Question

Divide (3x^4- 2x + 1) by (x² +2)

Answer 1

`3x^2-6-(2x-13)/(x^2+2)`

Explanation:

As the divisor is quadratic, use long division rather than synthetic division:

`\large \begin{array}{r}3x^2-6\phantom{)}\\x^2+2{\overline{\smash{\big)}\,3x^4-2x+1\phantom{)}}}\\{-~\phantom{(}\underline{(3x^4+6x^2)\phantom{-b)}}\\-6x^2-2x+1\phantom{)}\\-~\phantom{()}\underline{(-6x^2\phantom{))))}-12)\phantom{}}\\-2x+13\phantom{)}\\\end{array}`

Therefore:

`\textsf{Dividend}: \quad 3x^4-2x+1`

`\textsf{Divisor}: \quad x^2+2`

`\textsf{Quotient}: \quad 3x^2-6`

`\textsf{Remainder}: \quad -2x+13=-(2x-13)`

When dividing a polynomial, the result is the quotient plus the remainder over the divisor:

`\implies (3x^4-2x+1)/(x^2+2)=3x^2-6-(2x-13)/(x^2+2)`