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Polynomial functions

Algebra · level 8, lesson 1 of 8

A question to start

The equation

\[x^3 = 7x - 6\]

has three solutions, and all three are whole numbers. Can you find them? There is a formula for cubic equations (found in Italy in the 1500s), but it is long and messy, and you will not need it. By the end of this lesson you will find all three solutions with nothing but a short list of guesses and one division. (They are 1, 2 and \(-3\).)

The idea

Polynomial functions and their words

A polynomial function has the form

\[p(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0,\]

where \(n\) is a whole number and the \(a\)'s are numbers with \(a_n \ne 0\). The highest power \(n\) is the degree; \(a_n\) is the leading coefficient; \(a_n x^n\) is the leading term; \(a_0\) is the constant term (and the \(y\)-intercept, since \(p(0) = a_0\)). Degree 1 is linear, 2 quadratic, 3 cubic, 4 quartic, 5 quintic.

End behaviour: the leading term wins

What happens to \(p(x)\) when \(x\) is huge, positive or negative? The leading term decides. Take \(p(x) = x^3 - 100x^2\). At \(x = 1000\): \(x^3 = 10^9\) and \(100x^2 = 10^8\), so \(p = 9 \times 10^8\), positive and huge. Even a big coefficient on a lower power is beaten, because \(x^3 = x \times x^2\) and \(x\) itself keeps growing. In symbols,

\[p(x) = a_n x^n\left(1 + \frac{a_{n-1}}{a_n x} + \cdots + \frac{a_0}{a_n x^n}\right),\]

and every fraction in the bracket shrinks towards 0 as \(x\) grows, so the bracket gets close to 1.

So there are four cases, depending on whether the degree is even or odd and the leading coefficient positive or negative:

\(a_n > 0\) \(a_n < 0\)
even degree (like \(x^2\)) up on both ends down on both ends
odd degree (like \(x^3\)) down on the left, up on the right up on the left, down on the right

An odd-degree polynomial goes from one end to the opposite one, so it must cross the \(x\)-axis somewhere: every polynomial of odd degree has at least one real zero.

Zeros and multiplicity

If \(p(x) = -2(x - 1)^2(x + 3)\), the zeros are \(1\) and \(-3\). The factor \((x - 1)\) appears twice: we say 1 is a zero of multiplicity 2. The zero \(-3\) has multiplicity 1.

Multiplicity tells you what the graph does at the zero. Near \(x = r\), all the other factors are almost constant, so the graph looks like \(C(x - r)^m\) for some number \(C\):

multiplicity 1: crosses multiplicity 2: touches multiplicity 3: flattens
  • Odd multiplicity (1, 3, ...): \((x - r)^m\) changes sign at \(r\), so the graph crosses the axis. With multiplicity 3 it flattens as it crosses, like \(x^3\).
  • Even multiplicity (2, 4, ...): \((x - r)^m\) is never negative, so the graph touches the axis and turns back, like \(x^2\).

A polynomial of degree \(n\) has at most \(n\) zeros, and at most \(n - 1\) turning points (places where it changes from going up to going down). The zeros fact follows from the factor theorem below; the turning points fact is proved in calculus.

Dividing polynomials

Polynomials divide just like whole numbers: \(17 \div 5 = 3\) remainder 2 means \(17 = 5 \times 3 + 2\). For polynomials, dividing \(p(x)\) by \(d(x)\) gives a quotient \(q(x)\) and a remainder \(r(x)\) with a smaller degree than \(d(x)\):

\[p(x) = d(x)\,q(x) + r(x).\]

Long division of \(2x^3 - 3x^2 + 4x - 5\) by \(x - 2\), step by step:

  1. Divide the leading terms: \(2x^3 \div x = 2x^2\). Multiply: \(2x^2(x - 2) = 2x^3 - 4x^2\). Subtract from the first two terms: \((2x^3 - 3x^2) - (2x^3 - 4x^2) = x^2\). Bring down \(+4x\): now we have \(x^2 + 4x\).
  2. \(x^2 \div x = x\). Multiply: \(x(x - 2) = x^2 - 2x\). Subtract: \(6x\). Bring down \(-5\): \(6x - 5\).
  3. \(6x \div x = 6\). Multiply: \(6(x - 2) = 6x - 12\). Subtract: \(7\).

So the quotient is \(2x^2 + x + 6\) and the remainder is 7: \(2x^3 - 3x^2 + 4x - 5 = (x - 2)(2x^2 + x + 6) + 7\).

Synthetic division is the same calculation with only the numbers, for dividing by \(x - a\). Write \(a\) (here 2) and the coefficients. Bring down the first; then repeatedly multiply by \(a\) and add to the next column:

\(a = 2\) \(2\) \(-3\) \(4\) \(-5\)
multiply by 2 \(4\) \(2\) \(12\)
add \(2\) \(1\) \(6\) \(\mathbf{7}\)

The bottom row gives the quotient's coefficients, \(2, 1, 6\), and the last number is the remainder, 7.

The remainder theorem

Look again: the remainder was 7, and \(p(2) = 16 - 12 + 8 - 5 = 7\). That is no coincidence.

Remainder theorem. When \(p(x)\) is divided by \(x - a\), the remainder is \(p(a)\).

Proof. Divide: \(p(x) = (x - a)q(x) + r\), where \(r\) is a number (its degree is less than 1). Put \(x = a\): \(p(a) = 0 \cdot q(a) + r = r\). ∎

The factor theorem

Factor theorem. \(x - a\) is a factor of \(p(x)\) exactly when \(p(a) = 0\).

Proof. By the remainder theorem, \(p(x) = (x - a)q(x) + p(a)\). If \(p(a) = 0\), then \(p(x) = (x - a)q(x)\), so \(x - a\) is a factor. If \(x - a\) is a factor, then \(p(a) = 0 \cdot q(a) = 0\). ∎

Each zero gives a factor, and the degrees of the factors add up to \(n\), so a degree-\(n\) polynomial can have at most \(n\) zeros.

Which numbers to try: the rational root theorem

If \(p(x)\) has whole-number coefficients and \(x = \dfrac{c}{d}\) (in lowest terms) is a zero, then \(c\) divides the constant term \(a_0\), and \(d\) divides the leading coefficient \(a_n\).

Why (for whole-number zeros). If \(c\) is a zero, then \(a_n c^n + \cdots + a_1 c + a_0 = 0\), so \(a_0 = -c\left(a_n c^{n-1} + \cdots + a_1\right)\). The bracket is a whole number, so \(c\) divides \(a_0\). The fraction case works the same way after multiplying through by \(d^n\).

So for \(x^3 - 7x + 6\), any whole-number zero divides 6: try \(\pm1, \pm2, \pm3, \pm6\). That short list is the secret to the opening puzzle.

Worked examples

1. Reading a polynomial. Describe \(y = -2(x - 1)^2(x + 3)\).

  • Degree \(2 + 1 = 3\); leading term \(-2 \cdot x^2 \cdot x = -2x^3\). Odd degree, negative: up on the left, down on the right.
  • Zeros: 1 (multiplicity 2, touches) and \(-3\) (multiplicity 1, crosses).
  • \(y\)-intercept: \(-2(1)(3) = -6\).
  • So: coming down from the top left, it crosses at \(-3\), dips below the axis, rises to touch the axis at 1 (passing through \((0, -6)\) on the way), then falls away to the bottom right.

2. Synthetic division and the remainder theorem. Find the remainder when \(x^4 - 3x + 1\) is divided by \(x + 1\).

  • \(x + 1 = x - (-1)\), so the remainder is \(p(-1) = 1 + 3 + 1 = 5\).

3. The opening puzzle. Solve \(x^3 - 7x + 6 = 0\).

  • Try divisors of 6. \(p(1) = 1 - 7 + 6 = 0\), so \(x - 1\) is a factor.
  • Synthetic division by 1, with coefficients \(1, 0, -7, 6\) (the 0 holds the place of the missing \(x^2\)):
\(a = 1\) \(1\) \(0\) \(-7\) \(6\)
multiply by 1 \(1\) \(1\) \(-6\)
add \(1\) \(1\) \(-6\) \(\mathbf{0}\)
  • So \(x^3 - 7x + 6 = (x - 1)(x^2 + x - 6) = (x - 1)(x - 2)(x + 3)\).
  • The solutions are \(x = 1\), \(x = 2\) and \(x = -3\).
x y −3 −2 −1 1 2 −10 6 10 y = x³ − 7x + 6

4. A fraction as a zero. Factor \(2x^3 + 3x^2 - 8x + 3\).

  • Candidates: \(\pm1, \pm3, \pm\tfrac12, \pm\tfrac32\). Try \(p(1) = 2 + 3 - 8 + 3 = 0\). ✓
  • Divide by \(x - 1\): the bottom row is \(2, 5, -3\) with remainder 0, so the quotient is \(2x^2 + 5x - 3\).
  • \(2x^2 + 5x - 3 = (2x - 1)(x + 3)\). So \(p(x) = (x - 1)(2x - 1)(x + 3)\), with zeros 1, \(\tfrac12\) and \(-3\).

5. Finding a missing coefficient. For which \(k\) is \(x - 2\) a factor of \(x^3 + kx^2 - 4x + 12\)?

  • Need \(p(2) = 0\): \(8 + 4k - 8 + 12 = 0\), so \(4k = -12\), so \(k = -3\).
  • Then \(x^3 - 3x^2 - 4x + 12 = x^2(x - 3) - 4(x - 3) = (x - 3)(x - 2)(x + 2)\). ✓

Try it

  1. Describe the end behaviour of (a) \(y = -x^4 + 3x\) (b) \(y = 5x^3 - x\).
  2. For \(y = x(x - 2)^2(x + 1)\), give the degree, the zeros with their multiplicities, and what the graph does at each.
  3. Divide \(x^3 - 4x^2 + x + 6\) by \(x - 2\), and use the result to factor it fully.
  4. What is the remainder when \(x^{100} - 1\) is divided by \(x + 1\)?
  5. Find the remainder when \(2x^3 + x^2 - 5\) is divided by \(x - 3\).
  6. Factor \(x^3 + 2x^2 - 5x - 6\) fully.
  7. Find the cubic with zeros \(-1\), 2 and 4 whose graph passes through \((0, 16)\).
  8. Divide \(x^3 + 2x - 1\) by \(x^2 + 1\) (long division). What are the quotient and remainder?
Answers
  1. (a) Even degree, negative leading coefficient: down on both ends. (b) Odd degree, positive: down on the left, up on the right.
  2. Degree \(1 + 2 + 1 = 4\). Zeros: 0 (multiplicity 1, crosses), 2 (multiplicity 2, touches), \(-1\) (multiplicity 1, crosses). Leading coefficient 1, even degree: up at both ends.
  3. Synthetic division by 2 on \(1, -4, 1, 6\): bottom row \(1, -2, -3\), remainder \(0\). So \(x^3 - 4x^2 + x + 6 = (x - 2)(x^2 - 2x - 3) = (x - 2)(x - 3)(x + 1)\).
  4. \(p(-1) = (-1)^{100} - 1 = 1 - 1 = 0\). So \(x + 1\) is a factor of \(x^{100} - 1\).
  5. \(p(3) = 54 + 9 - 5 = 58\).
  6. \(p(-1) = -1 + 2 + 5 - 6 = 0\). Synthetic division by \(-1\) on \(1, 2, -5, -6\): bottom row \(1, 1, -6\), remainder
  7. So \((x + 1)(x^2 + x - 6) = (x + 1)(x - 2)(x + 3)\).
  8. \(y = a(x + 1)(x - 2)(x - 4)\). At \(x = 0\): \(a(1)(-2)(-4) = 8a = 16\), so \(a = 2\): \(y = 2(x + 1)(x - 2)(x - 4)\).
  9. \(x^3 \div x^2 = x\); \(x(x^2 + 1) = x^3 + x\); subtract from \(x^3 + 2x - 1\): \(x - 1\). That has degree 1, less than 2, so stop. Quotient \(x\), remainder \(x - 1\): \(x^3 + 2x - 1 = (x^2 + 1)x + (x - 1)\).

Common mistake

Leaving out missing powers in synthetic division. \(x^3 - 7x + 6\) has no \(x^2\) term, so its coefficients are \(1, 0, -7, 6\), not \(1, -7, 6\). Without the 0 every column slides over and the answer is wrong.

Using the wrong sign of \(a\). Dividing by \(x + 3\) means \(a = -3\). The remainder theorem gives \(p(-3)\), not \(p(3)\).

Where it leads