$A(x)$=$ax^{2}$ + $bx$ + $c$
= $ax^{2}$ +$(5a+c)x$ + $c$ ($vì$ $b$=$5a$+$c$)
=$ax^{2}$ + $5ax$ + $cx$ + $c$
$A(1)$=$a$ + $5a$ + $2c$
$A(-3)$=$9a$ - $15a$ - $3c$ + $c$
=$9a$ - $15a$ - $2c$
$A(1)$+$A(-3)$=$a$ + $5a$ + $2c$+$9a$ - $15a$ - $2c$
=($a$+ $5a$+$9a$ - $15a$)+($2c$-$2c$)
=0
$do$ $đó$ $A(1)$=$-A(-3)$
$hay$ $A(1)$.$A(-3)$=$-A(-3)$.$A(-3)$=-$A(-3)$²≤0 $với$ $mọi$ $x$