6.11: Solving Polynomial Equations

Definitions and Theorems

A polynomial equation of degree three or higher is solved by rewriting it so that one side is zero, factoring the nonzero side completely, and applying the Zero-Product Property to each factor. The definitions and theorems below supply the tools for the factoring step when the factors are not immediately obvious.

Definition: Polynomial Equation

A polynomial equation in one variable is an equation that can be written in the form\[a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 = 0,\nonumber\]where the coefficients \(a_n, a_{n-1}, \ldots, a_0\) are constants with \(a_n \neq 0\), and \(n\) is a nonnegative integer. The number \(n\) is the degree of the equation.

Definition: Root of a Polynomial Equation

A number \(r\) is a root (or solution) of a polynomial equation if substituting \(x = r\) produces a true statement. Equivalently, if \(p(x)\) denotes the polynomial on the left side, then \(r\) is a root exactly when \(p(r) = 0\).

Theorem: Zero-Product Property

If a product of factors equals zero, then at least one of the factors equals zero. That is, for real numbers or expressions \(a\) and \(b\), if \(ab = 0\), then \(a = 0\) or \(b = 0\) (or both).

Theorem: Factor Theorem

Let \(p(x)\) be a polynomial and let \(r\) be a real number. Then \((x - r)\) is a factor of \(p(x)\) if and only if \(p(r) = 0\).

Theorem: Rational Root Theorem

Let\[a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 = 0\nonumber\]be a polynomial equation with integer coefficients, where \(a_n \neq 0\) and \(a_0 \neq 0\). If this equation has a rational root \(\dfrac{p}{q}\), written in lowest terms, then \(p\) is an integer factor of the constant term \(a_0\), and \(q\) is an integer factor of the leading coefficient \(a_n\).

A Strategy for Higher-Degree Equations

When a polynomial does not factor by inspection, combine the theorems above into a single procedure. Use the Rational Root Theorem to list every candidate \(\dfrac{p}{q}\); test candidates until one, say \(x = r\), satisfies the equation; conclude by the Factor Theorem that \((x - r)\) is a factor, and divide it out to obtain a quotient of lower degree; then repeat the process on that quotient until what remains is quadratic or fully factored.

Removing Rational Roots First

The Rational Root Theorem locates only rational roots. After every rational root has been removed and the remaining quotient is quadratic, apply the techniques for quadratic equations—such as the quadratic formula—to recover any irrational or non-real roots.

Examples

Example \(\PageIndex{1}\): Factoring out a common factor

Solve \(x^3 = 9x\).

Answer

Move every term to one side so that the equation is set equal to zero, then factor out the greatest common factor \(x\) and the resulting difference of squares:\[\begin{aligned} x^3 - 9x &= 0 \\ x(x^2 - 9) &= 0 \\ x(x - 3)(x + 3) &= 0. \end{aligned}\nonumber\]By the Zero-Product Property, each factor may be set equal to zero:\[x = 0, \qquad x - 3 = 0, \qquad x + 3 = 0.\nonumber\]The solutions are\[x = -3, \qquad x = 0, \qquad x = 3.\nonumber\]

Caution: Do Not Divide by the Variable

In Example \(\PageIndex{1}\), dividing both sides of \(x^3 = 9x\) by \(x\) would leave \(x^2 = 9\) and silently discard the root \(x = 0\). Always move all terms to one side and factor instead of dividing by an expression that may equal zero.

Example \(\PageIndex{2}\): Factoring by grouping

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

Answer

Group the four terms in pairs, factor each pair, and then factor out the common binomial:\[\begin{aligned} (x^3 + 2x^2) + (-9x - 18) &= 0 \\ x^2(x + 2) - 9(x + 2) &= 0 \\ (x + 2)(x^2 - 9) &= 0 \\ (x + 2)(x - 3)(x + 3) &= 0. \end{aligned}\nonumber\]Setting each factor equal to zero gives the solutions\[x = -3, \qquad x = -2, \qquad x = 3.\nonumber\]

Example \(\PageIndex{3}\): Grouping with a leading coefficient

Solve \(2x^3 - 3x^2 - 8x + 12 = 0\).

Answer

Group and factor each pair; the common binomial \(2x - 3\) then factors out:\[\begin{aligned} (2x^3 - 3x^2) + (-8x + 12) &= 0 \\ x^2(2x - 3) - 4(2x - 3) &= 0 \\ (2x - 3)(x^2 - 4) &= 0 \\ (2x - 3)(x - 2)(x + 2) &= 0. \end{aligned}\nonumber\]From \(2x - 3 = 0\) we obtain \(x = \dfrac{3}{2}\); the remaining factors give \(x = 2\) and \(x = -2\). The solutions are\[x = -2, \qquad x = \dfrac{3}{2}, \qquad x = 2.\nonumber\]

Example \(\PageIndex{4}\): Using the Rational Root Theorem

Solve \(x^3 - 6x^2 + 11x - 6 = 0\).

Answer

The leading coefficient is \(1\) and the constant term is \(-6\). By the Rational Root Theorem, every rational root \(\dfrac{p}{q}\) has \(p\) dividing \(6\) and \(q\) dividing \(1\), so the candidates are\[\pm 1, \quad \pm 2, \quad \pm 3, \quad \pm 6.\nonumber\]Testing \(x = 1\) gives \(1 - 6 + 11 - 6 = 0\), so \(x = 1\) is a root. By the Factor Theorem, \((x - 1)\) is a factor; dividing it out leaves the quotient \(x^2 - 5x + 6\). Therefore\[(x - 1)(x^2 - 5x + 6) = 0,\nonumber\]and factoring the quadratic gives\[(x - 1)(x - 2)(x - 3) = 0.\nonumber\]The solutions are\[x = 1, \qquad x = 2, \qquad x = 3.\nonumber\]

Example \(\PageIndex{5}\): A leading coefficient other than one

Solve \(2x^3 + 3x^2 - 8x + 3 = 0\).

Answer

The leading coefficient is \(2\) and the constant term is \(3\). Each rational root \(\dfrac{p}{q}\) has \(p\) dividing \(3\) and \(q\) dividing \(2\), so the candidates are\[\pm 1, \quad \pm 3, \quad \pm \dfrac{1}{2}, \quad \pm \dfrac{3}{2}.\nonumber\]Testing \(x = 1\) gives \(2 + 3 - 8 + 3 = 0\), so \(x = 1\) is a root. By the Factor Theorem, \((x - 1)\) is a factor; dividing it out leaves the quotient \(2x^2 + 5x - 3\). Therefore\[(x - 1)(2x^2 + 5x - 3) = 0,\nonumber\]and factoring the quadratic gives\[(x - 1)(2x - 1)(x + 3) = 0.\nonumber\]The solutions are\[x = -3, \qquad x = \dfrac{1}{2}, \qquad x = 1.\nonumber\]

Example \(\PageIndex{6}\): Combining every technique

Solve \(x^5 - 5x^4 + 5x^3 + 5x^2 - 6x = 0\).

Answer

First factor out the greatest common factor \(x\):\[x(x^4 - 5x^3 + 5x^2 + 5x - 6) = 0.\nonumber\]One solution is \(x = 0\); the remaining roots come from the quartic factor. For that quartic the leading coefficient is \(1\) and the constant term is \(-6\), so the Rational Root Theorem gives candidates \(\pm 1, \pm 2, \pm 3, \pm 6\). Testing \(x = 1\) gives \(1 - 5 + 5 + 5 - 6 = 0\), so \((x - 1)\) is a factor; dividing it out leaves\[x^3 - 4x^2 + x + 6.\nonumber\]Testing \(x = -1\) in this cubic gives \(-1 - 4 - 1 + 6 = 0\), so \((x + 1)\) is a factor; dividing it out leaves the quotient \(x^2 - 5x + 6\), which factors as \((x - 2)(x - 3)\). Assembling every factor,\[x(x - 1)(x + 1)(x - 2)(x - 3) = 0.\nonumber\]The solutions are\[x = -1, \qquad x = 0, \qquad x = 1, \qquad x = 2, \qquad x = 3.\nonumber\]

---

Sources

Several parts of this text use modifications from the following sources:

• Wikipedia article: "Zero-product property"
• Wikipedia article: "Factor theorem"
• Wikipedia article: "Rational root theorem"

All of these sources are released under the Creative Commons Attribution-Share-Alike License 4.0.