Question

What are all the possible rational zeros for `f(x)=2x^4-13x^3-34x^2+65x+120` and how do you find all zeros?

Answer 1

"Possible" rational zeros:

`+-1/2, +-1, +-3/2, +-2, +-5/2, +-3, +-4, +-5, +-6, +-15/2, +-8, +-10, +-12, +-15, +-20, +-24, +-30, +-40, +-60, +-120`

Actual zeros:

`-3/2`, `8`, `+-sqrt(5)`

Explanation:

Given:

`f(x) = 2x^4-13x^3-34x^2+65x+120`

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 `120` 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, +-2, +-5/2, +-3, +-4, +-5, +-6, +-15/2, +-8, +-10, +-12, +-15, +-20, +-24, +-30, +-40, +-60, +-120`

That's quite a few possibilities to try, so let's try a few easier ones to evaluate and deduce where the zeros might be...

`f(-2) = 2(color(blue)(-2))^4-13(color(blue)(-2))^3-34(color(blue)(-2))^2+65(color(blue)(-2))+120`

`color(white)(f(-2)) = 32+102-136-130+120 = -12 < 0`

`f(-1) = 2+13-34-65+120 = 36 > 0`

So `f(x)` has a zero in `(-2, -1)`. Is it the only possible rational one?

`f(-3/2) = 2(color(blue)(-3/2))^4-13(color(blue)(-3/2))^3-34(color(blue)(-3/2))^2+65(color(blue)(-3/2))+120`

`color(white)(f(-3/2)) = 2(81/16)+13(27/8)-34(9/4)-65(3/2)+120`

`color(white)(f(-3/2)) = 1/8(81+351-612-780+960) = 0`

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

`2x^4-13x^3-34x^2+65x+120 = (2x+3)(x^3-8x^2-5x+40)`

The remaining cubic can be factored by grouping:

`x^3-8x^2-5x+40 = (x^3-8x^2)-(5x-40)`

`color(white)(x^3-8x^2-5x+40) = x^2(x-8)-5(x-8)`

`color(white)(x^3-8x^2-5x+40) = (x^2-5)(x-8)`

`color(white)(x^3-8x^2-5x+40) = (x^2-(sqrt(5))^2)(x-8)`

`color(white)(x^3-8x^2-5x+40) = (x-sqrt(5))(x+sqrt(5))(x-8)`

So the other three zeros are `+-sqrt(5)` and `8`