How do you show that `x^2+2x-3` and `2x^3-3x^2+7x-6` have a common linear factor?

Let's consider the general case, since that gives the principle we will use:

Suppose `P_1(x)` and `P_2(x)` are polynomials with a common polynomial factor `P(x)`.

Then we can long divide `P_1(x)` by `P_2(x)` to find a quotient polynomial `Q_1(x)` and remainder polynomial `R_1(x)` with degree less than `P_2(x)`:

`P_1(x) = Q_1(x)P_2(x) + R_1(x)`

Then since `P_1(x)` and `P_2(x)` are both multiples of `P(x)`, `R_1(x)` must also be a multiple of `P(x)` and has lower degree than `P_2(x)`. Note that scalar factors are not important to us in this context. If `P(x)` is a factor then any non-zero scalar multiple of it is too (and vice versa).

So we can find the GCF of `P_1(x)` and `P_2(x)` by the following method:

• Divide the polynomial of higher (or equal) degree by the one of lower degree to give a quotient and remainder.

• If the remainder is `0` then the divisor polynomial is the GCF.

• Otherwise repeat with the remainder and the divisor polynomial.

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So in our example:

`2x^3-3x^2+7x-6 = (x^2+2x-3)(2x-7)+27x-27`

That is:

`(2x^3-3x^2+7x-6)/(x^2+2x-3) = 2x-7" "` with remainder `27x-27`

Note that `27x-27 = 27(x-1)`, so for tidiness, let's divide by `27` before proceeding.

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

That is:

`(x^2+2x-3)/(x-1) = x+3" "` with no remainder

So the GCF is `(x-1)`.

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Footnote

Alternatively I could have factored both of the polynomials and simply identified the common factor.

The main reason I did not is that the method used above has the advantage of not requiring you to factor either of the polynomials.