Question

Find the quotient. Justify your answer. x^5+2x^4-7x^2-19x+15/x^2+2x+5

Answer 1

`(x^5+2x^4-7x^2-19x+15)/(x^2+2x+5) = x^3-5x+3`

Explanation:

To find the quotient we need to apply long division on
`(x^5+2x^4-7x^2-19x+15)/(x^2+2x+5)`, as follows:

1. Divide
   `(x^5+2x^4-7x^2-19x+15)/(x^2+2x+5)`

Divide the leading coefficients of the numerator
`x^5+2x^4-7x^2-19x+15` and the divisor
`x^2+2x+5`

`(x^5)/(x^2)=x^3`

Quotient =
`x^3`

Multiply
`x^2+2x+5` by
`x^3` =
`x^5+2x^4+5x^3`

Subtract
`x^5+2x^4+5x^3` from
`x^5+2x^4-7x^2-19x+15` to get new remainder

Remainder =
`-5x^3-7x^2-19x+15`

Therefore

`(x^5+2x^4-7x^2-19x+15)/(x^2+2x+5)=x^3+(-5x^3-7x^2-19x+15)/(x^2+2x+5)`

2. Divide
`(-5x^3-7x^2-19x+15)/(x^2+2x+5)`

Divide the leading coefficients of the numerator
`-5x^3-7x^2-19x+15` and the divisor
`x^2+2x+5`

`(-5x^3)/(x^2)=-5x`

Quotient =
`-5x`

Multiply
`x^2+2x+5` by
`-5x` =
`-5x^3-10x^2-25x`

Subtract
`-5x^3-10x^2-25x` from
`-5x^3-7x^2-19x+15` to get new remainder

Remainder =
`3x^2+6x+15`

Therefore

`(-5x^3-7x^2-19x+15)/(x^2+2x+5)=-5x+(3x^2+6x+15)/(x^2+2x+5)`

`(x^5+2x^4-7x^2-19x+15)/(x^2+2x+5)=x^3-5x+(3x^2+6x+15)/(x^2+2x+5)`

3. Divide
`(3x^2+6x+15)/(x^2+2x+5)`

Divide the leading coefficients of the numerator
`3x^2+6x+15` and the divisor
`x^2+2x+5`

`(3x^2)/(x^2)=3`

Quotient = 3

Multiply
`x^2+2x+5` by 3 =
`\:3x^2+6x+15`

Subtract
`\:3x^2+6x+15` from
`3x^2+6x+15` to get new remainder

Remainder = 0

Therefore

`(3x^2+6x+15)/(x^2+2x+5)=3`

`(x^5+2x^4-7x^2-19x+15)/(x^2+2x+5) = x^3-5x+3`