Question

Using synthetic division, I solved the equation below, however, I'm stuck trying to write it in the following format `" dividend"= "quotient" * "divisor"+ "remainder"`? `x^3-3x^2+7x-1 -: x-3` thanks in advance

Answer 1

`x^3-3x^2+7x-1=(x^2+7)(x-3)+20`

Explanation:

`(x^3-3x^2+7x-1)/(x-3)=(x^2(x-3)+7x-21+20)/(x-3) =`

`x^2+(7(x-3)+20)/(x-3)=x^2+7+20/(x-3)`

Now we multiply through by `(x-3)` giving

`x^3-3x^2+7x-1=(x^2+7)(x-3)+20`

Answer 2

`color(blue)(""(x^3-3x^2+7x-1))=color(lime)(""(1x^2+0x+7)) * color(red)(""(x-3))+color(magenta)(20)`

Explanation:

Let's start with the terms: `color(blue)("dividend")` and `color(red)("divisor")`

In the expression: `color(blue)(x^3-3x^2+7x-1) div color(red)(x-3)`
`color(blue)(x^3-3x^2+7x-1)` is the `color(blue)("dividend")`
and
`color(red)(x-3)` is the `color(red)("divisor")`

Performing synthetic division:
#{: (color(white)("xxx"),,color(gray)(x^3),color(gray)(x^2),color(gray)(x^1),color(gray)(x^0)), (," | ",1,-3,7,-1), (ul(+color(white)("X")),ul(" | "),ul(color(white)("xxx")),ul(3),ul(0),ul(21)), (xx3," | ",color(lime)1,color(lime)0,color(lime)7,color(magenta)20), (,,color(gray)(x^2),color(gray)(x^1),color(gray)(x^0),color(gray)("Remainder")) :}#

`color(lime)1`, `color(lime)0`, and `color(lime)7` are the coefficients of the `color(lime)("quotient")`
and
`color(magenta)(20)` is the `color(magenta)("Remainder")`

Therefore in the form
`color(white)("XXX")color(blue)("dividend")=color(lime)("quotient") * color(red)("divisor") + color(magenta)("remainder")`
we have
`color(white)("XXX")color(blue)(""(x^3-3x^2+7x-1))=color(lime)(""(1x^2+0x+7)) * color(red)(""(x-3))+color(magenta)(20)`