Question

Разложите на множители квадратный трехчлен x^2+21x+80

Answer 1

$x^{2}+21x+80=0\\\\D=21^{2}-4*80=441-320=121=11^{2}\\x_{1}=\frac{-21+11}{2}=-5\\\\x_{2}=\frac{-21-11}{2}=-16\\\\x^{2}+21x+80=(x+5)(x+16)$

Answer 2

x²+21x+80

a)x1+x2=-21, x1x2=80, x1=-5,x2=-16

 x²+21x+80=(x+5)(x+16)

b)D=b²-4ac, b=21, a=1, c=80

  D=441-320=121, √D=√121=11

x1=(-21+11)/2=-5 , x2=(-21-11)/2=-32/2=-16

x²+21x+80=(x+5)(x+16)