Question

(7y-1)(y в квадрате-5у+1),5(х+2)(х+3)​

Answer 1

$(7y - 1) \times ( {y}^{2} - 5y + 1) = \\ = 7 {y}^{3} - 35 {y}^{2} + 7y - {y}^{2} + 5y - 1 = \\ 7 {y}^{3} - 36 {y}^{2} + 12y - 1$
$5 \times (x + 2) \times (x + 3) = \\ = (5x + 10) \times (x + 3) = \\ = 5 {x}^{2} + 25x + 30$