Question

How do you factor 6c³ + 11c² -10c?

Answer 1

`6c^3+11c^2-10c = c(6c^2+11x-10)`

`=c(2c+5)(3c-2)`

I first noticed that all the terms were divisible by `c`, so separated that out as a factor.

I then looked for a factorisation of the form:

`6c^2+11x-10 = (2c+a)(3c+b)`

`= 6c^2+(3a+2b)c+ab`

So I looked for `a` and `b` such that `3a+2b = 11` and `ab=-10`

I could tell that `a` must be odd, because otherwise both `2c` and `a` would be even and all the resulting coefficients would be even.

So `a=+-1` or `a=+-5`.

In order to get `3a+b=11` then the best choice to try seemed to be `a=5`, which forces `b=(-10)/a=(-10)/5=-2`.

So try `(2c+5)(3c-2)` - works.