How do you factor 6c^2 + 17c-14=0?

1 Answer
KillerBunny
Oct 28, 2015

6c^2+17c-14 = 6(c+7/2)(c-2/3)

Explanation:

First of all, try to solve it using the classical formula

\frac{-b\pm\sqrt{b^2-4ac}}{2a}

and in your case a=6, b=17 and c=-14.

So, the formula becomes

\frac{-17\pm\sqrt{17^2-4*6*(-14)}}{2*6}= \frac{-17\pm\sqrt{289-(-336)}}{12}

= \frac{-17\pm\sqrt{625}}{12} = \frac{-17\pm25}{12}

Which leads us to the two solutions

c_1 = (-17-25)/12 = -7/2 and
c_2 = (-17+25)/12 = 2/3.

Now, use the fact that a polynomial can be written in terms of his solutions, as the product of factors of the form x-x_n, where x_n is the n-th solutions, multiplied by the coefficient of the higher power. In your case,

6c^2+17c-14 = 6(c+7/2)(c-2/3)

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