How can I find a common factor of two polynomials?

Question

I have some lecture notes here and they magically find (x-1) to be a common factor of $x^3-3x+2$ and $x^3-x^2-x+1$.

I'm wondering whether I'm missing something or the lecture notes are just weak?

Thanks!

Answer 1

There is a greatest common divisor for polynomials. The only caution I would add is that this need not have integer coefficients; it is by nature a polynomial with rational coefficients.

$$ \gcd(x^3 - 3 x + 2, x^3 - x^2 - x + 1) $$ first minus second replaces second $$ \gcd(x^3 - 3 x + 2, x^2 -2 x + 1) $$ first minus $(x+2)$ times second replaces first $$ \gcd(0, x^2 -2 x + 1) $$ $$ x^2 - 2 x + 1 $$

Answer 2

We have:

$$ x^3-3x+2=x^3-x-2x+2=x(x^2-1)-2(x-1)= $$ $$ =x(x-1)(x+1)-2(x-1)=(x-1)[x(x+1)-2]=(x-1)(x^2+x-2) $$ and: $$ x^3-x^2-x+1=x^2(x-1)-1(x-1)=(x-1)(x^2-1)=(x-1)^2(x+1) $$

Answer 3

We have

$$P(x)=x^3-3x+2$$ $$=x^3-x-2(x-1)$$

$$=(x-1)(x^2+x+1)-2(x-1)$$

$$=(x-1)(x^2+x-1)$$ $$=(x-1)(x-\frac{-1-\sqrt{5}}{2})(x-\frac{-1+\sqrt{5}}{2})$$

and $$Q(x)=x^3-x^2-x+1$$ $$=x^2(x-1)-(x-1)$$

$$=(x-1)(x^2-1)=(x-1)^2(x+1)$$

from here, we see that $(x-1)$, is the only common factor.

Answer 4

What you are looking for is the Polynomial Remainder Theorem. This theorem says that $f(a) = 0$ if and only if $x - a$ is a factor of $f(x)$, for polynomial $f(x)$.

The fact that both polynomials approach $0$ as $x \to 1$ suggests that they both share the common zero $x = 1$, and so must share the common factor $(x - 1)$. To keep evaluating the limit, you would factor $(x - 1)$ out of each with polynomial long or synthetic division.