Question

How do you divide `(2v^3-20v^2+56v-46)div(v-6)` using synthetic division?

Answer 1

The remainder is `=2` and the quotient is `=2v^2-8v+8`

Explanation:

Let's perform the synthetic division

`color(white)(aaaa)` `6` `color(white)(aaaaa)` `|` `color(white)(aaaa)` `2` `color(white)(aaaaaa)` `-20` `color(white)(aaaaaa)` `56` `color(white)(aaaaa)` `-46`
`color(white)(aaaaaaaaaaaa)`_________

`color(white)(aaaa)` `color(white)(aaaaaaa)` `|` `color(white)(aaaa)` `color(white)(aaaaaaaa)` `12` `color(white)(aaaa)` `-48` `color(white)(aaaaaaa)` `48`
`color(white)(aaaaaaaaaaaa)`________

`color(white)(aaaa)` `color(white)(aaaaaaa)` `|` `color(white)(aaaa)` `2` `color(white)(aaaaaa)` `-8` `color(white)(aaaaaaa)` `8` `color(white)(aaaaaaaa)` `color(red)(2)`

The remainder is `=2` and the quotient is `=2v^2-8v+8`

`(2v^3-20v^2+56v-46)/(v-6)=2v^2-8v+8+2/(v-6)`

Answer 2

`2v^2 -8v+8 " remainder " 2`

Or `" "2v^2 -8v+8 + 2/(v-6)`

Explanation:

In synthetic division, you use only the coefficients, making sure that all powers of `v` are considered:

`color(red)(2)v^3 color(red)(-20)v^2 color(red)(+56)v^1 color(red)(-46)v^0`

Make `v-6=0" "rarr v = color(blue)(6)" "larr` this is written 'outside'.

`color(white)(xxx.x)|color(red)(2) " "color(red)(-20)" " color(red)(+56)" " color(red)(-46)`
`color(white)(xx.xx)|darr`
`color(white)(xx.x)color(blue)(6)|ul(color(white)(xxxxx)12" "-48" "48`
`color(white)(xxxxxx)color(red)(2)" "-8" "+8" "2larr` the remainder

Steps are as follows:

Bring down the `color(red)(2)`

Multiply `color(blue)(6) xx color(red)(2) = 12`

Add `-20+12 = -8`

Multiply `color(blue)(6) xx -8 = -48`

Add `+56-48 = +8`

Multiply `color(blue)(6) xx 8 =48`

Add `-46+48 =+2`

What do the numbers tell us?

`2v^3div v= 2v^2" "larr` this is the first term in the quotient.

The numbers represent the co-efficients of the following terms:

`" "color(red)(2)" "-8" "+8" "2larr` the remainder

`" "2v^2" "-8v" "+8" remainder" 2`

This is the quotient.