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:

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

$= 1/(3c+1)$

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