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

Apr 22, 2015

$6 {c}^{2} + 29 c - 42$

We are looking for factors, $m$ and $n$, of $6$
and
factors, $p$ and $q$, of $\left(- 42\right)$
such that
$m q + n p = 29$

• this is because $\left(m c + p\right) \cdot \left(n c + q\right) = m n {c}^{2} + \left(m q + n p\right) c + p q$

factors of 6 $= \left\{\begin{matrix}1 & 6 \\ 2 & 3\end{matrix}\right\}$ 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
$\left(p , q\right) = \left(6 , 1\right)$
and
$\left(m , n\right) = \left(- 7 , 6\right)$
that satisfy the requirement for the middle term

$6 {c}^{2} + 29 c - 42 = \left(6 c - 7\right) \left(1 c + 6\right)$

Apr 22, 2015

Multiply $6 \times - 42 = - 252$

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 $6 {c}^{2} + 29 c - 42$ split the $29 c$ into $- 7 c + 36 c$
(or $36 c - 7 c$ -- either will work)

$6 {c}^{2} + 29 c - 42 = 6 {c}^{2} + - 7 c + 36 c - 42$

$\textcolor{w h i t e}{\text{ssssssssssssssss}}$ $= \left(6 {c}^{2} - 7 c\right) + \left(36 c - 42\right)$

$\textcolor{w h i t e}{\text{ssssssssssssssss}}$ $= c \left(6 c - 7\right) + 6 \left(6 c - 7\right)$

$\textcolor{w h i t e}{\text{ssssssssssssssss}}$ $= \left(c + 6\right) \left(6 c - 7\right)$