

8/1=8

-1/2

1/2

I didn't write any number twice like -8/2 is just -4 which I already wrote

The list is -1,1,-2,2,-4,4,-8,8,-1/2,1/2

where are the choices?