Polynomials including cubics (HL)

Polynomials are the building blocks of algebra.

Definition
Arithmetics
Fundamental theorem of algebra
Real polynomials (complex conjugate root theorem)
Graphs of polynomials
- zeros
- $y$ -intercept
Vieta’s formulas
Strategies for solving a cubic equation

Definition

A polynomial in $x$ is a sum of some natural number power of $x$ . The degree or order $n$ of a polynomial is the greatest exponent, where the coefficient is not zero.

\text{polynomial: }{\sum_{k=0}^n a_kx^k = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0}

Note: A a degree $n$ polynomial has up to $n+1$ non-zero terms.

$a_k$ , the coefficient, is also denoted by $\left[x^k\right]$ .

Arithmetics

See also polynomial division

Arithmetics of polynomials are largely similar to that of whole numbers, except instead of power of $10$ , we use power of $x$ .

You should be able to add, subtract, multiply, and divide polynomials by hand.

It helps to align terms of the same power.

Example: Expand $(-2x^2 + 5)(3x^2 - x + 4)$ .

\begin{matrix} -6x^4 & +2x^3 & -8x^2 & & \\ & & 15x^2 & -5x & +20 \\ \hline -6x^4 & +2x^3 & + 7x^2 & -5x & +20 \end{matrix}

Answer is $-6x^4 + 2x^3 + 7x^2 - 5x + 20\qed$

Fundamental theorem of algebra

Theorem: For $n \geq 1$ , an $n$ -degree single-variable polynomial, with real or complex coefficients, has exactly $^\dag n$ complex roots.

$^\dag$ Roots are allowed to have multiplicity, meaning counting a root multiple times. It sounds like cheating at first, but there are some practical uses. As mentioned in More rational functions, the multiplicity of the roots of the denominator affect behaviors around vertical asymptotes. Multiplicity also affects graphing.

As an example, the function

x \mapsto (x - 1)^3(x^2 + 4) (x +3)^2

is a degree $7$ polynomial. It has $7$ zeros: $1$ with a multiplicity of $3$ , $-3$ with a multiplicity of $2$ , along with $2\i$ and $-2\i$ each with a multiplicity of $1$ .

Real polynomials (complex conjugate root theorem)

Polynomials with real coefficients have roots that are either real, or exist as complex conjugate pairs with equal multiplicity. For example if ${-\sqrt 3 + 2\i}$ is a root, then so is ${-\sqrt3 - 2\i}$ .

Graphs of polynomials

The range of odd-degree polynomials (eg linear, cubic) is all real numbers, whereas even-degree (eg quadratic, quartic) polynomials have a global maximum or global minimum.

Leading coefficient $a_n$	Shape of odd-degree polynomials	Shape of even-degree polynomials
$a_n > 0$	/	$\cup$
$a_n < 0$	\	$\cap$

The shapes at very positive and negative values of $x$ are called the end behaviors of a function.

For a polynomial function $x \mapsto x^n, n \in \mathbb{Z}^+$ , it is odd if $n$ is odd, and even when $n$ is even.

zeros

The graph near $x$ -intercepts, or zeros, of a function relate to their multiplicity. Near a zero, the function looks like a polynomial of degree same as the multiplicity of the zero. Eg a zero with a multiplicity of $2$ will appear like a quadratic near the root.

For real zeros, odd multiplicity goes through the $x$ -axis, while even multiplicity touches the $x$ -axis then bounce back.

$y$ -intercept

The roots are insufficient to uniquely identify a function. You also need a point off the $x$ -axis. Typically $(0, f(0))$ is used.

Vieta’s formulas

Vieta’s formulas provide the general relationships between the coefficients of a polynomial and its roots. They are particularly useful in finding the sum and product of the roots.

\text{sum of roots: }(-1)^1\frac{a_{n-1}}{a_n} = -\frac{a_{n-1}}{a_n}

\text{product of roots: }(-1)^n\frac{a_0}{a_n}

$a_n$ is the leading coefficient, for the term of the greatest exponent.

Vieta’s formulas allow you to obtain the sum and product of a polynomials roots without having all the coefficients.

The formulas can also be used to a build a polynomial with specific sum or product of roots.

Example: Find the sum and product of roots of ${2x^3 + 3x^2 = 5x + 7}$ .

We first rewrite the equation in the standard form of a polynomial:

2x^3 + 3x^2 - 5x - 7 = 0

\text{sum of roots} = (-1)\cdot\frac{3}{2} = -\frac 32 \qed

\text{product of roots} = (-1)^3\cdot\frac{-7}{2} = \frac 72 \qed

Here we used $(-1)^3$ because it is a cubic. You would use $(-1)^n$ for the product of roots of an $n$ -degree polynomial.

Strategies for solving a cubic equation

While there is a cubic formula, the cubics on exam are relatively simple. If you are asked to solve a cubic, there will be at least one rational root, if not an integer root. The following are strategies to identity this rational root.

rational root theorem

The rational root theorem can limit down to about a dozen or two candidates.

Theorem: For all simplified rational roots $\frac{p}{q},\, p, q \in \mathbb Z$ , of a polynomial with integer coefficients, $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient, possibly with a negative sign.

For example a polynomial $4x^3 + mx^2 + nx + 27, \, m, n \in \mathbb Z$ , the candidate rational roots are

\pm 1, \pm\frac 12, \pm \frac 14, \pm 3, \pm \frac32, \pm \frac 34, \pm 9, \pm \frac92, \pm \frac 94, \pm 27, \pm \frac{27}{2}, \pm \frac {27}{4}

The rational root theorem does not guarantee the existence of a rational root. Check out the roots of $2x^3+3x^2-5x-7=0$ .

Descartes’ rule of signs

Theorem: Write the polynomial in descending powers, omitting terms with coefficients zero. The number of sign (positive or negative) changes between consecutive written terms is the upper limit of the number of positive roots. The actual number of positive roots is either this limit, or lower by an even number.

sign changes	number of positive roots (incl. multiplicity)
$3$	$3$ or $1$
$2$	$2$ or $0$
$1$	$1$
$0$	$0$

Corollary: Applying Descartes’ rule of signs on $f(-x)$ returns the number of negative roots of $f$ .

intermediate value theorem

This relates to the “completeness” of real numbers.

Theorem: For a continuous function $f$ defined over the interval $[a, b]$ , the range contains $[f(a), f(b)]$ and possibly additional intervals.

Corollary: If $f(a) < 0$ and $f(b) > 0$ , there is a root between $a$ and $b$ .

However, if you choose values that are too far away, and you get the same sign for $y$ , the IVT does not imply there is no root in between.

Furthermore, you know the end behaviors, so you can also apply IVT between any value and the end behavior.

after finding a root

When you have successfully guessed a rational root, perform polynomial division or apply Vieta’s formulas to yield a quadratic, on which the quadratic formula can be used to find the remaining two roots.

Here’s an example using all the techniques above. Usually the cubic is (a lot) simpler than this, or one of the roots will be given.

Example: Solve $-2x^3 - 7x^2 + 12 > 0$ .

In general, solving an inequality requires first solving an equation. Define

f(x) = -2x^3 - 7x^2 + 12

The rational root theorem limits the possible rational roots to $\displaystyle \pm \frac12, \pm 1, \pm\frac{3}{2},\pm2, \pm3, \pm4, \pm6, \pm12$ .

As a rule of thumb, we first explore ${f(1) = 3}$ , ${f(0) = 12}$ , ${f(-1) = 7}$ .

Applying Descartes’ rule of signs, coefficients of $f(x)$ have one sign change (from negative to positive) meaning exactly one positive real root, which from intermediate value theorem means it is greater than $1$ . And $f(-x) = 2x^3 - 7x^2 + 12$ with two sign changes can mean either zero or two negative real roots.

Next, $f(2) = -16 - 28 + 12 = -32$ . The IVT limits the only possible positive rational root to $\frac32$ , though $f\left(\frac32\right) \neq 0$ means it’s time to move on to negative rational roots.

Sure enough $f(-2) = 0$ .

Vieta’s formulas on $f$ say that the roots of the cubic add to $-\frac72$ , and multiply to $6$ . With $-2$ identified as a root, the two remaining roots must add to $-\frac32$ and multiply to $-3$ .

Then the remaining two roots must be solutions to $\displaystyle x^2 + \frac32 x - 3$ or ${2x^2 + 3x - 6}$ .

Applying quadratic formula results in the zeros:

\frac{-3 \pm \sqrt{57}}{4}

Note that $\displaystyle -2 = \frac{-8}{4} = \frac{-3 - \sqrt{25}}{4} > \frac{-3 - \sqrt{57}}{4}$ . From above, we also know the other root must be the positive root. Knowing cubics with negative leading coefficient looks like ”\”, we have the intervals of positive $f(x)$ as

x \in \left]-\infty, \frac{-3 - \sqrt{57}}{4}\right[ \cup \left]-2, \frac{-3 + \sqrt{57}}{4}\right[ \qed

The above example was chosen to explore various thought processes you could go through, as opposed to randomly checking values. Sure, it’s not as fast as randomly checking $-2$ on your first try, but working through the problem somewhat systematically can lower the average time to the solution.