Question

How do you factor by grouping `6c^2+29c-42`?

Answer 1

`6c^2+29c-42`

We are looking for factors, `m` and `n`, of `6`
and
factors, `p` and `q`, of `(-42)`
such that
`mq+np=29`

• this is because `(mc+p)*(nc+q) = mnc^2+(mq+np)c +pq`

factors of 6 `= {(1,6),(2,3)}` plus their negatives
factors of (-42) # = {(6,-7),(3,-14),(2,-21),(1,-42)} and their negatives

A bit of testing combinations comes up with the pairs
`(p,q) = (6,1)`
and
`(m,n)=(-7,6)`
that satisfy the requirement for the middle term

`6c^2+29c-42 = (6c-7)(1c+6)`

Answer 2

Multiply `6 xx -42 = -252`

Now we look for factor of `-252` that add to get `29`

Because the product is negative and the sum is positive, we want the smaller number to be negative and the larger positive:
-1 252 does not add up to 29
-2 126 does not add up to 29
-3 84
-4 63
-5 is not a factor
-6 42

-7 36 does add up to 29

In `6c^2 + 29c -42` split the `29c` into `-7c+36c`
(or `36c-7c` -- either will work)

`6c^2 + 29c -42 = 6c^2 + -7c + 36c -42`

`color(white)"ssssssssssssssss"` `=(6c^2 -7c) + (36c -42)`

`color(white)"ssssssssssssssss"` `=c(6c -7) + 6(6c - 7)`

`color(white)"ssssssssssssssss"` `=(c+ 6)(6c - 7)`