How do you divide #(x^2-5x-12)div(x-3)#?

1 Answer
Sep 21, 2016

#(x^2-5x-12)div(x-3) = (x-2) " rem "-18#

This can also be written as

#(x^2-5x-12)div(x-3) = (x-2) - 18/(x-3)#

Explanation:

The algebraic long division follows the same method as arithmetic long division...

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

  1. Write the dividend in the 'box' making sure that the indices are in descending powers of x.

  2. Divide the first term in divisor into the term in the dividend with the highest index. Write the answer at the top,

  3. Multiply it by BOTH terms of the divisor at the side

  4. Subtract - remember to change signs

  5. Bring down the next term

Repeat steps 2 to 5

#color(white)(xxxxxxxxxx)color(red)(x)color(blue)(-2) " "rem -18#
#color(white)(x)x-3 |bar( x^2 -5x -12)" "larr x^2divx = color(red)(x)#
#color(white)(xxxxxx)ul(color(red)(x^2-3x))" "darr" "larr# subtract (change signs)
#color(white)(xxxxxxxx) -2x-12""larr# bring down the -12,#-2x div x = color(blue)(-2)#
#color(white)(xxxxxxxx.)ul(color(blue)(-2x+6))" "larr# subtract (change signs)
#color(white)(xxxxxxxxxxx)-18 " "larr#remainder

#(x^2-5x-12)div(x-3) = (x-2) " rem "-18#

This can also be written as

#(x^2-5x-12)div(x-3) = (x-2) - 18/(x-3)#