D8 - Polynomial inequalities

Cubic and quartic signs

Can three roots, together with one known interval sign, tell you where a cubic is positive? Consider $p(x)=(x+2)(x-1)(x-3)$. At $x=0$, its three factors have signs positive, negative, negative, so their product is positive. We can extend that conclusion throughout $-2<x<1$: none of the factors changes sign inside that interval.

A polynomial is continuous. It cannot change from positive to negative without passing through zero. Once you have found all its real roots, they split the real line into intervals on which the polynomial has a constant sign. A test value in each interval therefore determines the sign throughout that interval.

Factorise before choosing test values. Finding a few roots on a calculator is insufficient unless you know that the factorisation accounts for the remaining polynomial too. A factor such as $x^2+1$ has no real roots and stays positive, so it introduces no extra boundary.

Now add just one factor, $x-4$. Its sign reverses the old product to the left of 4 and leaves it unchanged to the right. This turns the cubic into a quartic without requiring a new solving method.

The leading term gives a useful first sign: a cubic with positive leading coefficient goes down on the far left and up on the far right; a quartic with positive leading coefficient goes up at both ends. A negative leading coefficient reverses those directions. The actual roots still determine the intervening signs.

Roots alone cannot determine positivity: $p(x)$ and $-p(x)$ have the same roots and multiplicities, but opposite signs wherever they are nonzero. You also need one known interval sign, which can come from a test value or the sign of the leading coefficient.

Try another quartic, with all four roots different.

Repeated roots and sign changes

Two identical factors change the picture. The square $(x-1)^2$ is positive on both sides of 1 and zero at 1. It cannot reverse the sign of the remaining factors when you pass through its root.

If $p(x)=(x-r)^m q(x)$, where $q(r)\ne0$, then $r$ has multiplicity $m$. Close to $r$, the factor $q(x)$ keeps its sign. An odd power $(x-r)^m$ changes sign across $r$; an even power does not. This proves the crossing rule rather than asking you to memorise a sketch.

The next pair changes only the exponent at 1. Notice that the degree changes too, from three to four.

Replace the squared factor by a cube. Now both roots have odd multiplicity and both must reverse the sign.

The graph below shows why “repeated” alone does not determine whether a curve touches or crosses.

[DIAGRAM: asset_name: Double and triple roots have different sign behaviour; asset_slug: aqa7367-root-multiplicity; caption: A double root touches the axis while a triple root crosses it; description: Two coordinate panels compare y equals (x plus 2) times (x minus 1) squared with y equals (x plus 2) times (x minus 1) cubed. Both roots minus 2 and 1 are labelled. The cubic crosses at minus 2 and touches from above at 1. The quartic crosses at both roots, flattening as it crosses at 1. Positive and negative regions are visibly distinguished by the graph's position relative to the horizontal axis.]

A useful surprise is that a polynomial inequality can have a single isolated solution alongside an interval. Dividing by a squared factor and forgetting its zero would lose exactly that feature.

Keep the zero of the square when answering the next question.

Exact solution sets and endpoints

After finding the signs, express the answer as a set of inputs. The inequality symbol determines which roots belong: $>$ and $<$ exclude zero-valued boundaries; $\ge$ and $\le$ include them. Polynomial expressions have no denominator restrictions.

For example, 0 belongs to $[-1,1]$, written $0\in[-1,1]$; 2 does not, written $2\notin[-1,1]$. A single allowed value is a singleton such as $\{1\}$, while no allowed values gives the empty set $\varnothing$. If an extra domain $D$ is supplied, intersect your solution set with $D$. Thus an interval is a subset of $\mathbb R$, not a list of only its integer inputs.

Recall a few set symbols when a question restricts the domain:

For $\varepsilon=\{-2,-1,0,1,2\}$ and $A=\{-1,0,1\}$, we have $n(A)=3$, $A\subset\varepsilon$ and $A'=\{-2,2\}$. Complements depend on the stated universal set. Restricting $[-\sqrt2,\sqrt2]$ to $\mathbb N$ leaves only $\{1\}$; restricting it to $\mathbb Z$ leaves $\{-1,0,1\}$.

The symbol $\equiv$ can mark an identity: $x^2-1\equiv(x-1)(x+1)$ holds for every real $x$, unlike an equation whose solutions you must find. In modular arithmetic, $a\equiv b\pmod n$ instead means that $a-b$ is an integer multiple of the positive integer $n$, for example $8\equiv2\pmod3$. Also, $y\propto x$ means $y=kx$ for a fixed constant $k$; it does not mean $y=x$.

Exact roots matter: $\sqrt2\approx1.414$, but replacing it by 1.414 changes the exact boundary. Keep the surd in your final interval.

Change only $\ge$ to $>$. The interval signs stay the same; only endpoint membership changes.

This answers the opening question: the complete root list, each root's multiplicity and one interval sign determine every interval sign. You do not need to calculate the turning points to solve the inequality.

Use the same quadratic-in-a-square idea with a new quartic.

Quick Recap

• Factorise completely enough to find every real root.
• An odd-multiplicity root reverses the sign; an even-multiplicity root does not.
• Strict inequalities exclude roots; non-strict inequalities may include isolated roots.
• Use exact endpoints and join separate allowed intervals with $\cup$.

Yes: the roots and their multiplicities, together with one known interval sign, determine where a cubic is positive without finding its turning points.