# How do you factor 18c^2-3c-10?

Apr 16, 2016

$18 {c}^{2} - 3 c - 10 = \left(6 c - 5\right) \left(3 c + 2\right)$

#### Explanation:

We can use an AC method:

Look for a pair of factors of $A C = 18 \cdot 10 = 180$, with difference $B = 3$.

The pair $12 , 15$ works.

Then use this pair to split the middle term and factor by grouping:

$18 {c}^{2} - 3 c - 10$

$= 18 {c}^{2} + 12 c - 15 c - 10$

$= \left(18 {c}^{2} + 12 c\right) - \left(15 c + 10\right)$

$= 6 c \left(3 c + 2\right) - 5 \left(3 c + 2\right)$

$= \left(6 c - 5\right) \left(3 c + 2\right)$

$\textcolor{w h i t e}{}$
Alternative method

Alternatively, we can find this factorisation by completing the square then using the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = \left(12 c - 1\right)$ and $b = 9$ as follows:

To make the arithmetic easier, first multiply by $8$, remembering to divide by $8$ at the end:

$8 \left(18 {c}^{2} - 3 c - 10\right)$

$= 144 {c}^{2} - 24 c - 80$

$= \left(12 {c}^{2}\right) - 2 \left(12 c\right) - 80$

$= {\left(12 c - 1\right)}^{2} - 1 - 80$

$= {\left(12 c - 1\right)}^{2} - 81$

$= {\left(12 c - 1\right)}^{2} - {9}^{2}$

$= \left(\left(12 c - 1\right) - 9\right) \left(\left(12 c - 1\right) + 9\right)$

$= \left(12 c - 10\right) \left(12 c + 8\right)$

$= \left(2 \left(6 c - 5\right)\right) \left(4 \left(3 c + 2\right)\right)$

$= 8 \left(6 c - 5\right) \left(3 c + 2\right)$

So dividing both ends by $8$ we find:

$18 {c}^{2} - 3 c - 10 = \left(6 c - 5\right) \left(3 c + 2\right)$

Why did I multiply by $8$ first?

If I multiplied by $2$ then that would make the leading term $36 {c}^{2} = {\left(6 c\right)}^{2}$, but then the completed square would have involved ${\left(6 c - \frac{1}{2}\right)}^{2} = 36 {c}^{2} - 6 c + \frac{1}{4}$. To avoid the messiness of fractions, multiply by an extra factor of ${2}^{2} = 4$ first.