How do you factor 6c³ + 11c² -10c?

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How do you factor 6c³ + 11c² -10c?

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Answer 1 · 2015-05-23T12:34:29

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

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

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

I then looked for a factorisation of the form:

$6 c^{2} + 11 x - 10 = (2 c + a) (3 c + b)$ $6c^2+11x-10 = (2c+a)(3c+b)$

$= 6 c^{2} + (3 a + 2 b) c + a b$ $= 6c^2+(3a+2b)c+ab$

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

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

So $a = \pm 1$ $a=+-1$ or $a = \pm 5$ $a=+-5$ .

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

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

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How do you factor 6c³ + 11c² -10c?

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