The method is very easy, but the process is a bit difficult to explain.

Follow the colours.

#(5x^3-2x^2+5x-16)div(x-1) = ?????????#

#" (dividend) " div " (divisor)" = ("quotient")#

#color(magenta)("step 1:")# The dividend must be in descending powers of x.

#color(white)(xxxxxxxxxx)5x^3-2x^2+5x-16#

#color(white)(xxxxxxxx) rArr color(magenta)(5" -2 +5 -16") #

Write the numerical coefficients in the #color(magenta)("top row")#.

(If there are any powers missing, leave a space or fill in a zero).

#color(orange)("Step 2")#: Make the divisor = 0. # " " (x-1) = 0 rArr x = color(orange)(1) " this goes outside"#

**Step 3** : Begin the division:

#"Bring down the " color(brown)( 5 ) " to below the line"#

#"multiply " color(orange)(1) xx color(brown)(5) = color(red)(5)#

#"Add "-2+color(red)(5) = color(blue)(+3)#

#"multiply " color(orange)(1) xx color(blue)(+3) = color(blue)(3)#

#"Add " 5 color(blue)( +3) = color(olive)(+8)#

#"multiply " color(orange)(1) xxcolor(olive)(8) = color(olive)(8)#

#"Add " -16 +color(olive)(8) = color(teal)(-8)#

That's it Folks!

#color(white)(xxxxx) | color(brown)(5)" "-2" " 5 " "-16 color(magenta)(" step 1 top row")#

#color(white)(x....x)color(orange)(1) ""| darr " "color(red)(5) " "color(blue)(3) " "color(olive)(8)#

#color(white)(xxxxxx) ul(" ")#

#color(white)(xxxxxxx) color(brown)(5) " "color(blue)(+3) " "color(olive)(+8)" "color(teal)(-8) larr " the remainder!"#

#color(white)(xxxx.xx)uarr " "uarr" "uarr#

#color(white)(xxxxxxx) x^2 " "x^1 " "x^0#

We have now found the numerical coefficients of the terms in the quotient (answer)

We divided an expression with #x^3# by an expression with #x#,

so the first term will be #x^3/x = x^2#

The quotient is #5x^2 +3x +8 -8 #

The last value is the remainder. In this case it is #color(teal)(-8)#

This means that #(x-1)# is not a factor of #5x^3-2x^2+5x-16#

#(5x^3-2x^2+5x-16) div(x-1) = 5x^2 +3x+8 " rem -8"#