A cubic $f(x)=ax^{3}+bx^{2}+cx+d$ with $a\ne0$ always has one, two, or three distinct real roots, and its derivative decides which. This editable handout develops the test from end behaviour rather than quoting a formula: it explains why the leading term controls both ends, why the sign of $b^{2}-3ac$ decides whether turning points exist at all, and why the product of the two stationary values then settles the count. It includes five worked examples, two parameter problems, ten exercises with answer space, and a separate teacher key.
Why every cubic has at least one real root
For $x\ne0$,
$$ f(x)=ax^{3}\left(1+\frac{b}{a}\cdot\frac1x+\frac{c}{a}\cdot\frac1{x^{2}} +\frac{d}{a}\cdot\frac1{x^{3}}\right). $$
Every term in the bracket except the $1$ carries a power of $x$ in its denominator, so the bracket approaches $1$ as $|x|$ grows and $f$ behaves like $ax^{3}$ far from the origin. Concretely, with $M=\max(|b|,|c|,|d|)/|a|$, any $x$ with $|x|>1+M$ makes the cubic term outweigh the other three, so $f(x)$ takes the sign of $ax^{3}$ and every real root lies in $[-1-M,,1+M]$.
The two ends therefore carry opposite signs. Since $f$ is continuous, the Intermediate Value Theorem gives at least one real root, and a cubic has at most three. The remaining question — one, two or three — is decided between the ends.
The derivative test
Differentiating gives $f’(x)=3ax^{2}+2bx+c$, whose discriminant is $4b^{2}-12ac$. Write
$$ \delta=b^{2}-3ac . $$
If $\delta\le0$ then $f’$ never changes sign, $f$ is strictly monotone, and there is exactly one real root. (When $\delta=0$ the cubic is $f(x)=a(x-x_{0})^{3}+f(x_{0})$ with $x_{0}=-b/(3a)$: a translated copy of $ax^{3}$, still strictly monotone.)
If $\delta>0$ then $f’$ has two distinct zeros, the stationary points
$$ p=\frac{-b-\sqrt{\delta}}{3a},\qquad q=\frac{-b+\sqrt{\delta}}{3a}, $$
and the graph rises, falls, and rises again (or the reverse when $a<0$). Each of the three monotone branches meets the horizontal axis at most once, so the count depends only on where the axis sits relative to the two turning values:
| Condition | Distinct real roots |
|---|---|
| $\delta\le0$ | 1 |
| $\delta>0$ and $f(p),f(q)>0$ | 1 |
| $\delta>0$ and $f(p),f(q)=0$ | 2 |
| $\delta>0$ and $f(p),f(q)<0$ | 3 |
The product is symmetric in $p$ and $q$, so there is never any need to work out which stationary point is the maximum — a step that flips with the sign of $a$ and is a common source of error.
Worked example
Take $f(x)=x^{3}-3x+1$. Then $\delta=0-3(1)(-3)=9>0$, and $f’(x)=3x^{2}-3$ vanishes at $x=\pm1$. The stationary values are $f(-1)=3$ and $f(1)=-1$, whose product is $-3$. The axis passes between the turning values, so there are three distinct real roots; the sign changes at $f(-2)=-1$, $f(0)=1$ and $f(2)=3$ place them in $(-2,-1)$, $(0,1)$ and $(1,2)$.
Changing only the constant term to $f(x)=x^{3}-3x+5$ leaves the stationary points at $\pm1$ but raises the values to $7$ and $3$. The product is now positive and the same cubic shape has just one real root — the standard warning that two turning points do not by themselves produce three roots.
Choosing a parameter
Because the derivative of $x^{3}-3x+k$ does not involve $k$, the stationary points stay at $\pm1$ and the stationary values are $k+2$ and $k-2$. Their product $k^{2}-4$ is negative exactly when $-2<k<2$, so the cubic has three distinct real roots on that interval, two when $k=\pm2$, and one when $|k|>2$. The handout works a second parameter problem, $x^{3}-6x^{2}+9x+k$, where the repeated-root condition $k(k+4)=0$ produces the factorisations $x(x-3)^{2}$ and $(x-1)^{2}(x-4)$.
Representative exercises
- Count the real roots of $x^{3}-12x+1$, $x^{3}+3x^{2}+3x+7$, $2x^{3}-9x^{2}+12x-5$ and $-x^{3}+6x^{2}-9x+2$.
- Find every $k$ for which $x^{3}-12x+k$ has exactly one real root.
- Find every $k$ for which $x^{3}-3x^{2}+k$ has three distinct real roots.
- Find every $a$ for which $x^{3}-3x^{2}+a$ has a repeated root, and give the roots in each case.
- Show that $x^{3}+px+q$ has exactly one real root whenever $p\ge0$.
- Refute the claim that two real zeros of $f’$ force three real roots.
- Explain why a cubic cannot have exactly two distinct real roots unless one of them is repeated.
The teacher key solves all ten exercises, records the intervals containing each root, lists the errors this topic reliably produces, and supplies a compact scoring guide.
Classroom use and adaptation
The handout is designed as a single-sitting resource for a strong class that has met the first derivative test: the first two sections are direct instruction, the worked examples are for annotation, and the exercises with the recording table provide independent practice. A class that has not met derivatives can keep everything except the derivation in Section 2, replacing it with the completed square of $f’$.
Everything is editable: the school identity and page furniture, the choice and number of exercises, the amount of answer space, and whether the three-panel figure is kept, redrawn for another family, or removed. The student handout and the teacher key are separate files, so the key can be exported and distributed on its own.
References and teaching notes
The two Common Core standards cited, HSA-APR.B.3 and HSF-IF.C.7, cover identifying zeros of polynomials and graphing them with their end behaviour; OpenStax Calculus Volume 1 is cited for the Intermediate Value Theorem and the first derivative test.
The derivative argument belongs to a first course in calculus, even though the polynomial and graphing ideas also connect to Algebra II. This handout counts distinct real roots and locates them by sign changes; solving cubics in radicals and numerical root-finding are intentionally left for a later lesson.