Polynomial long division

When you reach for it

Multiplying polynomials takes two factors and gives their product. Division runs the other way: you know a product and one factor, and you want the other. A box’s volume and its floor area give its height. One known factor of a cubic leaves a quadratic, which you can factor further. Division also rewrites a fraction of polynomials whose top has the higher degree as a polynomial plus a fraction whose top has the lower degree, the way \displaystyle \frac{17}{5} becomes \displaystyle 3\frac{2}{5}.

A single-term divisor goes term by term. A longer one needs long division, laid out as long division of whole numbers, one power at a time.

The procedure

Dividing by a monomial

Dividing by d is multiplying by \displaystyle \frac{1}{d}, so the distributive property splits a sum over a single term:

Each piece is then a quotient of monomials: divide the coefficients and subtract the exponents, since \displaystyle \frac{x^m}{x^n} = x^{m - n} by the Laws of exponents.

Dividing by a polynomial

Long division of 4729 by 13 starts at the biggest place. 13 goes into 47 three times, so it goes into 4700 three hundred times, and what is left is passed on. Written as a chain of facts:

Substituting each line into the one above gives 4729 = 13 \cdot 363 + 10: quotient 363, remainder 10. The rounds stop when what is left is smaller than 13.

Now 4729 = 4 \cdot 10^3 + 7 \cdot 10^2 + 2 \cdot 10 + 9 is a polynomial in 10, and a polynomial in x has the same shape with x in place of 10. Its division runs on the same plan, and is in one way easier. Each round looks only at the leading term of what is left and the leading term of the divisor. Their quotient is the next term of the quotient. It is chosen so that, multiplied by the divisor, it reproduces that leading term exactly. Multiply the whole divisor by it and subtract the product from what is left: the leading terms cancel by design, so what is left drops in degree. No guessing how many times 13 fits, and no carrying: a coefficient may be any number, while a digit must stay between 0 and 9.

In the written layout you subtract only from the terms the product touches and then bring down the next term. That is bookkeeping: subtracting from the whole of what is left gives the same thing.

Why the answer is right. At every stage this statement is true for every x:

It is true at the start, with quotient 0 and all of p(x) left. Each round adds a term t to the quotient and takes d(x) \cdot t away from what is left, which changes nothing in total. The rounds stop when what is left is 0 or has degree below the divisor’s, because then its leading term divided by the divisor’s leading term would need a negative power of x. What is left is then the remainder. The result, with quotient q(x) and remainder r(x):

Dividing through by d(x) gives the fraction form, true wherever d(x) \neq 0:

To check a division, multiply back: expand d(x)\,q(x) + r(x) and compare it with p(x), and see that r has lower degree than d. A remainder of 0 means d(x) is a factor of p(x).

Placeholders. The layout needs one column per power. Before the first round, write both polynomials in descending powers and fill any missing power with a zero coefficient, so 6x^3 - x^2 + 4 becomes 6x^3 - x^2 + 0x + 4. Its value is unchanged, and the 0x holds the column where the x terms of later products land.

When the divisor is x - c, Synthetic division is a shorthand for exactly these rounds that keeps only the coefficients.

Worked examples

Boundaries

The divisor is not zero. Dividing by the zero polynomial is undefined, as dividing by the number 0 is. The fraction form \displaystyle \frac{p}{d} = q + \frac{r}{d} holds only where d(x) \neq 0; the product form p = dq + r holds for every x.

A single-term divisor splits the top, never the bottom. \displaystyle \frac{a + b}{d} = \frac{a}{d} + \frac{b}{d}, but \displaystyle \frac{d}{a + b} does not split. And when a term of the top has a lower power than the divisor, the result is not a polynomial: \displaystyle \frac{x^2 + 3}{x} = x + \frac{3}{x}. Long division says the same thing as quotient x, remainder 3.

The rounds stop on degree, not on size. While what is left has degree at least the divisor’s, another round is due. When it drops below, stop, even if the remainder contains x (Example 4). If the dividend’s degree is already below the divisor’s, the quotient is 0 and the remainder is the dividend.

The quotient and remainder are unique. Any route to a q and r with p = dq + r and \deg r < \deg d reaches the same pair (Exercise 23), which is why multiplying back is a complete check.

Fractions can appear. When the divisor’s leading coefficient is not 1, the quotient may have fractional coefficients even though every given coefficient is an integer (Exercise 14).

Where this goes next

Synthetic division runs the same rounds for a divisor x - c with only the coefficients written. The The remainder and factor theorems predicts the remainder on dividing by x - c without dividing at all, and turns a zero remainder into a zero of the polynomial.

Rational expressions, in the next unit, are fractions of polynomials, and \displaystyle q + \frac{r}{d} is their mixed-number form. Far out along the x-axis \displaystyle \frac{r}{d} shrinks toward 0, so the graph follows q; when q has degree 1, its line is a slant asymptote. In calculus, dividing before integrating splits a fraction into pieces that can each be integrated.

Exercises

Dividing by a monomial

1. \dfrac{18x^5 - 12x^3 + 6x^2}{6x^2}
2. \dfrac{20a^4b^3 + 15a^2b^2}{5a^2b}
3. \dfrac{9y^3 - 6y^2 + 4y}{3y}

Dividing by a binomial

Give the quotient and the remainder.

4. (x^2 + 9x + 20) \div (x + 4)
5. (2x^3 + 3x^2 - 8x + 3) \div (x + 3)
6. (x^3 - 4x^2 + 5x - 7) \div (x - 2)
7. (6x^2 + 7x - 5) \div (3x - 1)
8. (4x^3 - 8x^2 + 3x + 5) \div (2x - 3)

Missing powers

9. (x^3 + 2x - 12) \div (x - 2)
10. (2x^4 - 5x^2 + 3) \div (x + 1)
11. (x^4 - 3x + 1) \div (x - 1)

Divisors of degree two

12. (x^4 + x^3 - x^2 + 2x - 3) \div (x^2 + x - 2)
13. (3x^3 - 2x^2 + 4x - 1) \div (x^2 - 2)
14. (2x^3 + x + 5) \div (2x^2 - x + 1)

Problems

15. Write \dfrac{x^2 + 5x + 1}{x + 2} as a polynomial plus a fraction whose top has lower degree than its bottom.
16. Wairimu’s greenhouse room is a box with volume x^3 + 7x^2 + 14x + 8 cubic meters and height x + 1 meters. Find the floor area, factor it to give the floor’s two sides, and find all three dimensions and the volume when x = 2.
17. A rectangular pane for Elfriede’s greenhouse roof has area 6x^2 + x - 12 square decimeters and width 2x + 3 decimeters. Find its length.
18. Jaroslav wants x - 1 to divide x^3 + 2x^2 - 5x + k exactly. Divide to find the remainder in terms of k, then find k.
19. Given that x - 1 is a factor of x^3 - 2x^2 - 5x + 6, factor it completely.

Reasoning

20. Explain why dividing any polynomial by a divisor of degree 1, such as 3x - 5, always leaves a remainder that is a number.
21. Severine divides a cubic by x - 3 and stops when what is left is x + 14. Explain why she is not finished, and find how her quotient and remainder change after the next round.
22. A polynomial of degree 5 is divided by one of degree 2. What is the degree of the quotient, and what can the remainder be? Explain.
23. Suppose p = d\,q_1 + r_1 and p = d\,q_2 + r_2, where each of r_1, r_2 is 0 or has degree below \deg d. Prove that q_1 = q_2 and r_1 = r_2.
24. Divide 4x^3 + 7x^2 + 2x + 9 by x + 3. Then put x = 10 into your quotient and remainder and compare with 4729 = 13 \cdot 363 + 10. Explain why both statements are true and why they look different.