Question

What are all the possible rational zeros for `f(x)=2x^4-7x^3-2x^2-7x-4` and how do you find all zeros?

Answer 1

`f(x)` has zeros `-1/2`, `4`, `+-i`

Explanation:

`f(x) = 2x^4-7x^3-2x^2-7x-4`

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 `-4` and `q` a divisor of the coefficient `2` of the leading term.

That means that the only possible rational zeros are:

`+-1/2, +-1, +-2, +-4`

Before we embark on trying these, note that the coefficients `2, -2, -4` almost cancel out if we reverse the sign of the middle one and the coefficients `-7, -7` would cancel out if the sign of one were opposite to the other. the only way that could heppen would be if `x^2 = -1`. So let's try `x=i` first...

`f(i) = 2i^4-7i^3-2i^2-7x-4 = 2+7i+2-7i-4 = 0`

So `x=i` is a zero and `(x-i)` a factor.

Since the coefficients of `f(x)` are all Real, `x=-i` is also a zero and `(x+i)` a factor.

Hence `(x-i)(x+i) = x^2+1` is a factor of `f(x)`...

`2x^4-7x^3-2x^2-7x-4 = (x^2+1)(2x^2-7x-4)`

We can factor the remaining quadratic using an AC method:

Find a pair of factors of `AC = 2*4 = 8` which differ by `B=7`.

The pair `8, 1` works.

Use this pair to split the middle term and factor by grouping:

`2x^2-7x-4 = (2x^2-8x)+(x-4)`

`color(white)(2x^2-7x-4) = 2x(x-4)+1(x-4)`

`color(white)(2x^2-7x-4) = (2x+1)(x-4)`

Hence the other two zeros are:

`x=-1/2` and `x = 4`