# Is it possible to factor y= 6x^2 - 27x - 15 ? If so, what are the factors?

Feb 27, 2016

$y = 6 {x}^{2} - 27 x - 15 = 3 \left(2 x + 1\right) \left(x - 5\right)$

#### Explanation:

Separate out the common factor $3$, then factor by grouping as follows:

$y = 6 {x}^{2} - 27 x - 15$

$= 3 \left(2 {x}^{2} - 9 x - 5\right)$

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

$= 3 \left(2 x \left(x - 5\right) + 1 \left(x - 5\right)\right)$

$= 3 \left(2 x + 1\right) \left(x - 5\right)$

Note that I found the split of the middle term $- 9 x = - 10 x + x$ using an AC method: Look for a pair of factors of $A C = 2 \cdot 5 = 10$ which differ by $B = 9$. The pair $10$, $1$ works since $10 \times 1 = 10$ and $10 - 1 = 9$.