Unit 8, lesson 63

Polynomial Division and Useful Rewrites

How to divide one polynomial by another, write the result as divisor times quotient plus remainder, check it by multiplying back, and keep track of where the fraction form is valid.

  • Theory part 125
  • Practice part 126
  • Rewriting Rational Functions

The fraction x2+1x−1\frac{x^2 + 1}{x - 1} looks like one piece, but it is really x+1x + 1 plus a small correction 2x−1\frac{2}{x - 1}. The second form shows at once how the graph behaves for large xx, and it is the form that integration of rational functions starts from. Polynomial division is the tool that finds such rewrites, and multiplying back is how you know the rewrite is right.

Key terms

Degree

Definition A polynomial in xx is an expression anxn+an−1xn−1+⋯+a1x+a0a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 with real coefficients and whole number exponents. If it is not the zero polynomial, its degree is the largest exponent nn with an≠0a_n \ne 0, and ana_n is its leading coefficient. The zero polynomial is given no degree. A nonzero constant has degree 00.

In plain words The degree is the highest power of xx that actually appears. It decides which polynomial wins when xx is large.

Example The polynomial 4x3−x5+24x^3 - x^5 + 2 has degree 55 and leading coefficient −1-1, since the terms do not have to be written in order. The polynomial (x+1)2−x2=2x+1(x + 1)^2 - x^2 = 2x + 1 has degree 11, because the x2x^2 terms cancel.

Division algorithm

Definition Let PP and DD be polynomials with DD not the zero polynomial. Then there are unique polynomials QQ and RR such that

P(x)=D(x) Q(x)+R(x),P(x) = D(x)\,Q(x) + R(x),

where either RR is the zero polynomial or the degree of RR is less than the degree of DD. Here PP is the dividend, DD is the divisor, QQ is the quotient and RR is the remainder.

In plain words It is the same as dividing whole numbers, as in 17=5⋅3+217 = 5 \cdot 3 + 2. You find how many times the divisor fits, and what is left over must be smaller than the divisor, measured here by degree.

Example Take dividend P(x)=2x3−3x2+4x−5P(x) = 2x^3 - 3x^2 + 4x - 5 and divisor D(x)=x−2D(x) = x - 2. Division gives quotient 2x2+x+62x^2 + x + 6 and remainder 77, so

2x3−3x2+4x−5=(x−2)(2x2+x+6)+7.\begin{aligned} &2x^3 - 3x^2 + 4x - 5 \\[0.4em] &\quad = (x - 2)(2x^2 + x + 6) + 7. \end{aligned}

The remainder 77 has degree 00, less than the degree 11 of the divisor, as required.

Polynomial long division

Definition Polynomial long division finds QQ and RR by repeating one step. Write both polynomials in descending powers, with zero coefficients for missing powers. Divide the leading term of the current dividend (at first PP itself, later whatever is left) by the leading term of DD, add that term to the quotient, multiply it by DD, and subtract the product from the current dividend, lining up like powers. Stop as soon as what remains is zero or has degree less than the degree of DD. What remains is RR.

In plain words Each step kills the highest power that is left. Keep going until the leftover is too small to divide again.

Example Divide P(x)=x4+2x3−x+3P(x) = x^4 + 2x^3 - x + 3 by D(x)=x2+1D(x) = x^2 + 1. Writing the missing powers with zero coefficients keeps like powers in the same column,

P(x)=x4+2x3+0x2−x+3,D(x)=x2+0x+1.\begin{gathered} P(x) = x^4 + 2x^3 + 0x^2 - x + 3, \\[0.4em] D(x) = x^2 + 0x + 1. \end{gathered}

The first quotient term is x4x2=x2\frac{x^4}{x^2} = x^2. Subtracting x2D(x)x^2 D(x) power by power gives

(x4+2x3+0x2−x+3)−(x4+0x3+x2)=2x3−x2−x+3.\begin{aligned} &(x^4 + 2x^3 + 0x^2 - x + 3) \\[0.2em] &\quad - (x^4 + 0x^3 + x^2) \\[0.4em] &\quad = 2x^3 - x^2 - x + 3. \end{aligned}

The next quotient term is 2x3x2=2x\frac{2x^3}{x^2} = 2x. Subtracting 2x D(x)2x \, D(x) gives

(2x3−x2−x+3)−(2x3+0x2+2x)=−x2−3x+3.\begin{aligned} &(2x^3 - x^2 - x + 3) \\[0.2em] &\quad - (2x^3 + 0x^2 + 2x) \\[0.4em] &\quad = -x^2 - 3x + 3. \end{aligned}

The next quotient term is −x2x2=−1\frac{-x^2}{x^2} = -1. Subtracting −1⋅D(x)-1 \cdot D(x) gives

(−x2−3x+3)−(−x2+0x−1)=−3x+4.\begin{aligned} &(-x^2 - 3x + 3) - (-x^2 + 0x - 1) \\[0.4em] &\quad = -3x + 4. \end{aligned}

What remains has degree 11, less than 22, so the process stops. The quotient is x2+2x−1x^2 + 2x - 1 and the remainder is −3x+4-3x + 4.

Remainder

Definition The remainder RR in P=DQ+RP = DQ + R is either the zero polynomial or a polynomial of degree less than the degree of DD. For a linear divisor x−cx - c the remainder is the constant P(c)P(c), by setting x=cx = c in the identity. The divisor DD is a factor of PP exactly when the remainder is zero.

In plain words The remainder is the part the divisor cannot absorb. When it is zero, the division comes out even and you have found a factorization.

Example For P(x)=2x3−3x2+4x−5P(x) = 2x^3 - 3x^2 + 4x - 5 and divisor x−2x - 2, the remainder should be P(2)P(2), and indeed

P(2)=16−12+8−5=7.P(2) = 16 - 12 + 8 - 5 = 7.

For x3+1x^3 + 1 divided by x+1x + 1 the remainder is (−1)3+1=0(-1)^3 + 1 = 0, so x3+1=(x+1)(x2−x+1)x^3 + 1 = (x + 1)(x^2 - x + 1).

Polynomial identity and checking by multiplying back

Definition A polynomial identity is an equation between two polynomials that holds for every real input. Two polynomials are identical exactly when their coefficients of each power agree. The division result P=DQ+RP = DQ + R is a polynomial identity, so it holds at all real xx, including the roots of DD. To check a division, expand DQ+RDQ + R and compare it with PP coefficient by coefficient.

In plain words The division answer is not an approximation. Multiply the divisor by the quotient, add the remainder, and you must get the dividend back exactly.

Example Check the division of x4+2x3−x+3x^4 + 2x^3 - x + 3 by x2+1x^2 + 1. Expanding gives

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

Adding the remainder −3x+4-3x + 4 gives x4+2x3−x+3x^4 + 2x^3 - x + 3, which is PP. As a spot check at x=1x = 1, the dividend is 55 and the right side is 2⋅2+1=52 \cdot 2 + 1 = 5.

Proper rational expression

Definition A rational expression P(x)D(x)\frac{P(x)}{D(x)}, with PP and DD polynomials and DD not the zero polynomial, is proper when PP is the zero polynomial or the degree of PP is less than the degree of DD. Otherwise it is improper. Dividing both sides of P=DQ+RP = DQ + R by D(x)D(x) writes an improper expression as a polynomial plus a proper one,

P(x)D(x)=Q(x)+R(x)D(x),D(x)≠0.\begin{gathered} \frac{P(x)}{D(x)} = Q(x) + \frac{R(x)}{D(x)}, \\[0.4em] D(x) \ne 0. \end{gathered}

In plain words A proper fraction has a smaller top than bottom, measured by degree. Wherever it is defined, any rational expression is a polynomial part plus a proper part.

Example The expression x2+1x−1\frac{x^2 + 1}{x - 1} is improper, since the top has degree 22 and the bottom degree 11. Dividing gives x2+1=(x−1)(x+1)+2x^2 + 1 = (x - 1)(x + 1) + 2, so

x2+1x−1=x+1+2x−1,x≠1.\begin{gathered} \frac{x^2 + 1}{x - 1} = x + 1 + \frac{2}{x - 1}, \\[0.4em] x \ne 1. \end{gathered}

For large xx the correction 2x−1\frac{2}{x - 1} is small, so the graph runs close to the line y=x+1y = x + 1.

Excluded values

Definition An excluded value of a rational expression P(x)D(x)\frac{P(x)}{D(x)} is a real number cc with D(c)=0D(c) = 0. The rewrite Q(x)+R(x)D(x)Q(x) + \frac{R(x)}{D(x)} equals P(x)D(x)\frac{P(x)}{D(x)} only for xx with D(x)≠0D(x) \ne 0. This stays true when R=0R = 0 and the rewrite looks like a plain polynomial, because the original fraction is still undefined at the roots of DD.

In plain words The identity P=DQ+RP = DQ + R holds everywhere, but the fraction form inherits the forbidden inputs of the original denominator, and cancelling does not remove them.

Example Since x3+1=(x+1)(x2−x+1)x^3 + 1 = (x + 1)(x^2 - x + 1),

x3+1x+1=x2−x+1,x≠−1.\frac{x^3 + 1}{x + 1} = x^2 - x + 1, \qquad x \ne -1.

At x=−1x = -1 the right side gives 33, but the left side is 00\frac{0}{0} and undefined. The graph of the left side is the parabola with a hole at (−1,3)(-1, 3). Meanwhile the identity x3+1=(x+1)(x2−x+1)x^3 + 1 = (x + 1)(x^2 - x + 1) is perfectly true at x=−1x = -1, where both sides equal 00.

Check yourself

  1. Divide x3−4x+1x^3 - 4x + 1 by x+2x + 2 and verify by multiplying back.
Show the answer

The quotient is x2−2xx^2 - 2x and the remainder is 11. Check by expanding,

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

The remainder also matches P(−2)=−8+8+1=1P(-2) = -8 + 8 + 1 = 1.

  1. Divide 3x3+x2+23x^3 + x^2 + 2 by x2−x+1x^2 - x + 1 and write the result as a polynomial plus a proper rational expression.
Show the answer

The quotient is 3x+43x + 4 and the remainder is x−2x - 2, of degree 1<21 < 2. The quadratic x2−x+1x^2 - x + 1 has no real roots, so there are no excluded values and

3x3+x2+2x2−x+1=3x+4+x−2x2−x+1.\begin{aligned} \frac{3x^3 + x^2 + 2}{x^2 - x + 1} &= 3x + 4 \\[0.4em] &\quad + \frac{x - 2}{x^2 - x + 1}. \end{aligned}

Multiplying back, (x2−x+1)(3x+4)=3x3+x2−x+4(x^2 - x + 1)(3x + 4) = 3x^3 + x^2 - x + 4, and adding x−2x - 2 gives 3x3+x2+23x^3 + x^2 + 2.

  1. Is it correct to write x2−1x−1=x+1\frac{x^2 - 1}{x - 1} = x + 1 with no condition?
Show the answer

No. The identity x2−1=(x−1)(x+1)x^2 - 1 = (x - 1)(x + 1) holds for all xx, but the left fraction is undefined at x=1x = 1. The correct statement is x2−1x−1=x+1\frac{x^2 - 1}{x - 1} = x + 1 for x≠1x \ne 1.

How this lesson connects