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"
