# How do you find the product (5y-4)(3y-1)?

Apr 7, 2017

$\left(5 y - 4\right) \times \left(3 y - 1\right) = 5 y \left(3 y - 1\right) - 4 \left(3 y - 1\right)$ Now multiply across the parenthesis
$5 y \left(3 y - 1\right) - 4 \left(3 y - 1\right) = 15 {y}^{2} - 5 y - 12 y + 4$ combining like terms gives
$15 {y}^{2} + \left(- 5 y - 12 y\right) + 4 = {15}^{2} - 17 y + 4$