# How do you factor 10t^2 - 19t + 6?

Mar 10, 2018

#### Answer:

$\left(5 t - 2\right) \left(2 t - 3\right)$

#### Explanation:

$10 {t}^{2} - 19 t + 6$

1) multiply the coeff$\text{ } {x}^{2}$ and the constant.

$\textcolor{red}{10} {t}^{2} - 19 t + \textcolor{red}{6}$

$10 \times 6 = 60$

2) find the factors of this number that add to the middle term ie.$\text{ } - 19$

$- 15 , - 4$

3) now split the middle term with these factors and then factorise by grouping

.$10 {t}^{2} \textcolor{b l u e}{- 19 t} + 6$

$10 {t}^{2} \textcolor{b l u e}{- 15 t - 4 t} + 6$

$\left(10 {t}^{2} - 15 t\right) - \left(4 t - 6\right)$

$5 t \left(2 t - 3\right) - 2 \left(2 t - 3\right)$

$\left(5 t - 2\right) \left(2 t - 3\right)$