8. Polynomial Factorization and Graphs

Learning goals and preparation

A polynomial graph is constrained by its factors, multiplicities, leading term and continuity. This chapter connects those structures instead of treating graph sketching as guesswork. You will derive the remainder and factor theorems, factor selected real polynomials, and build a justified sign-and-shape description. Chapter 1 already derives quadratic roots and vertices; Chapter 7 supplies domains and graph transformations as functions.

8.1 Structure and degree

A real polynomial is a finite expression

$$P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0,$$

with real coefficients and nonnegative integer exponents. If $a_n\ne0$, its degree is $n$ and $a_n$ is its leading coefficient. The zero polynomial has no degree under the convention used here. A nonzero constant has degree zero. Expressions such as $x^{-1}$ or $\sqrt x$ are not polynomials in $x$.

For nonzero polynomials $P,Q$, the leading term of their product is the product of the leading terms, so $\deg(PQ)=\deg P+\deg Q$. In a sum the leading terms may cancel: $(x^3+x)+(-x^3+2)=x+2$ has degree one, not three.

Some useful identities are

$$u^2-v^2=(u-v)(u+v),$$ $$u^3-v^3=(u-v)(u^2+uv+v^2),$$ $$u^3+v^3=(u+v)(u^2-uv+v^2).$$

They are verified by multiplication. A common factor should be taken out before attempting more elaborate methods. Over the reals, $x^2+1$ has no linear factor because its value is positive for every real $x$. Factorization always has a coefficient field in the background; this chapter works over the reals.

8.2 Division, remainders, and roots

For a polynomial $P$ and nonzero polynomial $D$, division produces polynomials $Q,R$ such that

$$P=DQ+R,\qquad R=0\text{ or }\deg R<\deg D.$$

The algorithm repeatedly subtracts the multiple of $D$ that removes the current highest-degree term. Each subtraction decreases that degree, so it stops after finitely many steps. Uniqueness follows because two such representations would give $D(Q_1-Q_2)=R_2-R_1$. If $Q_1-Q_2$ were nonzero, the left degree would be at least $\deg D$, while the right degree would be smaller, an impossibility.

For $D=x-c$, the remainder is constant. Substituting $x=c$ gives $P(c)=R$. This proves the remainder theorem. In particular $P(c)=0$ if and only if $x-c$ divides $P$, the factor theorem.

Example 1 — find a root and retain the quotient. For

$$P(x)=x^3-2x^2-5x+6,$$

$P(1)=0$. Division by $x-1$ yields $x^2-x-6$, so

$$P(x)=(x-1)(x-3)(x+2).$$

The three distinct roots are $-2,1,3$. Expanding the result verifies the factorization and catches sign errors. Guessing the roots alone would not yet justify the complete product.

A nonzero polynomial of degree $n$ has at most $n$ distinct real roots. To prove this, factor out one real root using the factor theorem; all other distinct roots are roots of the quotient, whose degree is one less. Induction on degree finishes the argument, beginning with a nonzero constant, which has no root. This is an upper bound, not a promise of $n$ real roots.

For an integer-coefficient polynomial with leading coefficient $a_n$ and constant term $a_0\ne0$, any rational root $p/q$ in lowest terms with $q>0$ must have $p$ dividing $a_0$ and $q$ dividing $a_n$. Indeed, multiply $P(p/q)=0$ by $q^n$; reducing the resulting integer equation modulo $p$ gives $p\mid a_0q^n$, hence $p\mid a_0$ because $p,q$ are coprime. Reducing modulo $q$ similarly gives $q\mid a_np^n$, hence $q\mid a_n$. This restricts candidates; it does not guarantee any candidate is a root.

8.3 Multiplicity and sign changes

A root $c$ has multiplicity $m\ge1$ when

$$P(x)=(x-c)^mQ(x),\qquad Q(c)\ne0.$$

Because $Q$ is continuous and nonzero at $c$, its sign is constant in a sufficiently small neighborhood. If $m$ is odd, $(x-c)^m$ changes sign through $c$ and so does $P$. If $m$ is even, it does not change sign and the graph touches the axis without crossing there. Higher multiplicity also creates greater flattening: differentiating gives a factor $(x-c)^{m-1}$, so for $m\ge2$ the slope at the root is zero. “Crossing” and “nonzero slope” are not equivalent: $x^3$ crosses at zero with horizontal tangent.

Example 2 — a sign chart with a repeated root. For $P(x)=(x+2)(x-1)^2$, the square factor is positive except at $1$. Thus $P<0$ on $(-\infty,-2)$, $P>0$ on $(-2,1)$ and $(1,\infty)$, with zeros at $-2,1$. Therefore $P(x)\le0$ holds on $(-\infty,-2]\cup\{1\}$. A sign chart must not discard the isolated zero at a touching root.

8.4 End behavior from the leading term

For a nonzero degree-$n$ polynomial,

$$\frac{P(x)}{a_nx^n}=1+\frac{a_{n-1}}{a_nx}+\cdots+\frac{a_0}{a_nx^n}\longrightarrow1 \quad\text{as }x\to\pm\infty.$$

Each reciprocal power tends to zero. Consequently, sufficiently far in either direction, $P$ has the same sign and unbounded magnitude as its leading term when $n\ge1$. Positive-leading even-degree graphs rise at both ends; negative-leading even-degree graphs fall at both ends. Positive-leading odd-degree graphs fall left and rise right; negative-leading odd-degree graphs do the reverse. A nonzero constant is a horizontal line, not an unbounded tail.

Continuity fills an important logical gap. A sign change between two inputs forces a root between them by the intermediate value theorem. However, a root with even multiplicity does not force a sign change, so sign-change searches can miss roots.

Example 3 — a justified cubic sketch. For $P(x)=(x+2)(x-1)^2=x^3-3x+2$:

• The intercepts are $(-2,0)$, $(1,0)$ and $(0,2)$.
• The root $-2$ is simple and crosses; the root $1$ is double and touches.
• The leading term is $x^3$, so the graph falls left and rises right.
• Using the power rule from Chapter 4, $P'(x)=3(x-1)(x+1)$. Its signs are $+,-,+$ on the intervals separated by $-1,1$.
• Hence the graph increases to $(-1,4)$, decreases to $(1,0)$, and increases afterward. These are a local maximum and local minimum respectively, not global extrema on all of $\mathbb R$.

No plotting software is needed to establish those features. A drawing should be consistent with them, not used as their proof.

8.5 Transformations and nonunique sketches

For $y=A P(B(x-h))+k$ with $A,B\ne0$, the horizontal shift is $h$ and vertical shift is $k$. The horizontal length scale is $1/|B|$, not $|B|$; a negative $B$ reflects about the shifted vertical axis. A negative $A$ reverses vertical orientation. Keeping the transformation as a composition helps preserve the order.

Example 4 — transformed fourth power. For $Q(x)=16-(x-2)^4$, the maximum is $16$ at $x=2$, since a fourth power is nonnegative. The graph is symmetric about $x=2$, has zeros $0,4$, and falls to negative infinity at both ends. Its range is $(-\infty,16]$. The zeros are simple: near either zero the value changes sign. The exponent four refers to the behavior around the center $x=2$, not to multiplicity of the roots $0,4$.

The sign change alone proves odd multiplicity, not simplicity: $x^3$ is a counterexample to that inference. For this particular $Q$, use the exact factorization

$$Q(x)=\bigl(4-(x-2)^2\bigr)\bigl(4+(x-2)^2\bigr)=x(4-x)\bigl(4+(x-2)^2\bigr).$$

At $x=0$, the other factor is $(4-0)(4+4)=32\ne0$; at $x=4$, write $Q(x)=-(x-4)x(4+(x-2)^2)$, whose other factor is $-32\ne0$. The definition of multiplicity therefore gives exactly one at both roots.

Roots alone do not uniquely determine a polynomial. Both $(x-1)(x+2)$ and $5(x-1)(x+2)$ have the same roots, while $(x-1)^2(x+2)$ changes the multiplicity and the degree. Leading coefficient, multiplicities and additional values distinguish them. Similarly, a correct sign chart does not by itself locate every turning point.

8.6 Graded exercises

Foundation

1. Identify which are polynomials and state their degrees when applicable: $3x^4-2$, $x+1/x$, $7$, the zero polynomial.
2. Factor $x^4-16$ completely over the reals.
3. Find the remainder when $2x^3+x-5$ is divided by $x+1$.

Core

4. Factor $x^3-4x^2+x+6$, given that $x=-1$ is a root.
5. For $P(x)=-(x+1)^2(x-3)^3$, give root multiplicities, crossing/touching behavior and both tails.
6. Solve $(x+1)^2(x-3)\ge0$.
7. A quadratic has roots $-2,3$ and takes value $12$ at $0$. Find it.
8. Find every rational root of $2x^3-3x^2-8x+12$ and factor it over the reals.

Synthesis

9. For $P(x)=x^3-3x$, locate intercepts and stationary points, classify the latter, and state the tails.
10. Show that a nonzero polynomial of odd degree has at least one real root, specifying the continuity fact used. Must an even-degree polynomial have a real root?
11. Construct a degree-five polynomial with a double root at $-1$, a triple root at $2$, and value $8$ at $0$.
12. A student cancels $x-1$ in $R(x)=(x-1)(x+2)/(x-1)$ and draws the entire line $y=x+2$. Explain the missing graph feature and whether $x=1$ is a root of the original function.

8.7 Full solutions

1. $3x^4-2$ is degree four. $x+1/x$ is not a polynomial in $x$. The nonzero constant $7$ is degree zero. The zero polynomial has no degree under this chapter's convention.
2. $x^4-16=(x^2-4)(x^2+4)=(x-2)(x+2)(x^2+4)$. The final quadratic has no real root and no real linear factor.
3. The divisor is $x-(-1)$, so the remainder is $P(-1)=-2-1-5=-8$.
4. Divide by $x+1$ to get $x^2-5x+6=(x-2)(x-3)$. Thus the polynomial is $(x+1)(x-2)(x-3)$.
5. The root $-1$ has multiplicity two and touches; $3$ has multiplicity three and crosses with zero slope. The degree is five with leading coefficient $-1$: left tail tends to $+\infty$ and right tail to $-\infty$.
6. Away from $-1$, the square factor is positive, so the sign is that of $x-3$. Include the zero at $-1$ as well as all $x\ge3$. The answer is $\{-1\}\cup[3,\infty)$.
7. Write $P(x)=A(x+2)(x-3)$. Then $P(0)=-6A=12$, so $A=-2$ and $P(x)=-2x^2+2x+12$.
8. Group: $x^2(2x-3)-4(2x-3)=(2x-3)(x^2-4)=(2x-3)(x-2)(x+2)$. All roots are $3/2,2,-2$, and degree three excludes any additional distinct roots.
9. $P=x(x^2-3)$ gives intercepts at $-\sqrt3,0,\sqrt3$. Since $P'=3(x^2-1)$, stationary inputs are $-1,1$. The derivative is positive outside $[-1,1]$ and negative inside, so $(-1,2)$ is a local maximum and $(1,-2)$ a local minimum. The cubic leading term makes the tails $-\infty$ left and $+\infty$ right.
10. The leading term has opposite signs at sufficiently large positive and negative inputs. Choose finite endpoints with those opposite signs. A polynomial is continuous on the closed interval between them, so the intermediate value theorem gives a zero. Even degree need not have a root: $x^2+1>0$ everywhere. It may have roots, as $x^2-1$ does.
11. The required form is $A(x+1)^2(x-2)^3$. Its value at zero is $-8A=8$, so $A=-1$. The factors have the exact multiplicities because the other factor is nonzero at each root.
12. The original domain excludes $1$, so the line has a hole at $(1,3)$. The point $1$ cannot be a root of a function that is not defined there. Its only root is $-2$. A limit at the hole exists and equals $3$, as Chapter 9 develops.

Summary and transfer

Division explains the remainder theorem and factor theorem. Factors provide roots and multiplicities; odd multiplicity crosses and even multiplicity touches. The leading term controls the tails, while derivatives, when used, locate additional turning behavior. Continuity permits intermediate-value conclusions, but a sign-change search does not detect every repeated root. A rational expression that simplifies to a polynomial can still have a different domain and graph.