Question

How do you long divide `(6x^2+5x-4) div (2x+3)`?

Answer 1

The answer is`=color(red)(3x-2)+2/(color(blue)(2x+3))`

Explanation:

Let's do the long division

`color(white)(aaaa)` `6x^2+5x-4` `color(white)(aaaa)` `|` `color(blue)(2x+3)`

`color(white)(aaaa)` `6x^2+9x` `color(white)(aaaaaaaa)` `|` `color(red)(3x-2)`

`color(white)(aaaaaa)` `0-4x-4`

`color(white)(aaaaaaaa)` `-4x-6`

`color(white)(aaaaaaaaaaa)` `0+2`

The remainder is `=2`

The quotient is `=color(red)(3x-2)`

`(6x^2+5x-4)/(color(blue)(2x+3))=color(red)(3x-2)+2/(color(blue)(2x+3))`

Answer 2

`(6x^2+5x-4)div(2x+3) = (3x-2) " rem "2`

This can also be written as

`(6x^2+5x-4)div(2x+3) = (3x-2) +2/(2x+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 the divisor into the term in the dividend with the highest index. Write the answer at the top,

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

4. Subtract (remember to change the signs)

5. Bring down the next term

Repeat steps 2 to 5

`color(white)(xxxxxxxxxxx)color(red)(3x)color(blue)(-2)`
`color(white)(x)2x+3 |bar(6 x^2 +5x -4)" "larr 6x^2div2x = color(red)(3x)`
`color(white)(xxxxxxx)ul(color(red)(6x^2+9x))""darr" "larr` subtract
`color(white)(xxxxxxxxx) -4x-4""larr` bring down -4, `-4x div 2x = color(blue)(-2`
`color(white)(xxxxxxx.x.)ul(color(blue)(-4x-6))" "larr` subtract (change signs)
`color(white)(xxxxxxxxxxxxxx)2 " "larr`remainder

`(6x^2+5x-4)div(2x+3) = (3x-2) " rem "2`

This can also be written as

`(6x^2+5x-4)div(2x+3) = (3x-2) +2/(2x+3)`