Question

How do you divide `3x^3+5x^2+2x-12` by `x+3`?

Answer 1

`p(x)=3x^2-4x+14+((-54))/(x+3)`

Explanation:

Let `3x^3+5x^2+2x-12` be `p(x)`
Using long division of polynomial's
`(+)3x^2-4x+14`
`root(x+3)(3x^3+5x^2+2x-12)`
`((-)3x^3+9x^2)/(-4x^2+2x-12)`
`((-)-4x^2-12x)/(14x-12)`
`((-)14x+42)/-54`
`p(x)=Q(x)+(R)/D`
Where,`Q=`quotient
`R=`remainder
`D=`divisor
`=>p(x)=3x^2-4x+14+((-54))/(x+3)`

Answer 2

`3x^2-4x+14-54/(x+3)`

Explanation:

`"one way is to use the divisor as a factor in the numerator"`

`"consider the numerator"`

`color(red)(3x^2)(x+3)color(magenta)(-9x^2)+5x^2+2x-12`

`=color(red)(3x^2)(x+3)color(red)(-4x)(x+3)color(magenta)(+12x)+2x-12`

`=color(red)(3x^2)(x+3)color(red)(-4x)(x+3)color(red)(+14)(x+3)color(magenta)(-42)-12`

`=color(red)(3x^2)(x+3)color(red)(-4x)(x+3)color(red)(+14)(x+3)-54`

`"quotient "=color(red)(3x^2-4x+14)," remainder "=-54`

`rArr3x^3+5x^2+2x-12`

`=3x^2-4x+14-54/(x+3)`