How do you factor 6c³ + 11c² -10c?

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Answer 1 · 2015-05-23T12:34:29

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

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

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.