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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