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How to divide polynomials using long division

Divide, multiply, subtract, and repeat

Write both polynomials in descending powers, inserting zero coefficients for missing terms. Divide the leading terms, multiply the divisor by that quotient term, and subtract. Stop when the remainder has lower degree than the divisor. The identity is $P (x) = D (x) Q (x) + R (x)$ .

Problem

Use long division to simplify:

1. $(6 x^{4} - 20 x^{2} - 16) \div (2 x - 4)$ . 2. $(x^{5} + 6 x^{4} + 4 x^{2} - 13 x - 42) \div (x^{2} + 3)$ . 3. $(x^{3} - 64) \div (x - 4)$ .

Answer

3 x^{3} + 6 x^{2} + 2 x + 4

, with remainder

0

. 2.

x^{3} + 6 x^{2} - 3 x - 14 - \frac{4 x}{x ^{2} + 3}

, with polynomial remainder

- 4 x

. 3.

x^{2} + 4 x + 16

, with remainder

0

The first expression is undefined at x=2 and the third at x=4, even though their simplified quotients are polynomials. The second divisor never vanishes for real x.

Step-by-step solution

1. Divide a quartic by a linear polynomial

Start with $6 x^{4} + 0 x^{3} - 20 x^{2} + 0 x - 16$ . The first quotient term is $6 x^{4} / (2 x) = 3 x^{3}$ . Subtracting $(2 x - 4) (3 x^{3})$ leaves

12 x^{3} - 20 x^{2} - 16.

The next term is $6 x^{2}$ . Subtracting $(2 x - 4) (6 x^{2})$ leaves $4 x^{2} - 16$ . Next use $2 x$ ; subtraction leaves $8 x - 16$ . The last quotient term is $4$ , and the remainder is zero. Thus

6 x^{4} - 20 x^{2} - 16 = (2 x - 4) (3 x^{3} + 6 x^{2} + 2 x + 4) .

2. Divide by a quadratic and keep the remainder

Include the missing term $0 x^{3}$ . The successive quotient terms and remaining polynomials are:

- After $x^{3}$ : $6 x^{4} - 3 x^{3} + 4 x^{2} - 13 x - 42$ . - After $6 x^{2}$ : $- 3 x^{3} - 14 x^{2} - 13 x - 42$ . - After $- 3 x$ : $- 14 x^{2} - 4 x - 42$ . - After $- 14$ : $- 4 x$ .

The remainder has degree 1, below the divisor's degree 2, so division stops. Therefore

\frac{x ^{5} + 6 x ^{4} + 4 x ^{2} - 13 x - 42}{x ^{2} + 3} = x^{3} + 6 x^{2} - 3 x - 14 - \frac{4 x}{x ^{2} + 3} .

A nonzero remainder must not be dropped from the result.

3. Divide a difference of cubes

Write the dividend as $x^{3} + 0 x^{2} + 0 x - 64$ . The leading quotient term $x^{2}$ leaves $4 x^{2} - 64$ . The next term $4 x$ leaves $16 x - 64$ . The final term $16$ leaves zero:

x^{3} - 64 = (x - 4) (x^{2} + 4 x + 16) .

This also agrees with the difference-of-cubes formula.

4. Check the results

Multiply each divisor by its quotient and add the remainder. This must reproduce every coefficient of the original dividend. Keep any exclusions from the original denominator, even when the division is exact.

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