I have the following question:
Question: Find a linear factor that will fully factorise $x^3 - 2x^2 - 5x + 6$.
When I checked the solutions the linear factor is $x-1$. I do not understand how this linear factor is achieved.
2 Answers
There's a theorem which asserts that a polynomial $P(X)$ in $K[X]$ ($K$ a field) is divisible by $X-\alpha$ if and only if $\alpha$ is a root of $P(x)$.
The proof of this assertion relies on Euclidean division in the polynomial ring $K[X]$: there exists a unique polynomial $Q(X)$ and a unique constant $R\in K$ such that $$P(X)=Q(X)(X-\alpha)+R.$$ Further $R=P(\alpha)$.
So they checked $P(1)=1^-2\cdot 1^2-5\cdot 1+6=0$. Dividing, say, using Horner's scheme yields $$X^3-2X^2-5X+6=(X-1)(X^2-X-6).$$ To factorise further, you have to find the roots of the quadratic equation $\;x^2-x-6=0$. We can search for rational roots, before applying the resolution formula.
By the Rational root theorem, a rational root, if any, has to be a divisor of $6$, i.e. one among $\{\pm 1,\pm 2,\pm 3,\pm 6\}$. One readily checks $3$ is such a root, hence the other root is $-2$ (their product is equal to $-6$). Whence the factorisation $$X^3-2X^2-5X+6=(X-1)(X+2)(X-3).$$
One method to solve, is to use the Rational Root Theorem, with other theorems to find the factorization.
Rational Root Theorem: Let $f(x)$ be a polynomial$$f(x)=\sum_{k=0}^na_kx^k\tag1$$ Where $a_0,a_1,\ldots a_n$ are integers and $n\geq 1$. The roots of $(1)$ are of the form $\pm \frac pq$ where $p$ is a factor of the term $a_0$, and $q$ is a factor of $a_n$.
This is a really useful theorem for finding the roots of any polynomial, but since you're given a lot of possibilities, (try listing all of the factors of $972$), I have listed some more theorems you can employ to make the guesswork a bit easier for you:
Cauchy's Bound: Suppose $f(x)$ is a polynomial such that$$f(x)=\sum_{k=0}^{n}a_kx^k=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0\tag{2}$$With $n\geq 1$ and $a_n\neq 0$. Let $M$ be the largest of the numbers:$$\frac {|a_0|}{|a_n|},\frac {|a_1|}{|a_n|},\ldots,\frac {|a_{n-1}|}{|a_n|}\tag3$$All of the roots of $f$ lie in the interval $[-(M+1),M+1]$
Another one:
Descartes' Rule of Signs: Suppose $f(x)$ is a polynomial written in descending powers of $x$.
If $P$ denotes the number of variations of signs in $f(x)$, then the number of positive real zeroes (counting multiplicity) is one of the numbers $\{P,P-2,P-4,\ldots\}$
If $N$ denotes the number of variations of signs in the formula for $f(-x)$, then the number of negative real zeores is one of the numbers $\{N,N-2,N-4,\ldots\}$
Unfortunately, your polynomial is too "small" for these two theorems to be effective, but it's nice if you know them. In case you need to factor something like $x^4-13x^3-81x^2+215x+102=0$.
