How do you simplify #(c-6)/(3c^2-17c-6)#?

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Answer 1 · 2018-05-22T19:38:02

You break the denominator using middle term break up method

#(c-6)/(3c^2-17c-6)#

Let's break(factor) the denominator first using middle term break up method.
#3c^2-17c-6#

=#3c^2-18c+c-6#

=#3c(c-6)+1(c-6)#

We are taking common factors 3c from #3c^2-18c# and +1 from #c-6#
=#(c-6)(3c+1)#

Now let's put it in the math :)
#(c-6)/((c-6)(3c+1))#

The same factor (c-6) can be divided/cancelles which will result in 1.

#1/(3c+1)#

Here's your answer.

If you don't know about middle term break up do let me know

Answer 2 · 2018-05-22T19:41:12

#frac{c-6}{3c^2-17c-6} = frac{1}{3c+1}#

#frac{c-6}{3c^2-17c-6}#

Factor the denominator:

# = frac{c-6}{3c^2-18c+c-6}#

# = frac{c-6}{3c(c-6)+(c-6)}#

# = frac{c-6}{(3c+1)(c-6)}#

Simplify by cancelling:

# = frac{cancel((c-6))}{(3c+1)cancel((c-6))}#

# = 1/(3c+1)#