How do you divide #(x^2+7x-5)div(x-2)# using long division?

2 Answers
Oct 1, 2016

#(x^2+7x-5)div(x-2) = (x+9) " rem "13#
OR
#(x^2+7x-5)div(x-2) = (x+9) +13/(x-2)#
OR
#(x^2+7x-5) = (x-2)(x+9) +13#

Explanation:

Algebraic long division follows the same method as arithmetic long division...

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

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

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

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

Step 4. Subtract

Step 5. Bring down the next term

Repeat steps 2 to 5

#color(white)(xxxxxxxxxxxx)color(red)(x )color(blue)( + 9)" rem " 13#
#color(white)(xxx)x-2 |bar( x^2 +7x -5)" "larr x^2divx = color(red)(x)#
#color(white)(xxxxxx)ul(color(red)(-(x^2-2x)))color(white)(.)darr" "larr# subtract (change signs)
#color(white)(xxxxxxxxxxxx) 9x-5""larr# bring down the -5, #9x div x = color(blue)(9)#
#color(white)(xxxxxxxxxx)ul(color(blue)(-(9x-18))" "larr# subtract (change signs)
#color(white)(xxxxxxxxx.xxxxxx)13 " "larr#remainder

#(x^2+7x-5)div(x-2) = (x+9) " rem "13#

This can also be written as

#(x^2+7x-5)div(x-2) = (x+9) +13/(x-2)#

Or#" "(x^2+7x-5) = (x-2)(x+9) +13#

Dec 21, 2016

Alternative format of calculation

#x+9+13/(x-2)#

Explanation:

#" "x^2+7x-5#
#color(red)(x)(x-2)->" "ul(x^2-2x) larr" subtract"#
#" "0+9x-5#
#color(red)(9)(x-2)->" "ul(9x-18)larr" subtract" #
#" "color(red)(0+13 larr" remainder")#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#(x^2=7x-5)-:(x-2)" "=" "color(red)(x+9+)(color(red)(13))/(x-2)#