Question

divide (simplify like a mixed fraction)
p²+2p-22÷p+6

Answer 1

Let p² + 2p -22 =  (p+6) ( a p + b)  + c
   Where a, b  & c are some numbers.  Let us find a, b, and c
             So p² + 2 p - 22  =  a p² + ( b p + 6 a p ) + (6b + c)

  Equating corresponding terms on both sides, we get
               a  = 1          6a + b   = 2  =>    b = 2 - 6* 1 = -4
               6 b + c = -22    =>    c = -22 - 6 * -4 = 2

P² + 2 P - 22          (p+6) (p-4) + 2
----------------      =    -----------------          = p - 4  + 2 /  (p+6)
P + 6                          p+6

Answer 2

$w(p)=p^2+2p-22\\\\q(p)=p+6\\\\w(p):q(p)=s(p)+c\\\\c\ is\ equal\ w(-6)\\\\c=(-6)^2+2(-6)-22=36-12-22=2\\\\s(p)=ap+b\\\\w(p)=q(p)\times s(p)+c$

$p^2+2p-22=(p+6)(ap+b)+2\\\\p^2+2p-22=ap^2+bp+6ap+6b+2\\\\p^2+2p-22=ap^2+(b+6a)p+6b+2\\\Updownarrow\\a=1\ \wedge\ b+6a=2\ \wedge\ 6b+2=-22\\\\a=1\ \wedge\ b+6(1)=2\ \wedge\ 6b=-22-2\\\\a=1\ \wedge\ b=2-6\ \wedge\ 6b=-24\\\\a=1\ \wedge\ b=-4\ \wedge\ b=-4$

$Solution:(p^2+2p-22):(p+6)=p-4+\frac{2}{p+6}$