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^4+x^3-7x^2-3x+3`?

Answer 1

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

Explanation:

It iseasy to find
`x=-1`
By the rational root theórem we get
`x=1/2`
Dividing

`2x^4+x^3-7x^2-3x+3` by
`2x^2+x-1` we get
`x^2-3`

Answer 2

See explanation...

Explanation:

Given:

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

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 `3` 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`

In addition, note that the pattern of signs of the coefficients of `f(x)` is `+ + - - +`. With two changes of sign, Descartes' Rule of Signs tells us that `f(x)` has `2` or `0` positive real zeros.

Further, note that the pattern of signs of the coefficients of `f(-x)` is `+ - - + +`. With two changes of sign, Descartes' Rule of signs tells us that `f(x)` has `2` or `0` negative real zeros.

Trying each of the possible rational zeros in turn we find:

`f(color(blue)(1/2)) = 2(color(blue)(1/2))^4+(color(blue)(1/2))^3-7(color(blue)(1/2))^2-3(color(blue)(1/2))+3`

`color(white)(f(1/2)) = 2/16+1/8-7/4-3/2+3`

`color(white)(f(1/2)) = 0`

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

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

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

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

So the other zeros are `x=+-sqrt(3)` and `x=-1`

graph{2x^4+x^3-7x^2-3x+3 [-2.5, 2.5, -6, 5]}