$p^{2}x+py+8=0$
(1;−6)
$p^2\ast 1+p\ast (-6)+8=0$
$p^2-6p+8=0$
$(p^2-6p+9)-1=0$
$(p-3{)}^2-1^2=0$
(p − 3 − 1)(p − 3 + 1) = 0
(p − 4)(p − 2) = 0
p − 4 = 0
p = 4
или
p − 2 = 0
p = 2
Ответ: при p = 2 и p = 4

$p^{2}x+py+8=0$
(−1;2)
$p^2\ast (-1)+p\ast 2+8=0$
$-p^2+2p+8=0$ | * (−1)
$p^2-2p-8=0$
$p^2-4p+2p-8=0$
$(p^2-4p)+(2p-8)=0$
p(p − 4) + 2(p − 4) = 0
(p − 4)(p + 2) = 0
p − 4 = 0
p = 4
или
p + 2 = 0
p = −2
Ответ: −2; 4.