3.3 Real Zeros of Polynomials

3.3.1 Learning Objectives

By the end of this section, you should be able to:

list possible rational zeros using the Rational Zeros Theorem;
use synthetic division to identify rational zeros and factor polynomials;
predict possible positive and negative zeros using Descartes’ Rule of Signs;
use bounds and graphs to locate real zeros; and
solve polynomial equations exactly or approximately.

3.3.2 The Rational Zeros Theorem

For a polynomial

\[ P(x)=a_nx^n+\cdots+a_1x+a_0 \]

with integer coefficients, every rational zero written in lowest terms as \(p/q\) must satisfy:

\(p\) is a factor of the constant term \(a_0\); and
\(q\) is a factor of the leading coefficient \(a_n\).

The theorem produces candidates. Each candidate must still be tested.

Example 3.11 Listing Possible Rational Zeros

List the possible rational zeros of \(P(x)=2x^3-3x^2-8x+12\).

Solution

The factors of \(12\) are \(1,2,3,4,6,12\), and the factors of \(2\) are \(1,2\). After removing duplicates, the candidates are

\[ \boxed{\pm1,\ \pm2,\ \pm3,\ \pm4,\ \pm6,\ \pm12,\ \pm\frac12,\ \pm\frac32}. \]

Exercise 3.11 List the possible rational zeros of \(P(x)=3x^3+5x^2-2\).

3.3.3 Finding Rational Zeros

Test likely candidates with substitution or synthetic division. When the remainder is zero, the corresponding linear expression is a factor. Divide it out and continue with the lower-degree quotient.

Example 3.12 Finding All Rational Zeros

Find the zeros of \(P(x)=2x^3-3x^2-8x+12\).

Solution

Testing \(x=2\) by synthetic division gives

\[ \begin{array}{r|rrrr} 2 & 2 & -3 & -8 & 12\\ & & 4 & 2 & -12\\ \hline & 2 & 1 & -6 & 0 \end{array} \]

Thus,

\[ P(x)=(x-2)(2x^2+x-6). \]

Factor the quadratic:

\[ 2x^2+x-6=(2x-3)(x+2). \]

\[ \boxed{x=2,\quad x=\frac32,\quad x=-2} \]

Exercise 3.12 Find all zeros of \(P(x)=x^3-4x^2-x+4\).

3.3.4 Descartes’ Rule of Signs

A sign variation occurs when consecutive nonzero coefficients have different signs.

Descartes’ Rule of Signs states:

the number of positive real zeros equals the number of sign variations in \(P(x)\), or is less than it by an even integer;
the number of negative real zeros equals the number of sign variations in \(P(-x)\), or is less than it by an even integer.

Zeros are counted with multiplicity. The rule narrows the possibilities but does not give the zeros.

Example 3.13 Counting Possible Real Zeros

Use Descartes’ Rule of Signs for \(P(x)=x^4-3x^3+2x^2+5x-4\).

Solution

The signs of \(P(x)\) are \(+,-,+,+,-\), giving \(3\) variations. Therefore, there are \(3\) or \(1\) positive real zeros.

\[ P(-x)=x^4+3x^3+2x^2-5x-4 \]

has signs \(+,+,+,-,-\), giving \(1\) variation. Therefore, there is exactly \(1\) negative real zero.

\[ \boxed{\text{Positive zeros: }3\text{ or }1;\quad \text{negative zeros: }1} \]

Exercise 3.13 Use Descartes’ Rule of Signs on \(P(x)=2x^5+x^3-4x+7\).

3.3.5 Upper and Lower Bounds

Synthetic division can help bound all real zeros.

If synthetic division by a positive number \(c\) produces a bottom row whose entries are all non-negative, then \(c\) is an upper bound.
If synthetic division by a negative number \(c\) produces a bottom row with alternating non-negative and non-positive entries, allowing zeros, then \(c\) is a lower bound.

These tests help select a graphing window and limit the range of candidates.

Example 3.14 Finding an Upper Bound

Find a positive upper bound for the real zeros of \(P(x)=x^3-2x^2-5x+6\).

Solution

\[ \begin{array}{r|rrrr} 3 & 1 & -2 & -5 & 6\\ & & 3 & 3 & -6\\ \hline & 1 & 1 & -2 & 0 \end{array} \]

The bottom row contains a negative entry, so this test does not establish \(3\) as an upper bound. Try \(4\):

\[ \begin{array}{r|rrrr} 4 & 1 & -2 & -5 & 6\\ & & 4 & 8 & 12\\ \hline & 1 & 2 & 3 & 18 \end{array} \]

All entries are non-negative, so

\[ \boxed{4\text{ is an upper bound}}. \]

Exercise 3.14 Use synthetic division to show that \(5\) is an upper bound for \(P(x)=x^3-3x^2-4x+2\).

3.3.6 Graphical and Numerical Zeros

Some polynomial zeros are irrational and cannot be found using the Rational Zeros Theorem. Desmos can estimate them:

graph \(y=P(x)\);
choose a window containing the real zeros;
select each \(x\)-intercept; and
report approximations to suitable precision.

A sign change between \(P(a)\) and \(P(b)\) guarantees at least one real zero in \((a,b)\).

Example 3.15 Approximating Irrational Zeros

Approximate the real zeros of \(P(x)=x^3-2x-2\).

Solution

The rational candidates \(\pm1,\pm2\) are not zeros. Desmos shows one \(x\)-intercept near \(x=1.77\). Since

\[ P(1)=-3\quad\text{and}\quad P(2)=2, \]

the Intermediate Value Theorem confirms a zero between \(1\) and \(2\).

\[ \boxed{x\approx1.77} \]

Exercise 3.15 Use Desmos to approximate the real zeros of \(P(x)=x^3-3x+1\) to two decimal places.

3.3.7 Conceptual Takeaways

The Rational Zeros Theorem gives a finite list of candidates, not guaranteed zeros.
A zero remainder confirms both a zero and a linear factor.
Descartes’ Rule of Signs limits the possible numbers of positive and negative real zeros.
Bound tests and graphs help focus a search for zeros.
Polynomial zeros may be rational, irrational, or non-real.

3.3.8 Skills You Should Be Able to Do

List all possible rational zeros.
Test candidates and factor a polynomial completely over the real numbers.
Apply Descartes’ Rule of Signs.
Find upper and lower bounds for real zeros.
Use Desmos and sign changes to approximate real zeros.

3.3.9 Practice Problems with Solutions

List the possible rational zeros of \(P(x)=2x^3+5x^2-7x-3\).

Show Solution

Factors of the constant term are \(\pm1,\pm3\); factors of the leading coefficient are \(\pm1,\pm2\). Thus,

\[ \boxed{\pm1,\ \pm3,\ \pm\frac12,\ \pm\frac32}. \]
Find all zeros of \(P(x)=x^3-2x^2-5x+6\).

Show Solution

Testing candidates shows \(P(1)=0\). Dividing by \(x-1\) gives \(x^2-x-6\), so

\[ P(x)=(x-1)(x-3)(x+2). \]

Therefore, \(\boxed{x=-2,1,3}\).
Find all real zeros of \(P(x)=2x^3+x^2-8x-4\).

Show Solution

Factor by grouping:

\[ P(x)=x^2(2x+1)-4(2x+1)=(2x+1)(x^2-4). \]

Thus,

\[ \boxed{x=-\frac12,-2,2}. \]
Use Descartes’ Rule of Signs for \(P(x)=x^5-2x^4+x^2-6\).

Show Solution

The nonzero signs of \(P(x)\) are \(+,-,+,-\), giving \(3\) variations. There are \(3\) or \(1\) positive zeros.

\[ P(-x)=-x^5-2x^4+x^2-6 \]

has signs \(-,-,+,-\), giving \(2\) variations. There are \(2\) or \(0\) negative zeros.

\[ \boxed{\text{Positive: }3\text{ or }1;\quad \text{negative: }2\text{ or }0} \]
Show that \(3\) is an upper bound for the real zeros of \(P(x)=x^3-x^2-4x+2\).

Show Solution

Synthetic division with \(3\) produces the bottom row

\[ 1,\ 2,\ 2,\ 8. \]

All entries are non-negative, so \(\boxed{3\text{ is an upper bound}}\).
Show that \(-3\) is a lower bound for the real zeros of \(P(x)=x^3+x^2-4x-2\).

Show Solution

Synthetic division with \(-3\) gives

\[ 1,\ -2,\ 2,\ -8. \]

The signs alternate, so \(\boxed{-3\text{ is a lower bound}}\).
Approximate the real zeros of \(P(x)=x^3-x-1\) to two decimal places.

Show Solution

None of the rational candidates \(\pm1\) is a zero. A Desmos graph shows one real intercept:

\[ \boxed{x\approx1.32}. \]
A rectangular container has length \(x+2\) centimetres, width \(x-1\) centimetres, and height \(x\) centimetres. Its volume is \(40\text{ cm}^3\). Find its dimensions.

Show Solution

The volume equation is

\[ x(x+2)(x-1)=40. \]

Expanding gives \(x^3+x^2-2x-40=0\). Testing positive factors of \(40\) gives \(x=3\):

\[ 3(5)(2)=30, \]

so \(x=3\) is not correct. Testing \(x=4\) gives

\[ 4(6)(3)=72. \]

There is no integer solution. Using Desmos gives \(x\approx3.29\). Therefore, the dimensions are approximately

\[ \boxed{5.29\text{ cm by }2.29\text{ cm by }3.29\text{ cm}}. \]