13

ROOTS OF POLYNOMIALS

OF DEGREE GREATER THAN 2

The factor theorem

The fundamental theorem of algebra

A strategy for finding roots

The integer root theorem

Conjugate pairs

Proof of the factor theorem

Proof of the integer root theorem

IN THIS TOPIC we will see how to find the roots of a polynomial of degree greater than 2.  This will depend on the previous topic:  Synthetic division.

We saw in that topic what is called the factor theorem.

The Factor Theorem.   x − r  is a factor of a polynomial P(x) if and only if  r is a root of P(x).

This means that if a polynomial can be factored, for example, as follows:

P(x) = (x − 1)(x + 2)(x + 3)

then the theorem tells us that the roots are 1, −2, and −3.

Conversely, if we know that roots of a polynomial are  −2, 1, and 5, then the polynomial has the following factors:

(x + 2)(x − 1)(x − 5)

We will see how to prove the factor theorem below.

a)   Use the Factor Theorem to prove:  (x + 1) is a factor of x⁵ + 1.

To see the answer, pass your mouse over the colored area.
To cover the answer again, click "Refresh" ("Reload").
Do the problem yourself first!

−1 is a root of x⁵ + 1.  For, (−1)⁵ + 1 = −1 + 1 = 0.
Therefore, according to the Factor Theorem,
[x −(−1)] = (x + 1) is a factor.

b)   Use synthetic division to find the other factor.

Therefore, x⁵ + 1 = (x + 1)(x⁴ − x³ + x² − x + 1)

Following this same procedure, we could prove:

(x + a) is a factor of x⁵ + a⁵,

and completely generally:

(x + a) is a factor of xⁿ + aⁿ, where n is odd.

The Fundamental Theorem of Algebra

A polynomial P(x) of degree n has exactly n roots, real or complex.

If the leading coefficient of P(x)is 1, then the Factor Theorem allows us to conclude:

P(x) = (x − r_n)(x − r_{n − 1}). . . (x − r₂)(x − r₁)

Hence a polynomial of the third degree, for example, will have three roots.  And if they are all real, then its graph will look something like this:

For, the three roots are the three x-intercepts.

Problem 3.   Determine the polynomial with integer coefficients whose roots are −½, −2, −2, and sketch the graph.

Question.   If r is a root of a polynomial p(x), then upon dividing p(x) by x − r, what remainder do you expect?

0.  Because r being a root will mean that x − r is a factor
of p(x).

Problem 4.   Is x = 2 a root of this polynomial:

x⁶ − 3x⁵ + 3x⁴ − 3x³ + 3x² −3x + 2 ?

Example 2.   Find the three roots of

P(x) = x³ − 2x² − 9x + 18,

given that one root is 3.

Solution.  Since 3 is a root of P(x), then according to the factor theorem, x − 3 is a factor.  Therefore, on dividing P(x) by x − 3, we can find the other, quadratic factor.

We have

The three roots are:  2, −3, 3.

Again, since x − 3 is a factor of P(x), the remainder is 0.

Problem 5.   Sketch the graph of this polynomial,

y = x³ − 2x² − 5x + 6,

given that one root is −2.

We will prove this below.

This Integer Root Theorem is an instance of the more general Rational Root Theorem:

If the rational number r/s is a root of a polynomial whose coefficients are integers, then the integer r is a factor of the constant term, and the integer s is a factor of the leading coefficient.

Example.   What are the possible integer roots of  x³ − 4x² + 2x + 4?

Answer.   If there are integer roots, they will be factors of the constant term 4; namely:  ±1,  ±2,  ±4.

Now, is 1 a root?  To answer, we will divide the polynomial by x − 1, and hope for remainder 0.

The remainder is not 0.  1 is not a root.  Let's try −1:

The remainder again is not 0.  Let's try 2:

Yes!  2 is a root.  We have

x³ − 4x² + 2x + 4 = (x² − 2x − 2)(x − 2)

We can now find the roots of the quadratic by completing the square.  As we found in Topic 11:

x = 1 ±

Therefore, the three roots are

1 + ,  1 − ,  2.

Problem 6.

a)  What are the possible integer roots of this polynomial?

x³ − 2x² − 3x + 1

±1.  Because those are the only factors of the constant term.

b)  Does that polynomial have integer roots?

No, because neither 1 nor −1 will make that polynomial equal to 0.  Synthetic division by both ±1 does not give remainder 0.

Problem 7.   Factor this polynomial into a product of linear factors.

x³ + 2x² − 5x − 6

If the irrational number a +   is a root, then its conjugate  a −   is also a root.  (See Skill in Algebra, Lesson 28.)  And if the complex number a + bi is a root, then so is its conjugate, a − bi.

Example 6.   A polynomial P(x) has the following roots:

−2,  1 + ,  5i.

What is the smallest degree that P(x) could have?

 Answer.   5.  For, since 1 + is a root, then so is its conjugate, 1 − .  And since 5i is a root, so is its conjugate, −5i.

P(x) has at least these 5 roots:

−2,  1 ± ,   ±5i.

Problem 8.   Construct a polynomial that has the following root:

a)   2 +

Since 2 + is a root, then so is 2 − .  Therefore, according to the theorem of the sum and product of the roots (Topic 10), they are the roots of x² − 4x + 1.  .

b)   2 − 3i

Since 2 − 3i is a root, then so is 2 + 3i.  Again, according to the theorem of the sum and product of the roots, they are the roots of x² − 4x + 13.  See Topic 10, Example 7.

Problem 9.   Let f(x) = x⁵ + x⁴ + x³ + x² − 12x − 12.  One root is and another is −2i.

a)  Name the certain roots that f(x) has.

±, ±2i.

b)  What are the possible integer roots?

The factors of the constant term −12:

±1, ±2, ±3, ±4, ±6, ±12.

c)  Of those twelve possible roots, how many could f(x) actually have?

One.  This is a polynomial of degree 5, which has 5 roots, and we already know 4 of them.

Problem 10.   Is it possible for a polynomial of the 5th degree to have 2 real roots and 3 imaginary roots?

No, it is not. Since imaginary roots always come in pairs, then if there are any imaginary roots, there will always be an even number of them.

Consider the graph of a 5th degree polynomial with positive leading term.  When x is a large negative number, the graph is below the x-axis.  When x is a large positive number, it is above the x-axis. Therefore, the graph must cross the x-axis at least once.  Now, can you draw the graph so that it crosses the x-axis exactly twice?  No, you cannot.  A polynomial of odd degree must have an odd number of real roots.

First, if  (x − r)  is a factor of P(x), then P(r) will have the factor (r − r), which is 0.  This will make P(r) = 0.  This means that r is a root.

Conversely, if r is a root of P(x), then P(r) = 0.   But according to the remainder theorem, P(r) = 0 means  that upon dividing P(x) by x − r, the remainder is 0.  x − r, therefore, is a factor of P(x).

This is what we wanted to prove.

Proof of the integer root theorem

If an integer is a root of a polynomial whose coefficients are integers
and whose leading coefficient is ±1, then that integer is a factor of the constant term.

Let the integer r be a root of this polynomial:

P(x) = ±xⁿ + a_n−1xⁿ⁻¹ + a_n−2xⁿ⁻² + . . . + a₂x² + a₁x + a₀,

where the a's are integers.  Then, since r is a root,

P(r) = ±rⁿ + a_n−1rⁿ⁻¹ + a_n−2rⁿ⁻² + . . . + a₂r² + a₁r + a₀ = 0.

Transpose the constant term a₀, and factor r from the remaining terms:

r(±rⁿ⁻¹ + a_n−1rⁿ⁻² + . . . + a₂r + a₁) = −a₀

Now the a's are all integers; therefore the expression in parentheses is an integer, which, for convenience, we will call −q:

r(−q) = −a₀,

or,

rq = a₀.

Thus, the constant term a₀ can be factored as rq, if r and q are both integers.  Under those conditions, then, r is a factor of the constant term.

This is what we wanted to prove.

Next Topic:  Multiple roots

Table of Contents | Home