How do you simplify (c-6)/(3c^2-17c-6)c63c217c6?

2 Answers
May 22, 2018

You break the denominator using middle term break up method

Explanation:

(c-6)/(3c^2-17c-6)c63c217c6

Let's break(factor) the denominator first using middle term break up method.
3c^2-17c-63c217c6

=3c^2-18c+c-63c218c+c6

=3c(c-6)+1(c-6)3c(c6)+1(c6)

We are taking common factors 3c from 3c^2-18c3c218c and +1 from c-6c6
=(c-6)(3c+1)(c6)(3c+1)

Now let's put it in the math :)
(c-6)/((c-6)(3c+1))c6(c6)(3c+1)

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

1/(3c+1)13c+1

Here's your answer.

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

May 22, 2018

frac{c-6}{3c^2-17c-6} = frac{1}{3c+1}c63c217c6=13c+1

Explanation:

frac{c-6}{3c^2-17c-6}c63c217c6

Factor the denominator:

= frac{c-6}{3c^2-18c+c-6}=c63c218c+c6

= frac{c-6}{3c(c-6)+(c-6)}=c63c(c6)+(c6)

= frac{c-6}{(3c+1)(c-6)}=c6(3c+1)(c6)

Simplify by cancelling:

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

= 1/(3c+1)