Question

Use the Rational Zeros Theorem to find the possible zeros of the following polynomial function: `f(x)=33x^3-245x^2+407x-35`?

Answer 1

The possible rational zeros are:

`+-1/33, +-1/11, +-5/33, +-7/33, +-5/11, +-7/11, +-1/3, +-1, +-35/33, +-5/3, +-7/3, +-35/11, +-5, +-7, +-35/3, +-35`

Explanation:

Given:

`f(x) = 33x^3-245x^2+407x-35`

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 `-35` and `q` a divisor of the coefficient `33` of the leading term.

The divisors of `-35` are:

`+-1, +-5, +-7, +-35`

The divisors of `33` are:

`+-1, +-3, +-11, +-33`

So the possible rational zeros are:

`+-1, +-5, +-7, +-35`

`+-1/3, +-5/3, +-7/3, +-35/3`

`+-1/11, +-5/11, +-7/11, +-35/11`

`+-1/33, +-5/33, +-7/33, +-35/33`

or in increasing order of size:

`+-1/33, +-1/11, +-5/33, +-7/33, +-5/11, +-7/11, +-1/3, +-1, +-35/33, +-5/3, +-7/3, +-35/11, +-5, +-7, +-35/3, +-35`

Note that these are only the rational possibilities. The rational zeros theorem does not tell us about possible irrational or complex zeros.

Using Descartes' Rule of Signs, we can determine that this cubic has no negative zeros and `1` or `3` positive real zeros.

So the only possible rational zeros are:

`1/33, 1/11, 5/33, 7/33, 5/11, 7/11, 1/3, 1, 35/33, 5/3, 7/3, 35/11, 5, 7, 35/3, 35`

Trying each in turn, we find:

`f(1/11) = 33(color(blue)(1/11))^3-245(color(blue)(1/11))^2+407(color(blue)(1/11))-35`

`color(white)(f(1/11)) = (3-245+4477-4235)/121`

`color(white)(f(1/11)) = 0`

So `x=1/11` is a zero and `11x-1` a factor:

`33x^3-245x^2+407x-35 = (11x-1)(3x^2-22x+35)`

To factor the remaining quadratic we can use an AC method:

Find a pair of factors of `AC = 3*35=105` with sum `B=22`

The pair `15, 7` works.

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

`3x^2-22x+35 = (3x^2-15x)-(7x-35)`

`color(white)(3x^2-22x+35) = 3x(x-5)-7(x-5)`

`color(white)(3x^2-22x+35) = (3x-7)(x-5)`

So the other two zeros are:

`x=7/3" "` and `" "x = 5`