Pre-Calculus Polynomial and rational functions

Solving polynomial equations (Precalc-level techniques)

Q: What is the Rational Root Theorem?

Any rational root is plus or minus p over q, where p divides the constant term and q divides the leading coefficient.

Q: How many roots does a polynomial equation have?

Exactly n, the degree, counting multiplicity and complex roots, by the Fundamental Theorem of Algebra.

20 practice questions 0 video lessons Theory + worked examples

Solving Polynomial Equations

California Pre-Calculus • Standard A-APR.3 • Polynomial & Rational Functions

Solving Polynomial Equations is a topic in Polynomial & Rational Functions in the California Common Core State Standards. It is aligned to Standard A-APR.3, which requires students to identify zeros of polynomials and apply the Remainder and Factor theorems.

Solving a polynomial equation means finding every root by factoring — using grouping, the Rational Root Theorem, and synthetic division — with a degree-\(n\) equation having \(n\) roots counting complex ones.

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Practice questions

Every question with a fully worked solution.

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Theory

To solve a polynomial equation \(p(x)=0\), rewrite it as a product of factors and set each factor to zero (the zero-product property). The real solutions are exactly the \(x\)-intercepts of \(y=p(x)\).

Useful tools:

Common factor / grouping for four-term expressions.
Quadratic form: substitute \(u=x^2\) (or similar) to reduce the degree.
Rational Root Theorem: any rational root is \(\pm\dfrac{\text{factor of constant}}{\text{factor of leading coeff.}}\).
Synthetic division to test a candidate and peel off a factor.

Fundamental Theorem of Algebra: a degree-\(n\) polynomial has exactly \(n\) roots in the complex numbers, counting multiplicity. Some may be complex conjugate pairs.

Real solutions are the \(x\)-intercepts of the graph.

Synthetic division by \((x-3)\): remainder 0 confirms \(x=3\) is a root, leaving \(x^2-x-2\).

The Rational Root Theorem and the Factor Theorem:

\[\text{rational root}=\pm\dfrac{p}{q},\quad p\mid a_0,\ q\mid a_n;\qquad (x-r)\mid f \iff f(r)=0\]

rational roots are plus or minus p over q; x minus r divides f exactly when f of r is zero

Complex roots come in pairs. For a polynomial with real coefficients, if \(a+bi\) is a root then so is \(a-bi\).

How to solve a polynomial equation

Set the equation to \(0\) and factor out any common factor.
Try structure: grouping, or a \(u\)-substitution for quadratic form.
List rational-root candidates and test with synthetic division to find one factor.
Solve the leftover quotient (often a quadratic, via factoring or the quadratic formula).
Collect all roots, real and complex.

Example 1 — Factor by grouping

Solve \(x^3+2x^2-9x-18=0\).

Solution

Group in pairs and factor each pair.

\(x^2(x+2)-9(x+2)\)	\(=\)	\(0\)
\((x+2)(x^2-9)\)	\(=\)	\(0\)
\((x+2)(x-3)(x+3)\)	\(=\)	\(0\)

So \(x=-2,\ 3,\ -3\).

solutions x equals negative 2, 3, and negative 3

Example 2 — Quadratic form

Solve \(x^4-5x^2+4=0\).

Solution

Let \(u=x^2\); the equation becomes a quadratic in \(u\).

\(u^2-5u+4\)	\(=\)	\(0\)
\((u-1)(u-4)\)	\(=\)	\(0\)
\(u\)	\(=\)	\(1\ \text{ or }\ 4\)

Back-substitute \(u=x^2\):

\(x^2=1\)	\(\Rightarrow\)	\(x=\pm 1\)
\(x^2=4\)	\(\Rightarrow\)	\(x=\pm 2\)

solutions x equals plus or minus 1 and plus or minus 2

Example 3 — Rational Root Theorem + synthetic division

Solve \(x^3-4x^2+x+6=0\).

Solution

Possible rational roots are \(\pm\)(factors of 6): \(\pm1,\pm2,\pm3,\pm6\). Testing \(x=3\) by synthetic division gives remainder 0, so \((x-3)\) is a factor and the quotient is \(x^2-x-2\).

\((x-3)(x^2-x-2)\)	\(=\)	\(0\)
\((x-3)(x-2)(x+1)\)	\(=\)	\(0\)

So \(x=3,\ 2,\ -1\).

solutions x equals 3, 2, and negative 1

Example 4 — Complex roots

Solve \(x^2+9=0\), and state how many roots \(x^3-1=0\) has.

Solution

Isolate and take square roots, allowing imaginary values.

\(x^2\)	\(=\)	\(-9\)
\(x\)	\(=\)	\(\pm 3i\)

By the Fundamental Theorem of Algebra, \(x^3-1=0\) has exactly 3 roots (counting complex ones): one real, \(x=1\), and two complex.

x equals plus or minus 3 i; a degree-3 equation has 3 roots

Common pitfalls

Back-substitute after a \(u\)-substitution. Solving for \(u\) is only halfway; return to \(x\).

Don't lose complex roots. \(x^2=-9\) gives \(x=\pm 3i\), not “no solution.”

Rational Root Theorem lists candidates, not answers. You still must test each until a factor works.

Frequently asked questions

How do you solve a polynomial equation?

Set it to zero, factor it fully, and set each factor to zero. Use grouping, substitution, the Rational Root Theorem, and synthetic division as needed.

What is the Rational Root Theorem?

Any rational root of a polynomial equation is \(\pm\dfrac{p}{q}\), where \(p\) divides the constant term and \(q\) divides the leading coefficient.

How many roots does a polynomial equation have?

Exactly \(n\), the degree, counting multiplicity and complex roots (Fundamental Theorem of Algebra).

What is synthetic division used for?

To quickly test whether a candidate is a root and, if so, to divide out its factor, lowering the degree of the equation you still have to solve.

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