Originally Posted by
Archie Meade $\displaystyle 6x^3+17x^2=10-x$
$\displaystyle x^2(6x+17)=10-x$
$\displaystyle x^2=\frac{10-x}{6x+17}$
Positive integers are unhelpful here, but if $\displaystyle x=-1
$
$\displaystyle 1=\frac{11}{11}$
hence $\displaystyle x=-1$ is a solution, so $\displaystyle [x-(-1)]=(x+1)$ is a factor.
$\displaystyle (x+1)(ax^2+bx+c)=6x^3+17x^2+x-10$
a=6, c=-10
$\displaystyle ax^3+bx^2+cx+ax^2+bx+c=ax^3+(a+b)x^2+(b+c)x+c=6x^3 +17x^2+x-10$
Hence, a+b=17, b=17-a=17-6=11
or b+c=1, b-10=1, b=10+1=11
Using this technique, "a" and "c" are found immediately,
"b" can be found from either of the other two co-efficients using "a" or "c".
$\displaystyle 6x^3+17x^2+x-10=(x+1)\left(6x^2+11x-10\right)$
The quadratic is easier to factorise.
We get
$\displaystyle 6x^3+17x^2+11x-10=(x+1)(3x-2)(2x+5)$