Two integers are coprime if their `gcd` is 1. Similarly, two polynomials `p`(`t`) and `q`(`t`) are coprime if `gcd`(`p`(`t`), `q`(`t`)) = 1 (or any nonzero number). Specifically this means that

`c`(`t`) divides `p`(`t`) and `q`(`t`) ⇒ `c`(`t`) is a nonzero constant.

From this example, we find that the polynomials `t`^{4} + `t`^{3} + 3`t`^{2} + `t` + 1 and `t`^{2} -` t` + 1 are coprime.

For two complex polynomials `p`(`t`) and `q`(`t`), the following are equivalent.

`p`(`t`) and`q`(`t`) are coprime.- There are polynomials
`u`(`t`) and`v`(`t`)`, such that``u`(`t`)`p`(`t`) +`v`(`t`)`q`(`t`) = 1. `p`(`t`) and`q`(`t`) do not have common roots.

Proof Applying this result to the special case that `d`(`t`) = 1, we find that the first property implies the second.

Now assume `u`(`t`)`p`(`t`) + `v`(`t`)`q`(`t`) = 1 for some polynomials `u`(`t`) and `v`(`t`). If `r` is a of `p`(`t`), then we have `p`(`r`) = 0, so that `v`(`r`)`q`(`r`) = `u`(`r`)`p`(`r`) + `v`(`r`)`q`(`r`) = 1. In particular, this implies `q`(`r`) ≠ 0. This proves that any root of `p`(`t`) cannot be a root of `q`(`t`). This completes the proof that the second property implies the third.

Finally, if `p`(`t`) and `q`(`t`) are not coprime, then `gcd`(`p`(`t`), `q`(`t`)) = `d`(`t`) is a polynomial of positive degree. By the fundamental theorem of algebra, `d`(`t`) must have a complex root `r`. Since `d`(`t`) divides `p`(`t`) and `q`(`t`), by this result, `r` is also a common root of `p`(`t`) and `q`(`t`). This completes the proof that the third property implies the first.

How can we determine whether two given polynomials have common roots? Let

`p`(`t`) = `a`_{0} + `a`_{1}`t` + `a`_{2}`t`^{2} + ... + `a`_{m-1}`t`^{m-1} + `t ^{m}` = (

Then for `p`(`t`) and `q`(`t`) to have a common root means that the following expression, called the resultant, is zero.

`Res`(`p`, `q`) = ∏_{ij}(`r _{i}` -

= (

Note that the expression is symmetric in the roots `r`_{1}, `r`_{2}, ... `r _{m}` of

Example Let `p`(`t`) = `a`_{0} + `a`_{1}`t` + `t`^{2} and `q`(`t`) = `b`_{0} + `b`_{1}`t` + `t`^{2} be of second order. Then by `r`_{1} + `r`_{2} = - `a`_{1} and `r`_{1}`r`_{2} = `a`_{0}, we have

`Res`(`p`, `q`) = (`b`_{0} + `b`_{1}`r`_{1} + `r`_{1}^{2})(`b`_{0} + `b`_{1}`r`_{2} + `r`_{2}^{2})

= `b`_{0}^{2} + `b`_{0}`b`_{1}(`r`_{1} + `r`_{2})
+ `b`_{0}(`r`_{1}^{2} + `r`_{2}^{2}) + `b`_{1}^{2}`r`_{1}`r`_{2} + `b`_{1}`r`_{1}`r`_{2}(`r`_{1} + `r`_{2}) + `r`_{1}^{2}`r`_{2}^{2}

= `b`_{0}^{2} + `b`_{0}`b`_{1}(- `a`_{1}) + `b`_{0}((- `a`_{1})^{2} - 2`a`_{0}) + `b`_{1}^{2}`a`_{0} + `b`_{1}`a`_{0}(- `a`_{1}) + `a`_{0}^{2}

= `a`_{0}^{2} + `b`_{0}^{2}
- 2`a`_{0}`b`_{0}` + ``a`_{1}^{2}`b`_{0}` + ``a`_{0}`b`_{1}^{2} - `a`_{1}`b`_{0}`b`_{1} - `a`_{0}`a`_{1}`b`_{1}.