Unit 2, lesson 24

Polynomial Division

How to divide one polynomial by another with long division or synthetic division, what the quotient and remainder mean, and how to check a division by multiplying back.

  • Theory part 47
  • Practice part 48
  • Polynomial Functions

Dividing 1717 by 55 gives 33 with 22 left over, and you can check it because 5⋅3+2=175 \cdot 3 + 2 = 17. Polynomials divide the same way, with a quotient, a remainder and the same multiply back check. The result is often more useful than the fraction you started with, because it splits a complicated quotient of polynomials into a polynomial plus a small leftover piece.

Key terms

Dividend and divisor

Definition When a polynomial f(x)f(x) is divided by a polynomial d(x)d(x), the polynomial ff is the dividend and dd is the divisor. The divisor must not be the zero polynomial. Both are written in standard form, with the powers in decreasing order, and every power missing from the dividend gets the coefficient 00 so that like terms line up during the division.

In plain words The dividend is what gets divided and the divisor is what you divide by, just as in 17÷517 \div 5.

Example To divide x4−3x+1x^4 - 3x + 1 by x2−2x^2 - 2, write the dividend as

x4+0x3+0x2−3x+1x^4 + 0x^3 + 0x^2 - 3x + 1

and use the divisor x2−2x^2 - 2. The zero terms hold a place for each power, the way the digit 00 does in the number 10051005.

Quotient and remainder

Definition The division algorithm states that for polynomials ff and dd with real coefficients, with dd not the zero polynomial, there are unique polynomials qq and rr such that

f(x)=d(x) q(x)+r(x),r=0 or deg⁡r<deg⁡d.\begin{gathered} f(x) = d(x) \, q(x) + r(x), \\[0.4em] r = 0 \ \text{or} \ \deg r < \deg d. \end{gathered}

The polynomial qq is the quotient and rr is the remainder. The condition on the degree of rr is what makes qq and rr unique.

In plain words The quotient says how many times the divisor fits into the dividend, and the remainder is what is left over. The remainder must be 00 or have smaller degree than the divisor, just as a remainder after dividing by 55 must be less than 55.

Example Dividing 2x2+7x+82x^2 + 7x + 8 by x+2x + 2 gives the quotient 2x+32x + 3 and the remainder 22, because

(x+2)(2x+3)+2=(2x2+7x+6)+2=2x2+7x+8.\begin{aligned} &(x + 2)(2x + 3) + 2 \\[0.4em] &\quad = (2x^2 + 7x + 6) + 2 \\[0.4em] &\quad = 2x^2 + 7x + 8. \end{aligned}

The identity 2x2+7x+8=(x+2)(2x+1)+(2x+6)2x^2 + 7x + 8 = (x + 2)(2x + 1) + (2x + 6) is also true, but 2x+62x + 6 has degree 11, which is not less than the degree of x+2x + 2. So 2x+62x + 6 is not the remainder, and that division was stopped too early.

Long division

Definition Long division finds qq and rr by repeating one step, starting with the whole dividend as the current polynomial. Divide the leading term of the current polynomial by the leading term of dd to get the next term of the quotient. Multiply dd by that term and subtract the product from the current polynomial. The leading term cancels, so the degree drops with each step. Stop as soon as the current polynomial is 00 or has degree less than deg⁡d\deg d. That last polynomial is the remainder, and the sum of the terms found is the quotient.

In plain words It is the same routine as long division of whole numbers. Match the leading term, multiply, subtract, and repeat until what is left is too small to divide.

Example Divide f(x)=x4+2x3−x+5f(x) = x^4 + 2x^3 - x + 5 by d(x)=x2+1d(x) = x^2 + 1, writing the dividend as x4+2x3+0x2−x+5x^4 + 2x^3 + 0x^2 - x + 5.

StepNext quotient termSubtractLeft over
1x4÷x2=x2x^4 \div x^2 = x^2x2(x2+1)=x4+x2x^2(x^2 + 1) = x^4 + x^22x3−x2−x+52x^3 - x^2 - x + 5
22x3÷x2=2x2x^3 \div x^2 = 2x2x(x2+1)=2x3+2x2x(x^2 + 1) = 2x^3 + 2x−x2−3x+5-x^2 - 3x + 5
3−x2÷x2=−1-x^2 \div x^2 = -1−1(x2+1)=−x2−1-1(x^2 + 1) = -x^2 - 1−3x+6-3x + 6

The leftover −3x+6-3x + 6 has degree 11, less than 22, so the division stops. The quotient is q(x)=x2+2x−1q(x) = x^2 + 2x - 1 and the remainder is r(x)=−3x+6r(x) = -3x + 6. Multiplying back confirms it.

(x2+1)(x2+2x−1)+(−3x+6)=(x4+2x3+2x−1)+(−3x+6)=x4+2x3−x+5.\begin{aligned} &(x^2 + 1)(x^2 + 2x - 1) + (-3x + 6) \\[0.4em] &\quad = (x^4 + 2x^3 + 2x - 1) \\[0.4em] &\qquad + (-3x + 6) \\[0.4em] &\quad = x^4 + 2x^3 - x + 5. \end{aligned}

Synthetic division

Definition Synthetic division is a shortcut for dividing a polynomial f(x)=anxn+⋯+a1x+a0f(x) = a_n x^n + \dots + a_1 x + a_0 of degree n≥1n \ge 1 by a divisor of the form x−cx - c. List the coefficients an,…,a0a_n, \dots, a_0, with zeros for missing powers. The bottom row starts with bn−1=anb_{n - 1} = a_n, and each next entry is the next coefficient plus cc times the previous bottom entry,

bk−1=ak+c bkfor k=n−1,…,1,r=a0+c b0.\begin{gathered} b_{k - 1} = a_k + c \, b_k \\[0.4em] \text{for } k = n - 1, \dots, 1, \\[0.8em] r = a_0 + c \, b_0. \end{gathered}

Then the quotient is q(x)=bn−1xn−1+⋯+b1x+b0q(x) = b_{n - 1} x^{n - 1} + \dots + b_1 x + b_0 and the remainder is the constant rr. For a divisor x+cx + c use −c-c, since x+c=x−(−c)x + c = x - (-c). For a divisor ax−bax - b with a≠0a \ne 0, divide by x−bax - \frac{b}{a} to get a quotient Q(x)Q(x) and a remainder rr, and then divide Q(x)Q(x) by aa. The remainder stays the same, because

f(x)=(x−ba)Q(x)+r=(ax−b) Q(x)a+r.\begin{aligned} f(x) &= \left(x - \frac{b}{a}\right) Q(x) + r \\[0.8em] &= (ax - b) \, \frac{Q(x)}{a} + r. \end{aligned}

In plain words It is long division by x−cx - c with the powers of xx left out. Only the numbers are written, and multiplying by cc and adding takes the place of multiplying and subtracting.

Example Divide 2x3−5x2+72x^3 - 5x^2 + 7 by x−3x - 3, so c=3c = 3, with the coefficients 2,−5,0,72, -5, 0, 7.

Rowx3x^3x2x^2xxconstant
Coefficients22−5-50077
33 times the previous bottom entry663399
Bottom row2211331616

The quotient is 2x2+x+32x^2 + x + 3 and the remainder is 1616. Multiplying back gives

(x−3)(2x2+x+3)+16=(2x3−5x2−9)+16=2x3−5x2+7.\begin{aligned} &(x - 3)(2x^2 + x + 3) + 16 \\[0.4em] &\quad = (2x^3 - 5x^2 - 9) + 16 \\[0.4em] &\quad = 2x^3 - 5x^2 + 7. \end{aligned}

Zero remainder and factors

Definition The divisor dd divides ff, or is a factor of ff, when the remainder is 00, that is when f(x)=d(x) q(x)f(x) = d(x) \, q(x) for a polynomial qq. When the divisor is linear, x−cx - c, the remainder is always a constant, because it is 00 or has degree less than 11.

In plain words A zero remainder means the division comes out even, and you get a factorization for free. The next lesson shows a quick way to know the remainder before dividing.

Example Divide x3−8x^3 - 8 by x−2x - 2 with synthetic division, using the coefficients 1,0,0,−81, 0, 0, -8 and c=2c = 2. The bottom row is 1,2,4,01, 2, 4, 0, so the remainder is 00 and

x3−8=(x−2)(x2+2x+4).x^3 - 8 = (x - 2)(x^2 + 2x + 4).

Polynomial plus a proper fraction

Definition Dividing both sides of f=dq+rf = dq + r by dd gives

f(x)d(x)=q(x)+r(x)d(x)\frac{f(x)}{d(x)} = q(x) + \frac{r(x)}{d(x)}

at every xx with d(x)≠0d(x) \ne 0. When r≠0r \ne 0 the fraction r(x)d(x)\frac{r(x)}{d(x)} is proper, which means that its numerator has smaller degree than its denominator. The numbers where d(x)=0d(x) = 0 stay excluded on both sides.

In plain words Division splits a top-heavy fraction into a polynomial part and a leftover fraction. As ∣x∣\lvert x \rvert grows, a proper fraction shrinks toward 00, so far out the graph of f/df/d stays close to the graph of the polynomial qq.

Example Dividing x2+3x+5x^2 + 3x + 5 by x+1x + 1 gives the quotient x+2x + 2 and the remainder 33, so

x2+3x+5x+1=x+2+3x+1,x≠−1.\begin{gathered} \frac{x^2 + 3x + 5}{x + 1} = x + 2 + \frac{3}{x + 1}, \\[0.8em] x \ne -1. \end{gathered}

At x=99x = 99 the left side is 10103100=101.03\frac{10103}{100} = 101.03, while the line y=x+2y = x + 2 gives 101101. The gap of 3100\frac{3}{100} keeps shrinking as xx grows, which is the idea behind slant asymptotes.

Check yourself

  1. Use long division to divide 2x3+x2−4x+32x^3 + x^2 - 4x + 3 by x2−x+1x^2 - x + 1.
Show the answer

The first quotient term is 2x2x. Subtracting 2x(x2−x+1)=2x3−2x2+2x2x(x^2 - x + 1) = 2x^3 - 2x^2 + 2x leaves 3x2−6x+33x^2 - 6x + 3. The next term is 33, and subtracting 3(x2−x+1)3(x^2 - x + 1) leaves −3x-3x, which has degree less than 22. So the quotient is 2x+32x + 3 and the remainder is −3x-3x. As a check,

(x2−x+1)(2x+3)−3x=(2x3+x2−x+3)−3x=2x3+x2−4x+3.\begin{aligned} &(x^2 - x + 1)(2x + 3) - 3x \\[0.4em] &\quad = (2x^3 + x^2 - x + 3) - 3x \\[0.4em] &\quad = 2x^3 + x^2 - 4x + 3. \end{aligned}
  1. Use synthetic division to divide x4−3x2+2x−5x^4 - 3x^2 + 2x - 5 by x+2x + 2.
Show the answer

Here c=−2c = -2 and the coefficients are 1,0,−3,2,−51, 0, -3, 2, -5. The bottom row is 1,−2,1,0,−51, -2, 1, 0, -5. So the quotient is x3−2x2+xx^3 - 2x^2 + x and the remainder is −5-5.

  1. Divide 6x2+x−46x^2 + x - 4 by 2x−12x - 1.
Show the answer

Use synthetic division with c=12c = \frac{1}{2} on the coefficients 6,1,−46, 1, -4. The bottom row is 6,4,−26, 4, -2, so

6x2+x−4=(x−12)(6x+4)−2=(2x−1)(3x+2)−2.\begin{aligned} &6x^2 + x - 4 \\[0.4em] &\quad = \left(x - \frac{1}{2}\right)(6x + 4) - 2 \\[0.8em] &\quad = (2x - 1)(3x + 2) - 2. \end{aligned}

The quotient is 3x+23x + 2 and the remainder is −2-2.

How this lesson connects