Question

(2n^3+15n^2+11n-42)÷(n+6)

Answer 1

We have a polynomial division:

$\frac{2n^3+15n^2+11n-42}{n+6}$

To solve this we take the highest term and divide it by the highest term of the denominator. Then, multiply this result by the denominator and add and substract inthe numerator. Then, we can group in two terms: one exactly divisible by the numerator and the other as remainder. In this way we are left with a division of a lower-grade polynomial andn we can repeat the process until we have an irreductible remainder or the remainder is 0.

First step:

$\frac{2n^3+12n^2-12n^2+15n^2+11n-42}{n+6}=2n^2+\frac{3n^2+11n-42}{n+6}$

Second step:

$2n^2+\frac{3n^2+18n-18n+11n-42}{n+6}=2n^2+3n+\frac{-7n-42}{n+6}$

Third step:

$2n^2+3n+\frac{-7n-7\cdot6}{n+6}=2n^2+3n+\frac{-7(n+6)}{n+6}=2n^2+3n-7$

The result is:

$2n^2+3n-7$