Question

What are all the possible rational zeros for `f(x)=2x^3+x^2+8x-21` and how do you find all zeros?

Answer 1

`f(x)` has rational zero `x=3/2` and other zeros `x = -1+-sqrt(6)i`

Explanation:

`f(x) = 2x^3+x^2+8x-21`

By the rational roots theorem, any rational zeros of `f(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-21` and `q` a divisor of the coefficient `2` of the leading term.

That means that the only possible rational zeros are:

`+-1/2, +-1, +-3/2, +-3, +-7/2, +-7, +-21/2, +-21`

In addition note that the signs of the coefficients are in the pattern `+ + + -`. By Descartes' rule of signs, since there is one change of sign, there is exactly one positive Real zero.

So let's try the positive rational possibilities first:

`f(1/2) = 2(1/8)+(1/4)+8(1/2)-21 = -33/2`

`f(1) = 2+1+8-21 = -10`

`f(3/2) = 2(27/8)+(9/4)+8(3/2)-21 = (27+9+48-84)/4 = 0`

So `x=3/2` is a zero and `(2x-3)` a factor:

`2x^3+x^2+8x-21 = (2x-3)(x^2+2x+7)`

We can factor the remaining quadratic by completing the square and using the difference of squares identity:

`a^2-b^2 = (a-b)(a+b)`

with `a=(x+1)` and `b=sqrt(6)i` as follows:

`x^2+2x+7 = x^2+2x+1+6`

`color(white)(x^2+2x+7) = (x+1)^2-(sqrt(6)i)^2`

`color(white)(x^2+2x+7) = ((x+1)-sqrt(6)i)((x+1)+sqrt(6)i)`

`color(white)(x^2+2x+7) = (x+1-sqrt(6)i)(x+1+sqrt(6)i)`

Hence zeros:

`x = -1+-sqrt(6)i`