2.5. Dividing Polynomials
In Section P.6 we described a method, similar to long division, for dividing polynomials. In this section, we will discuss univariate polynomial division in greater detail. In particular, we will state a useful theorem and introduce a powerful method, known as synthetic division, for performing certain polynomial divisions.
In our two polynomial division examples of Section P.6, you might have noticed that the degree of the remainder was in each case less than the degree of the divisor. This is in fact a general result, which we state below without proof.
- Theorem 2.5.1: If p1(x) and p2(x) are univariate polynomials, then there exist unique polynomials q(x) and r(x) such that p1(x) = q(x) p2(x) + r(x) with deg(r) < deg(p2).
In the above expression, the polynomial q(x) is called the quotient of the polynomials p1(x) and p2(x) and the polynomial r(x) is known as the remainder of p1(x) and p2(x). The long polynomial division method described in Section P.6 finds q(x) and r(x). To refamiliarize you with this method, we present another example.
Example 1: Find the quotient and remainder of the polynomial p1(x) = 8x4 + 16x3 - 10x2 + 21x - 25 divided by the polynomial p2(x) = 2x2 + 3x - 1.
Solution: The polynomial division is shown below.
Thus we have q(x) = 4x2 + 2x - 6 and r(x) = 41x - 31, whence 8x4 + 16x3 - 10x2 + 21x - 25 = (4x2 + 2x - 6)(2x2 + 3x - 1) + 41x - 31, or equivalently,
There is a nice method known as synthetic division for dividing a polynomial by a linear polynomial. We illustrate this method with a couple examples:
- Example 1: Find the quotient and remainder of the polynomial p1(x) = x3 + 6x2 + 11x - 8 divided by the polynomial p2(x) = x + 2.
- Solution: The long polynomial division is shown below.
Thus we have (x3 + 6x2 + 11x - 8) / (x + 2) = x2 + 4x + 3 + 2 / (x + 2).
Synthetic division provides a shorthand for the same calculation. In synthetic division, no powers of the variable x are written down, just rows of numbers. On the top row, we write down the coefficients of the dividend, proceded to the left by the negative of the constant coefficient of the divisor divided by the linear coefficient, which in this case is 1. The result in this case is -2. This number is the root we are testing. Two rows down, we copy down the leading coefficient of the dividend. Then on the second row, we multiply this coefficient by the root -2 and write down the result one space to the right as shown. Then we add the corresponding coefficient in the top row, which is 6 in this case, and write down the result (4) in the third row. Once again, we multiply this number by the root, obtaining -8 in this case, which we write down in the second row one space to the right. We continue in this manner until each row is completed up to the rightmost coefficient of the dividend. The result is shown below.
Thus, once again we see that the quotient is x2 + 4x + 3 and the remainder is 2.