Question

How do you use the Rational Zeros theorem to make a list of all possible rational zeros, and use the Descarte's rule of signs to list the possible positive/negative zeros of `f(x)=-2x^3+19x^2-49x+20`?

Answer 1

See explanation...

Explanation:

Given:

`f(x) = -2x^3+19x^2-49x+20`

By the rational zeros 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 `20` 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, +-5/2, +-4, +-5, +-10, +-20`

The pattern of signs of the coefficients of `f(x)` is `- + - +`. With `3` changes, Descartes' Rule of Signs tells us that `f(x)` has `3` or `1` positive real zeros.

The pattern of signs of the coefficients of `f(-x)` is `+ + + +`. With no changes, Descartes' Rule of Signs tells us that `f(x)` has no negative real zeros.

Putting these together, we can deduce that the only possible rational zeros of `f(x)` are:

`1/2, 1, 2, 5/2, 4, 5, 10, 20`

Trying each in turn, we find:

`f(1/2) = -2(color(blue)(1/8))+19(color(blue)(1/4))-49(color(blue)(1/2))+20`

`color(white)(f(1/2)) = -1/4+19/4-98/4+80/4 = 0`

So `x=1/2` is a zero and `(2x-1)` a factor:

`-2x^3+19x^2-49x+20 = (2x-1)(-x^2+9x-20)`

`color(white)(-2x^3+19x^2-49x+20) = -(2x-1)(x-4)(x-5)`

So the three zeros are: `1/2`, `4` and `5`