Question

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

Answer 1

`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)`