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

Jan 28, 2017

Find their GCF, which turns out to be the linear polynomial $x - 1$

#### Explanation:

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

Suppose ${P}_{1} \left(x\right)$ and ${P}_{2} \left(x\right)$ are polynomials with a common polynomial factor $P \left(x\right)$.

Then we can long divide ${P}_{1} \left(x\right)$ by ${P}_{2} \left(x\right)$ to find a quotient polynomial ${Q}_{1} \left(x\right)$ and remainder polynomial ${R}_{1} \left(x\right)$ with degree less than ${P}_{2} \left(x\right)$:

${P}_{1} \left(x\right) = {Q}_{1} \left(x\right) {P}_{2} \left(x\right) + {R}_{1} \left(x\right)$

Then since ${P}_{1} \left(x\right)$ and ${P}_{2} \left(x\right)$ are both multiples of $P \left(x\right)$, ${R}_{1} \left(x\right)$ must also be a multiple of $P \left(x\right)$ and has lower degree than ${P}_{2} \left(x\right)$. Note that scalar factors are not important to us in this context. If $P \left(x\right)$ is a factor then any non-zero scalar multiple of it is too (and vice versa).

So we can find the GCF of ${P}_{1} \left(x\right)$ and ${P}_{2} \left(x\right)$ 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:

$2 {x}^{3} - 3 {x}^{2} + 7 x - 6 = \left({x}^{2} + 2 x - 3\right) \left(2 x - 7\right) + 27 x - 27$

That is:

$\frac{2 {x}^{3} - 3 {x}^{2} + 7 x - 6}{{x}^{2} + 2 x - 3} = 2 x - 7 \text{ }$ with remainder $27 x - 27$

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

${x}^{2} + 2 x - 3 = \left(x - 1\right) \left(x + 3\right)$

That is:

$\frac{{x}^{2} + 2 x - 3}{x - 1} = x + 3 \text{ }$ with no remainder

So the GCF is $\left(x - 1\right)$.

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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.