Question

How do you find all zeros with multiplicities of `f(x)=-2x^3+19x^2-49x+20`?

Answer 1

The zeros of `f(x)` are `1/2`, `4` and `5`, all with multiplicity `1`.

Explanation:

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

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 `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`

in addition, notice that the signs of the coefficients are in the pattern `- + - +`. So by Descartes' Rule of Signs, since there are `3` changes of sign, this cubic has `1` or `3` positive real zeros. Also the signs of the coefficients of `f(-x)` are in the pattern `+ + + +`. So with no changes of sign, this cubic has no negative real zeros.

So the only possible rational zeros are:

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

Trying each in turn we first find:

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

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

`color(white)(f(1/2)) = 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)`

To factor the remaining quadratic note that `4+5 = 9` and `4*5 = 20`.

Hence:

`-x^2+9x-20 = -(x-4)(x-5)`

So the remaining two zeros are `x=4` and `x=5`.

All of the zeros have multiplicity `1`.