Question

Suppose you use polynomial long division to divide p(x) by q(x), and after doing your calculations you end up with the polynomial $x^2-x+3$ as the quotient above the top line, and the polynomial 3x − 1 at the bottom as the remainder. Then$p\left(x\right)=\_\_\_and\frac{p\left(x\right)}{q\left(x\right)}=\_\_\_\_$

Short answer

The required values are

$p\left(x\right)=q\left(x\right)(x^2-x+3)+(3x+1)and\frac{p\left(x\right)}{q\left(x\right)}=(x^2-x+3)+\frac{(3x+1)}{q\left(x\right)}$

Step-by-step solution

Step 1. Given Information

The given data is that we usepolynomial long division to divide p(x) by q(x), and after doing your calculations you end up with the polynomial $x^2-x+3$as the quotient above the top line, and the polynomial 3x − 1 at the bottom as the remainder.

Step 2. Explanation

Suppose $\frac{p\left(x\right)}{q\left(x\right)}$is an improper rational function where p(x) and q(x) are polynomial functions with $deg\left(q\right(x\left)\right)=m<n$. Then,

$\frac{p\left(x\right)}{q\left(x\right)}=s\left(x\right)+\frac{r\left(x\right)}{q\left(x\right)}$

For some polynomial s(x) of degree $n-m$and some proper rational function $\frac{r\left(x\right)}{q\left(x\right)}$with $deg\left(r\left(x\right)\right)<m$

Thus, we get,

$p\left(x\right)=q\left(x\right)(x^2-x+3)+(3x+1)and\frac{p\left(x\right)}{q\left(x\right)}=(x^2-x+3)+\frac{(3x+1)}{q\left(x\right)}$