Polynomials and long division

Unit 4 · Polynomials

Degree, leading coefficient and the algebra of polynomials, then dividing one polynomial by another to get a quotient and a remainder.

What is a polynomial?

A polynomial in $x$ is an expression such as $$P(x) = 3x^4 - 2x^3 + x - 7,$$ built from powers $x^0, x^1, x^2, \dots$ (whole numbers only) multiplied by real-number coefficients and added. The numbers $3, -2, 0, 1, -7$ are the coefficients of $x^4, x^3, x^2, x, 1$ (note the missing $x^2$ term has coefficient $0$).

• The degree of $P$ is the highest power of $x$ that has a non-zero coefficient. Here $\deg P = 4$.
• The coefficient of that highest power is the leading coefficient ($3$ here). If it is $1$, the polynomial is monic.
• Degree $0$: constants. Degree $1$: linear, $ax+b$. Degree $2$: quadratic. Degree $3$: cubic. Degree $4$: quartic.
• $\sqrt{x}$, $\dfrac1x$ and $x^{-2}$ are not polynomials, because the powers are not whole numbers $\ge 0$.

Algebra of polynomials

Polynomials are added, subtracted and multiplied like ordinary algebra: collect like powers.

• For $P + Q$ add coefficients of equal powers. The degree is at most the larger of the two degrees (it can drop: $(x^2+x) + (-x^2+3)$ has degree $1$).
• For $PQ$ multiply every term of one by every term of the other. The leading terms multiply, so $\deg(PQ) = \deg P + \deg Q$.

Two polynomials are equal (identical) only if all their coefficients are equal. We use this in the identities lesson.

Example: Let $P = 2x^3 - x + 4$ and $Q = x^2 + 3x - 1$. Then $P + Q = 2x^3 + x^2 + 2x + 3$, and $PQ = 2x^3(x^2+3x-1) - x(x^2+3x-1) + 4(x^2+3x-1) = 2x^5 + 6x^4 - 2x^3 - x^3 - 3x^2 + x + 4x^2 + 12x - 4 = 2x^5 + 6x^4 - 3x^3 + x^2 + 13x - 4.$ The degree of $PQ$ is $5 = 3 + 2$, as expected.

Dividing polynomials

Dividing $P(x)$ (the dividend) by $D(x)$ (the divisor) gives a quotient $Q(x)$ and a remainder $R(x)$ with $$P(x) \equiv D(x),Q(x) + R(x), \qquad \deg R < \deg D \text{ (or } R = 0).$$ This is just like $17 = 5\times 3 + 2$ with remainder smaller than the divisor. If $R = 0$ we say $D$ divides $P$ exactly, or $D$ is a factor of $P$.

Long division, step by step

1. Write both polynomials in descending powers, putting $0$ for any missing power.
2. Divide the leading term of the dividend by the leading term of the divisor. That is the next term of the quotient.
3. Multiply the whole divisor by this term and subtract from the dividend. Be careful with signs.
4. Repeat with what is left. Stop when the left-over has lower degree than the divisor. That is the remainder.

Example: Divide $x^3 + 2x^2 - 5x + 1$ by $x - 2$.

• $x^3 \div x = x^2$. Subtract $x^2(x-2) = x^3 - 2x^2$: left $4x^2 - 5x + 1$.
• $4x^2 \div x = 4x$. Subtract $4x(x-2) = 4x^2 - 8x$: left $3x + 1$.
• $3x \div x = 3$. Subtract $3(x-2) = 3x - 6$: left $7$. The left-over $7$ has degree $0 < 1$, so it is the remainder. Quotient $x^2 + 4x + 3$, remainder $7$. Check: $(x-2)(x^2+4x+3) + 7 = x^3 + 2x^2 - 5x - 6 + 7 = x^3 + 2x^2 - 5x + 1$. Correct.

Example: Divide $x^4 - 1$ by $x^2 + x + 1$. Write the dividend as $x^4 + 0x^3 + 0x^2 + 0x - 1$.

• $x^4 \div x^2 = x^2$. Subtract $x^2(x^2+x+1) = x^4 + x^3 + x^2$: left $-x^3 - x^2 - 1$.
• $-x^3 \div x^2 = -x$. Subtract $-x(x^2+x+1) = -x^3 - x^2 - x$: left $x - 1$. The left-over has degree $1 < 2$, so stop. Quotient $x^2 - x$, remainder $x - 1$. Check: $(x^2+x+1)(x^2-x) = x^4 - x$, and $x^4 - x + (x - 1) = x^4 - 1$.

Exam tip: The two most common marks lost are a forgotten $0x^2$ term and a sign slip when subtracting. Write the subtraction line in full, and do the one-line check at the end. Examiners often give the quotient in the question, so a quick check also tells you if you are on the right track.

Matching a given quotient

Sometimes the quotient is given and you must find unknown constants. Expand $D(x)Q(x) + R$ and compare it with $P(x)$, or divide by the shorter route in the next lesson.