Question

Разложи на множители трёхчлен4y2+20y+24

Answer 1

$4y^2+20y+24=4(y^2+2y+3y+6)=4((y^2+2y)+(3y+6))= \ \ =4(y(y+2)+3(y+2)=4(y+2)(y+3)$

Ответ: $4(y+2)(y+3)$

Answer 2

4y²+20y+24=4(y²+5y+6)=4(y+3)(y+2)
y1+y2=-5 U y1*y2=6⇒y1=-3 U y2=-2