# How do you factor 6y^2-5y+1?

Aug 7, 2015

$\left(2 y - 1\right) \left(3 y - 1\right)$

#### Explanation:

When factorizing a quadratic function look for factors of the square coefficient (in this case $6$) and factors of the constant term (in this case $1$).

In this case the coefficient of the $y$ term is negative and the constant term is positive, so the constants in each of the factors must both be negative. Since the constant is $+ 1$, its factors must be $- 1$ and $- 1$.

Integer factors of $6$, independent of order, are

(1,6); (2,3); ("-"1,"-"6); ("-"2,"-"3)

Now look for the factor pair that when multiplied by $\left(- 1 , - 1\right)$ sums to $- 5$. Here this is $\left(2 , 3\right)$. These are the coefficients of the $y$ terms in each factor.

Hence, the factors are $\left(2 y - 1\right)$ and $\left(3 y - 1\right)$.

Note: Not all quadratic functions can be factorized.