08/14 8. Higher-Degree Polynomial Functions

Solving Higher-Degree Polynomial Equations

Higher-degree polynomial equations extend the factoring methods used for quadratics. A polynomial equation has the form

where $n$ is a nonnegative integer and $a_n\ne 0$. The degree is $n$.

A reliable solution process is:

1. Move every term to one side so the equation equals zero.

2. Factor the polynomial as completely as possible.

3. Apply the Zero-Product Property: if $AB=0$, then $A=0$ or $B=0$.

4. Solve each factor equation.

5. Check solutions in the original equation when necessary.

The Factor Theorem connects zeros with factors: for a polynomial $f(x)$, $f(k)=0$ exactly when $x-k$ is a factor. For example,

so the solutions of $x^3-4x^2-x+4=0$ are $x=4$, $x=1$, and $x=-1$.

Takeaway: Set the equation equal to zero, factor, and use each factor to identify a solution.

Complex Numbers and Polynomial Zeros

The imaginary unit is defined by $i=\sqrt{-1}$ and $i^2=-1$. A complex number has the form $a+bi$, with real part $a$ and imaginary coefficient $b$. Complex arithmetic uses the distributive property together with the substitution $i^2=-1$. For example,

The Fundamental Theorem of Algebra guarantees exactly $n$ complex zeros, counting multiplicity, for every nonconstant polynomial of degree $n$. Some of those zeros may be real. For example,

so its four zeros are $x=\pm\sqrt{2}$ and $x=\pm\sqrt{3}$.

For polynomials with real coefficients, the Complex Conjugate Theorem says that nonreal zeros occur in conjugate pairs. If $2+3i$ is a zero, then $2-3i$ is also a zero, and their factors combine to form

Takeaway: Include complex solutions when the degree requires them, and use conjugate pairs when coefficients are real.

Finding Rational Zeros Efficiently

When a polynomial has integer coefficients, the Rational Root Theorem narrows the search for rational zeros. If $\frac{p}{q}$ is a rational zero in lowest terms, then $p$ divides the constant term and $q$ divides the leading coefficient. The candidates are therefore formed as

These candidates are possibilities, not guaranteed zeros. Test them by substitution or synthetic division. For

a test of $x=2$ gives $f(2)=0$, so $x-2$ is a factor. Synthetic division and factoring produce

The solutions are therefore $x=2$, $x=\frac{3}{2}$, and $x=-2$.

A practical strategy is to list candidates, test a convenient candidate, divide out each confirmed factor, and solve the remaining factor. If a quadratic remains, use factoring or the quadratic formula; if its discriminant is negative, its solutions are nonreal.

Takeaway: Use the Rational Root Theorem to organize the search, but verify every candidate.

Interpreting Polynomial Graphs

A polynomial function is continuous: its graph has no breaks, holes, or vertical asymptotes. Its factored form reveals several important features. In

each $r_i$ is an x-intercept, and its exponent $m_i$ is its multiplicity. The y-intercept is $f(0)$.

The multiplicity determines the local behavior at an x-intercept:

• An odd multiplicity generally causes the graph to cross the x-axis.

• An even multiplicity causes the graph to touch the x-axis and turn around.

• A multiplicity greater than one often makes the graph appear flatter near the intercept.

End behavior is controlled by the degree and the sign of the leading coefficient:

• Even degree with a positive leading coefficient: both ends rise.

• Even degree with a negative leading coefficient: both ends fall.

• Odd degree with a positive leading coefficient: the left end falls and the right end rises.

• Odd degree with a negative leading coefficient: the left end rises and the right end falls.

For example, $f(x)=-2(x+1)^2(x-3)$ has degree $3$, a negative leading coefficient, a touch-and-turn zero at $x=-1$, and a crossing zero at $x=3$. Its y-intercept is $f(0)=6$. These features are enough to create a reliable sketch without plotting many points.

A degree-$n$ polynomial has at most $n-1$ turning points and at most $n$ real zeros.

Takeaway: Read intercepts, multiplicities, end behavior, and the y-intercept directly from the polynomial.

Solving Polynomial Inequalities

A polynomial inequality is solved by locating the real zeros and determining the sign of the polynomial on each interval. Use this procedure:

1. Move all terms to one side.

2. Factor completely when possible.

3. Find the real zeros, which divide the number line into intervals.

4. Test one value in each interval or determine the sign from the factors.

5. Include zeros for $\ge$ or $\le$; exclude them for $>$ or $<$.

6. Express the solution in interval notation.

For example,

factors as

The critical numbers are $x=-2$, $x=1$, and $x=2$. Testing the intervals gives a nonnegative product on $(-2,1)$ and $(2,\infty)$, and the endpoints are included because the inequality is non-strict. The solution is

Multiplicity predicts sign changes: crossing a zero of odd multiplicity changes the sign, while crossing a zero of even multiplicity does not. For instance, in $f(x)=(x-2)^2(x+1)$, the sign stays the same across $x=2$ and changes across $x=-1$.

Takeaway: Factor first, divide the number line at the zeros, apply the correct sign and endpoint rules, and write the result in interval notation.

Applying Polynomial Models

Higher-degree polynomials can represent quantities such as dimensions, volume, motion, and revenue. Algebraic solutions must be checked against the context: lengths may need to be positive, and nonreal values may not describe a physical quantity.

For a rectangular box with dimensions $x$, $x+1$, and $x+2$, a volume of $60$ cubic units gives

After expansion,

Testing $x=3$ gives zero, so

The quadratic factor has discriminant

so it has no real zeros. The only real solution is $x=3$, which gives dimensions $3$ units by $4$ units by $5$ units. The negative or nonreal possibilities are rejected because they cannot represent the box's dimensions.

Takeaway: Solve the equation completely, then retain only solutions that satisfy the domain and meaning of the application.