Upper and Lower Bounds on Zeros

The Rational Root Theorem gives a list of candidates. Descartes’ Rule tells you how many positive and negative zeros to expect. The bound theorems narrow the search further by establishing a range $[L, U]$ that contains all real zeros. Any candidate outside this range can be skipped entirely.

The Upper Bound Theorem

If you divide $f(x)$ by $(x - c)$ using synthetic division where $c > 0$, and all numbers in the bottom row are non-negative (no sign changes), then $c$ is an upper bound for the real zeros of $f$. No real zero of $f$ is greater than $c$.

Why It Works

When you perform synthetic division of $f(x)$ by $(x - c)$ with $c > 0$ and get all non-negative entries in the bottom row, the quotient polynomial has all non-negative coefficients. For any $x > c$, both the quotient $q(x)$ and the divisor $(x - c)$ are positive, so $f(x) = (x - c)q(x) + r > 0$ (where $r \geq 0$ is the non-negative remainder). Therefore $f(x) > 0$ for all $x > c$, meaning no zero exists beyond $c$.

Worked Example 1: Finding an Upper Bound

Find an upper bound for the zeros of $f(x) = x^4 - 5x^3 + 3x^2 + 7x - 10$.

Try $c = 5$:

$\begin{array}{c|ccccc} 5 & 1 & -5 & 3 & 7 & -10 \\ & & 5 & 0 & 15 & 110 \\ \hline & 1 & 0 & 3 & 22 & 100 \end{array}$

Bottom row: $1, 0, 3, 22, 100$ — all non-negative.

Conclusion: $5$ is an upper bound. No real zero is greater than $5$.

Can we find a tighter bound? Try $c = 4$:

$\begin{array}{c|ccccc} 4 & 1 & -5 & 3 & 7 & -10 \\ & & 4 & -4 & -4 & 12 \\ \hline & 1 & -1 & -1 & 3 & 2 \end{array}$

Bottom row: $1, -1, -1, 3, 2$ — contains negative values. So $4$ is NOT confirmed as an upper bound.

Since $c = 4$ failed and $c = 5$ succeeded, $5$ is the smallest integer upper bound for this polynomial.

The Lower Bound Theorem

If you divide $f(x)$ by $(x - c)$ using synthetic division where $c < 0$, and the numbers in the bottom row alternate in sign (counting zeros as either sign), then $c$ is a lower bound for the real zeros. No real zero of $f$ is less than $c$.

The alternation means: positive, negative, positive, negative, … or negative, positive, negative, positive, … (zeros can be assigned to either sign to continue the pattern).

Worked Example 2: Finding a Lower Bound

Find a lower bound for the zeros of $f(x) = x^4 - 5x^3 + 3x^2 + 7x - 10$.

Try $c = -2$:

$\begin{array}{c|ccccc} -2 & 1 & -5 & 3 & 7 & -10 \\ & & -2 & 14 & -34 & 54 \\ \hline & 1 & -7 & 17 & -27 & 44 \end{array}$

Bottom row: $1, -7, 17, -27, 44$ — alternating signs: $+, -, +, -, +$.

Conclusion: $-2$ is a lower bound. No real zero is less than $-2$.

Can we tighten it? Try $c = -1$:

$\begin{array}{c|ccccc} -1 & 1 & -5 & 3 & 7 & -10 \\ & & -1 & 6 & -9 & 2 \\ \hline & 1 & -6 & 9 & -2 & -8 \end{array}$

Bottom row: $1, -6, 9, -2, -8$ — the pattern is $+, -, +, -, -$. The last two entries are both negative (no alternation), so $-1$ is NOT confirmed as a lower bound.

So the lower bound is $-2$: all real zeros lie in $[-2, 5]$.

Complete Zero-Finding Strategy

Combining all the tools from this cluster:

Step 1: Degree and Leading Coefficient

Determine end behavior and the maximum number of zeros.

Step 2: Descartes’ Rule

Count possible positive and negative real zeros to build the possibilities table.

Step 3: Rational Root Theorem

List all $\frac{p}{q}$ candidates.

Step 4: Bound Theorems

Use synthetic division to find upper and lower bounds. Eliminate candidates outside the bounds.

Step 5: Test Remaining Candidates

Use synthetic division on the remaining candidates, reducing the polynomial’s degree each time you find a zero.

Step 6: Finish

When the polynomial is reduced to degree 2, use the quadratic formula.

Worked Example 3: Full Strategy in Action

Find all zeros of $f(x) = 2x^4 - x^3 - 14x^2 + 7x + 6$.

Step 1: Degree 4, leading coefficient $2 > 0$. Both ends go up.

Step 2 — Descartes’ Rule:

$f(x)$: signs $+, -, -, +, +$. Sign changes: $+\to -$ (1), $+\to +$ (no). Wait, let me recount: $+, -, -, +, +$. Changes: $+\to-$ (1), $-\to-$ (no), $-\to+$ (2), $+\to+$ (no). Two sign changes.

Positive zeros: 2 or 0.

$f(-x) = 2x^4 + x^3 - 14x^2 - 7x + 6$: signs $+, +, -, -, +$. Changes: $+\to+$ (no), $+\to-$ (1), $-\to-$ (no), $-\to+$ (2). Two sign changes.

Negative zeros: 2 or 0.

Step 3 — Candidates: $\frac{p}{q}$ where $p \mid 6$ and $q \mid 2$:

$\pm 1, \pm 2, \pm 3, \pm 6, \pm\frac{1}{2}, \pm\frac{3}{2}$

Step 4 — Bounds:

Test $c = 3$:

$\begin{array}{c|ccccc} 3 & 2 & -1 & -14 & 7 & 6 \\ & & 6 & 15 & 3 & 30 \\ \hline & 2 & 5 & 1 & 10 & 36 \end{array}$

All non-negative: $3$ is an upper bound. Eliminate $6$.

Test $c = -3$:

$\begin{array}{c|ccccc} -3 & 2 & -1 & -14 & 7 & 6 \\ & & -6 & 21 & -21 & 42 \\ \hline & 2 & -7 & 7 & -14 & 48 \end{array}$

Alternating: $+, -, +, -, +$. So $-3$ is a lower bound. Eliminate $-6$.

Remaining candidates: $\pm 1, \pm 2, \pm 3, \pm\frac{1}{2}, \pm\frac{3}{2}$.

Step 5 — Test:

$f(1) = 2 - 1 - 14 + 7 + 6 = 0$. Found $x = 1$.

$\begin{array}{c|ccccc} 1 & 2 & -1 & -14 & 7 & 6 \\ & & 2 & 1 & -13 & -6 \\ \hline & 2 & 1 & -13 & -6 & 0 \end{array}$

Reduced: $2x^3 + x^2 - 13x - 6$.

Test $x = -\frac{1}{2}$:

$\begin{array}{c|cccc} -\frac{1}{2} & 2 & 1 & -13 & -6 \\ & & -1 & 0 & 6.5 \end{array}$

$2, 0, -13, 0.5$ — remainder $0.5 \neq 0$. Not a zero.

Test $x = 2$:

$\begin{array}{c|cccc} 2 & 2 & 1 & -13 & -6 \\ & & 4 & 10 & -6 \\ \hline & 2 & 5 & -3 & -12 \end{array}$

Remainder $-12 \neq 0$.

Test $x = -2$:

$\begin{array}{c|cccc} -2 & 2 & 1 & -13 & -6 \\ & & -4 & 6 & 14 \\ \hline & 2 & -3 & -7 & 8 \end{array}$

Remainder $8 \neq 0$.

Test $x = 3$:

$\begin{array}{c|cccc} 3 & 2 & 1 & -13 & -6 \\ & & 6 & 21 & 24 \\ \hline & 2 & 7 & 8 & 18 \end{array}$

Remainder $18 \neq 0$.

Test $x = -3$:

$\begin{array}{c|cccc} -3 & 2 & 1 & -13 & -6 \\ & & -6 & 15 & -6 \\ \hline & 2 & -5 & 2 & -12 \end{array}$

Remainder $-12 \neq 0$.

Test $x = \frac{1}{2}$:

$\begin{array}{c|cccc} \frac{1}{2} & 2 & 1 & -13 & -6 \\ & & 1 & 1 & -6 \\ \hline & 2 & 2 & -12 & -12 \end{array}$

Remainder $-12 \neq 0$.

Test $x = \frac{3}{2}$:

$\begin{array}{c|cccc} \frac{3}{2} & 2 & 1 & -13 & -6 \\ & & 3 & 6 & -10.5 \\ \hline & 2 & 4 & -7 & -16.5 \end{array}$

Not a zero. Test $x = -\frac{3}{2}$:

$\begin{array}{c|cccc} -\frac{3}{2} & 2 & 1 & -13 & -6 \\ & & -3 & 3 & 15 \\ \hline & 2 & -2 & -10 & 9 \end{array}$

Not a zero. So the cubic $2x^3 + x^2 - 13x - 6$ has no rational zeros from our list (other than what we may have missed). Let us re-check $x = -\frac{1}{2}$ more carefully:

$\begin{array}{c|cccc} -\frac{1}{2} & 2 & 1 & -13 & -6 \\ & & -1 & 0 & 6.5 \\ \hline & 2 & 0 & -13 & 0.5 \end{array}$

Not zero. Let me try $x = -2$ again on the original: $f(-2) = 2(16) - (-8) - 14(4) + 7(-2) + 6 = 32 + 8 - 56 - 14 + 6 = -24 \neq 0$.

Try the original polynomial with $x = -1$: $f(-1) = 2 + 1 - 14 - 7 + 6 = -12 \neq 0$.

The remaining cubic $2x^3 + x^2 - 13x - 6$ must be solved numerically or has rational zeros we have not found. Let me verify by trying to factor directly: grouping $2x^3 + x^2 - 13x - 6$. Testing $x = 2$ gave remainder $-12$, $x = -2$ gave $8$. By the IVT, there is a zero between $2$ and $3$, and also between $-2$ and $-3$. These are irrational zeros. The quadratic formula finishes the job after finding one more rational zero or confirming there are none.

This example illustrates an important point: not all polynomials with integer coefficients have rational zeros, even when the Rational Root Theorem gives many candidates.

Real-World Application: Engineering Tolerance Bands

In manufacturing, a polynomial model describes how a measurement varies with a parameter $x$. If all real roots must lie within a tolerance band $[a, b]$, the bound theorems confirm this algebraically. For instance, if a vibration amplitude is modeled by:

$A(f) = f^3 - 12f^2 + 44f - 48$

where $f$ is frequency in Hz, the bound theorems can verify that all resonant frequencies (zeros) lie within the operational range.

Testing $c = 12$:

$\begin{array}{c|cccc} 12 & 1 & -12 & 44 & -48 \\ & & 12 & 0 & 528 \\ \hline & 1 & 0 & 44 & 480 \end{array}$

All non-negative: $12$ is an upper bound.

Testing $c = -1$:

$\begin{array}{c|cccc} -1 & 1 & -12 & 44 & -48 \\ & & -1 & 13 & -57 \\ \hline & 1 & -13 & 57 & -105 \end{array}$

Alternating signs: $-1$ is a lower bound.

All resonant frequencies lie in $[-1, 12]$. (In fact, this factors as $(f - 2)(f - 4)(f - 6)$, with zeros at $2$, $4$, $6$ Hz.)

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• Upper Bound Theorem: If synthetic division by $(x - c)$ with $c > 0$ gives all non-negative entries, no zero exceeds $c$
• Lower Bound Theorem: If synthetic division by $(x - c)$ with $c < 0$ gives alternating signs, no zero is less than $c$
• The bounds narrow the search interval — eliminate candidates outside $[L, U]$ immediately
• The complete strategy combines degree analysis, Descartes’ Rule, Rational Root Theorem, bounds, and synthetic division
• Start with small candidates ($\pm 1, \pm 2$) before testing larger ones
• When reduced to degree 2, use the quadratic formula to finish
• Not all polynomials have rational zeros — the tools tell you when to stop looking

Return to College Algebra for more topics in this section.