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2. Polynomial & Rational Functions
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Grades 9-12

2.04 Zeros of polynomial functions

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Lesson

Finding real zeros

Recall that the zeros of a polynomial function $P(x)$P(x) are the solutions of the equation $P(x)=0$P(x)=0. Each zero is an $x$x-intercept of the graph of $P(x)$P(x). A polynomial of degree $n>0$n>0 can have at most $n$n different real zeros.

We can classify these real zeros like we classify the real number system.  We say that a zero of a polynomial is a rational zero is a zero that is a rational number. If the zero is an irrational number, then we say that it is irrational.

Real zeros

The polynomial $P(x)=(x+5)(2x-1)$P(x)=(x+5)(2x−1) has rational zeros at $x=-5$x=−5 and $x=\frac{1}{2}$x=12​.

The polynomial $M(x)=x^2-2$M(x)=x2−2 has irrational zeros at $x=\pm\sqrt{2}$x=±√2.

It might seem difficult to find all of the zeros of a polynomial function of a high degree. However, there are some techniques that we can use to help narrow our approach.

 

Using the rational root theorem

If all of the coefficients of a polynomial are integers or rational numbers, then we can use a specific method to find all of the rational zeros if they exist. We call this method the rational root theorem. It is outlined below:

Rational root theorem

If $P(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0$P(x)=an​xn+an−1​xn−1+…+a1​x+a0​ is a polynomial with integer coefficients, degree $n>0$n>0, and $a_0\ne0$a0​≠0,  then every rational zero of $P(x)$P(x) will be of the form $\frac{c}{d}$cd​ where

  • $c$c and $d$d have no common factors (other than $1$1)
  • $c$c is a factor of $a_0$a0​
  • $d$d is a factor of $a_n$an​

Knowing that rational zeros will be of the form outlined above allows us to narrow the choices and find rational zeros more quickly. We can use the rational root theorem to list all the possible rational zeros. Then, we can use synthetic substitution to determine which ones are the actual zeros of the polynomial.

Practice questions

Question 1

Consider the polynomial $P\left(x\right)=2x^3-2x^2-9x-6$P(x)=2x3−2x2−9x−6.

  1. Determine whether the following statement is true or false:

    $\frac{3}{2}$32​ is a possible rational zero of $P\left(x\right)=2x^3-2x^2-9x-6$P(x)=2x3−2x2−9x−6

    True

    A

    False

    B

Question 2

Consider the polynomial $P\left(x\right)=6x^4+4x^3-3x^2-5x-4$P(x)=6x4+4x3−3x2−5x−4.

  1. Write down all the possible rational zeros. List the zeros on the same line, separated by a comma.

Question 3

Consider the polynomial equation $2x^3-x^2-9x-4=0$2x3−x2−9x−4=0.

  1. Write down all the possible rational zeros, according to the Rational Zeros Theorem.

  2. We have found $8$8 possible rational roots.

    Now let's start by testing the smallest possible root, $x=\frac{1}{2}$x=12​ by synthetic division

    $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
    $\editable{}$ $\editable{}$ $\editable{}$
    $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$

  3. We did not have a zero remainder on the right, which means $x=\frac{1}{2}$x=12​ is not a solution. Let's now test $x=-\frac{1}{2}$x=−12​.

    $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
    $\editable{}$ $\editable{}$ $\editable{}$
    $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$

  4. The zero remainder tells us that $-\frac{1}{2}$−12​ is a root of the equation $2x^3-x^2-9x-4=0$2x3−x2−9x−4=0, which means $x+\frac{1}{2}$x+12​ is a factor.

    Rewrite the equation in the form $(Ax+B)$(Ax+B)$(Cx^2+Dx+E)$(Cx2+Dx+E), such that $A$A and $B$B are integers.

  5. Now solve the quadratic $\left(x^2-x-4\right)=0$(x2−x−4)=0 to find the remaining two roots. Write your solutions on the same line, separated by a comma, or by using the ± symbol.

 

Using the intermediate value theorem

Since polynomials are continuous functions, we can also apply the intermediate value theorem to narrow the search for finding real zeros.

The intermediate value theorem

If $f(x)$f(x) is a function that is continuous over the interval $[a,b]$[a,b], then for any number $n$n between $f(a)$f(a) and $f(b)$f(b) there is a number $c$c such that $aa<c<b and $f(c)=n$f(c)=n.

One immediate implication of the intermediate value theorem is that it gives us a way to determine whether a continuous function has a zero in a given interval. Notice that if the sign of $f(a)$f(a) is different to the sign of $f(b)$f(b), then there must be some function value between $f(a)$f(a) and $f(b)$f(b) that is exactly equal to $0$0. There will then be a corresponding $x$x-value for this function value that we call the zero of the function.

Locating zeros of a function

If $f(x)$f(x) is a function that is continuous over the interval $[a,b]$[a,b], and if $f(a)$f(a) and $f(b)$f(b) have opposite signs, then there exists at least one value $c$c such that $aa<c<b and $f(c)=0$f(c)=0.

 

Exploration

A certain quantity varies with time according to the rule $x(t)=t^3-2t^2+1$x(t)=t3−2t2+1. It is clear that $x(1)=0$x(1)=0, but we wonder whether there are other values of $t$t such that $x(t)=0$x(t)=0.

We choose some values for $t$t. Say, $t=-1$t=−1, $t=0.5$t=0.5, $t=1.5$t=1.5, and $t=2$t=2, and we calculate $x(t)$x(t) in each case.

$t$t $-1$−1 $0.5$0.5 $1.5$1.5 $2$2
$x\left(t\right)$x(t) $-2$−2 $0.625$0.625 $-0.125$−0.125 $1$1

By the intermediate value theorem, we see that the function $x(t)$x(t) has a zero between $t=-1$t=−1 and $t=0.5$t=0.5, another between $0.5$0.5 and $1.5$1.5 (which we already knew about), and another between $t=1.5$t=1.5 and $t=2$t=2.

We conclude that there is at least three zeros in the interval between $t=-1$t=−1 and $t=2$t=2.

Practice questions

Question 4

Consider the polynomial $P\left(x\right)=4x^2-8x+2$P(x)=4x2−8x+2. Dylan would like to know if it has a real zero between $x=1$x=1 and $x=2$x=2.

  1. Find $P\left(1\right)$P(1).

  2. Find $P\left(2\right)$P(2).

  3. What conclusion can Dylan make about $P\left(x\right)$P(x) between $x=1$x=1 and $x=2$x=2 using the intermediate value theorem?

    There are no real zeros between $x=1$x=1 and $x=2$x=2.

    A

    Dylan cannot conclude anything about the zeros of $P\left(x\right)$P(x).

    B

    There is exactly one real zero between $x=1$x=1 and $x=2$x=2.

    C

    There is at least one real zero between $x=1$x=1 and $x=2$x=2.

    D

Question 5

Consider the polynomial $P\left(x\right)=4x^3-x^2+7x+7$P(x)=4x3−x2+7x+7. Sharon would like to know if it has a real zero between $x=-0.8$x=−0.8 and $x=-0.7$x=−0.7.

  1. Find $P\left(-0.8\right)$P(−0.8) to one decimal place.

  2. Find $P\left(-0.7\right)$P(−0.7) to one decimal place.

  3. What conclusion can Sharon make about $P\left(x\right)$P(x) between $x=-0.8$x=−0.8 and $x=-0.7$x=−0.7 using the intermediate value theorem?

    There is exactly one real zero between $x=-0.8$x=−0.8 and $x=-0.7$x=−0.7

    A

    She cannot conclude anything about the zeros of the function.

    B

    There is at least one real zero between $x=-0.8$x=−0.8 and $x=-0.7$x=−0.7

    C

    There are no real zeros between $x=-0.8$x=−0.8 and $x=-0.7$x=−0.7

    D

Question 7

Consider the polynomial $P\left(x\right)=2x^3-8x^2+6x+6$P(x)=2x3−8x2+6x+6. Yuri would like to know if it has a real zero between $x=2.5$x=2.5 and $x=2.6$x=2.6.

  1. Find $P\left(2.5\right)$P(2.5) to one decimal place.

  2. Find $P\left(2.6\right)$P(2.6) to one decimal place.

  3. What conclusion can Yuri make about $P\left(x\right)$P(x) between $x=2.5$x=2.5 and $x=2.6$x=2.6 using the intermediate value theorem?

    There is exactly one real zero between $x=2.5$x=2.5 and $x=2.6$x=2.6.

    A

    There is at least one real zero between $x=2.5$x=2.5 and $x=2.6$x=2.6.

    B

    There are no real zeros between $x=2.5$x=2.5 and $x=2.6$x=2.6.

    C

    He cannot conclude anything about the zeros of $P\left(x\right)$P(x).

    D

 

Using Descartes' rule of signs

Descartes' rule of signs gives us a method for determining how many positive and negative real roots a polynomial equation can have.

We assume the constant term of the polynomial in question is non-zero. We write the polynomial terms in order of descending powers. The coefficients of the terms can be positive or negative and we count how many changes of sign there are for the coefficients as we read through the terms of the polynomial.

Descartes' rule of signs

If P(x) is a polynomial with real coefficients and nonzero constant term, then 

  • The number of positive real zeros of the polynomial is equal to the number of variations of sign in $P(x)$P(x) or less than that number by an even integer.
  • The number of negative real zeros of the polynomial is equal to the number of variations of sign in $P(-x)$P(−x) or is less than that number by an even integer.

For example, if $P(x)=x^4-2x^3+2x^2-5x+4$P(x)=x4−2x3+2x2−5x+4 the signs of the coefficients are positive-negative-positive-negative-positive, making a total of four sign changes. 

Descartes rule says there can be no more than four positive real roots if there are four changes of sign. The maximum number of positive real roots is the number of changes of sign but the actual number of positive real roots is some multiple of two less than the maximum. So, in the example, there could be four, two or zero positive real roots.

The maximum number of negative real roots is the corresponding number of changes of sign in $P(-x)$P(−x). In the example, $P(-x)=x^4+2x^3+2x^2+5x+4$P(−x)=x4+2x3+2x2+5x+4. There are no sign changes. So, there can be no negative real roots.

 

Worked example

Question 8

What can we conclude about the number and sign of the zeros of the polynomial $P(x)=x^5-2x^4-10x^3+20x^2+9x-18$P(x)=x5−2x4−10x3+20x2+9x−18 ?

Think: There are three changes of sign along the list of coefficients. So, there are at most three positive real roots. We find that $P(-x)=-x^5-2x^4+10x^3+20x^2-9x-18$P(−x)=−x5−2x4+10x3+20x2−9x−18, showing two sign changes. Therefore, there are at most two negative real roots.

Do: By inspection, we see that $P(1)=0$P(1)=0. Therefore, $(x-1)$(x−1) must be a factor of $P(x)$P(x). We carry out the division algorithm to find the quotient polynomial.

$P(x)$P(x) $1$1 $-2$−2 $-10$−10 $20$20 $9$9 $-18$−18
$1$1 $1$1 $-1$−1 $-11$−11 $9$9 $18$18
sum $1$1 $-1$−1 $-11$−11 $9$9 $18$18 $0$0

Therefore, $P(x)=(x-1)(x^4-x^3-11x^2+9x+18)$P(x)=(x−1)(x4−x3−11x2+9x+18).

Reflect: You could check that $-1$−1 is a zero of the quotient polynomial. This means that you could perform the division algorithm with this root on the quotient polynomial to obtain a degree three polynomial factor. You may then be able to spot further roots and continue the division process to obtain a full factoring.

 

Practice questions

Question 9

Consider the polynomial $P\left(x\right)=2x^4+5x^3+8x^2-2x-2$P(x)=2x4+5x3+8x2−2x−2.

  1. According to Descartes' rule of signs, what are the possible numbers of positive real zeros for $P\left(x\right)$P(x)? If there are multiple answers, write them on the same line separated by a comma.

  2. Find $P\left(-x\right)$P(−x).

  3. According to Descartes' rule of signs, what are the possible numbers of negative real zeros for $P\left(x\right)$P(x)? If there are multiple answers, write them on the same line separated by a comma.

Question 10

Consider the polynomial $F\left(x\right)=x^7-3x^6+7x^5+8x^4+x^3-3x^2+7x+2$F(x)=x7−3x6+7x5+8x4+x3−3x2+7x+2.

  1. According to Descartes' rule of signs, what are the possible numbers of positive real zeros for $F\left(x\right)$F(x)?

  2. Find $F\left(-x\right)$F(−x).

  3. According to Descartes' rule of signs, what are the possible numbers of negative real zeros for $F\left(x\right)$F(x)?

Question 11

Consider the polynomial $P\left(x\right)=x^4-14x^2-3x+6$P(x)=x4−14x2−3x+6.

  1. Perform an appropriate synthetic division that would help determine whether $4$4 is a root of $P\left(x\right)$P(x):

    $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
        $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
      $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$

  2. Is $4$4 a root of the polynomial $P\left(x\right)$P(x)?

    Yes

    A

    No

    B

  3. Which of the following statements about the roots of $P\left(x\right)$P(x) can we conclude?

    $P\left(x\right)$P(x) has no real roots greater than $4$4.

    A

    There are no real roots of $P\left(x\right)$P(x).

    B

    $P\left(x\right)$P(x) has no real roots less than $4$4.

    C

    We do not know anything more about the roots of $P\left(x\right)$P(x).

    D

 

Finding complex zeros

Just as quadratic functions can have both real and imaginary zeros, so too can other polynomial functions. Although we have no formula for finding the zeros, the fundamental theorem of algebra states that at least one exists.

The fundamental theorem of algebra

If a polynomial $P(x)$P(x) has positive degree and complex coefficients, then $P(x)$P(x) has at least one complex zero.

A special corollary to the fundamental theorem of algebra is that the degree of a polynomial will be equal to its number of zeros in the complex number system.

Corollary

A polynomial function of degree $n$n has exactly $n$n zeros, including repeated zeros, in the complex number system.

 By extending the factor and remainder theorems to the system of complex numbers, and applying the above corollary, we can express every polynomial $P(x)$P(x) of positive degree as a product of polynomials of degree $1$1. We call this the linear factorization theorem.

Linear factorization theorem

If $P(x)$P(x) is a polynomial function of degree $n>0$n>0, then $P(x)$P(x) has exactly $n$n linear factors and

$P(x)=a(x-c_1)(x-c_2)\ldots(x-c_n)$P(x)=a(x−c1​)(x−c2​)…(x−cn​)

where $a$a is the leading coefficient and each number $c_1,c_2,\ldots,c_n$c1​,c2​,…,cn​ are the complex zeros (including repeated zeros) of $P(x)$P(x). 

Using the conjugate root theorem 

In the specific instance where the coefficients of a polynomial are real, the complex solutions to $P\left(z\right)=0$P(z)=0, should there be any, will always occur in conjugate pairs. 

This beautiful result is known as the conjugate root theorem. Specifically, it states that if $a+bi$a+bi is a root of $P\left(z\right)=0$P(z)=0, then so also is its conjugate $a-bi$a−bi. 

The conjugate root theorem

If a polynomial $P(x)$P(x) of degree $n>0$n>0 has real coefficients, and if $a+bi$a+bi is a zero of $P(x)$P(x), then the conjugate $a-bi$a−bi is also a zero of $P(x)$P(x).

We discuss a few examples to illustrate the point. 

Worked examples

Question 12

Think:  Consider the monic cubic function $y=f\left(x\right)$y=f(x) that has only one (and no other) real root at $x=a$x=a.

Do:  Then we know that it must have the form $y=\left(x-a\right)\left(x^2+qx+r\right)$y=(x−a)(x2+qx+r) where the contained quadratic polynomial expression is irreducible and hence has no real zeros itself.

However, if we admit complex numbers as candidates for zeros, we know that there are two other zeros that exist as a complex conjugate pair. 

In addition, suppose we also know that this function has a complex root of $x=2+i$x=2+i and a particular function value given by $f\left(0\right)=35$f(0)=35.  

Reflect:  With this new information, we immediately know two more things:

  1. It must have another complex root of $x=2-i$x=2−i. In other words, the function is expressible as $f\left(x\right)=\left(x-a\right)\left(x-2+i\right)\left(x-2-i\right)$f(x)=(x−a)(x−2+i)(x−2−i).
  2. By multiplying out these last two factors, we see that, equivalently,  $f\left(x\right)$f(x) is given by $f\left(x\right)=\left(x-a\right)\left(x^2-4x+5\right)$f(x)=(x−a)(x2−4x+5). Since $f\left(0\right)=35$f(0)=35, this implies that $-5a=35$−5a=35 and $a=-7$a=−7, and so the three roots are $2+i$2+i, $2-i$2−i, and $-7$−7.      

 

Question 13

Think:  How many real roots might a degree $6$6 polynomial equation with real coefficients have if it is know to have the complex roots $1-i$1−i, $2-i$2−i, and $3-i$3−i?

Do:  The clear answer is none.

Reflect: By the conjugate root theorem, the other three roots are $1+i$1+i, $2+i$2+i and $3+i$3+i. 

 
Question 14

Think:  The polynomial $P\left(x\right)=x^3-x^2-20x+50$P(x)=x3−x2−20x+50 has a zero of $-5$−5. Find the other zeros.

Do:  Here is one way we can proceed:

We can immediately write that $x^3-x^2-20x+50=\left(x+5\right)\left(x^2+px+q\right)$x3−x2−20x+50=(x+5)(x2+px+q) and from equating the constant terms on both sides, we know that $5q=50$5q=50, from which $q=10$q=10.

Equating the terms in $x$x on both sides reveals that $q+5p=-20$q+5p=−20 and since $q=10$q=10, we have that $p=-6$p=−6.

Thus the quadratic factor becomes $x^2-6x+10$x2−6x+10.

Reflect:  This could also have been deduced directly using polynomial division.

A quick calculation shows that the discriminant of the quadratic is $-4$−4, and so $P\left(x\right)$P(x) must only have exactly one real zero.

If we wish to find the complex roots, we could solve the equation   $x^2-6x+10=0$x2−6x+10=0. We can do this by completing squares: 

$x^2-6x+10$x2−6x+10 $=$= 0
$x^2-6x+9+1$x2−6x+9+1 $=$= 0
$\left(x-3\right)^2-i^2$(x−3)2−i2 $=$= 0
$\left(x-3-i\right)\left(x-3+i\right)$(x−3−i)(x−3+i) $=$= 0
$\therefore$∴    $x$x $=$= $3\pm i$3±i
     

 

Practice questions

Question 15

The polynomial $P\left(x\right)=x^3-7x^2+17x-15$P(x)=x3−7x2+17x−15 has a zero at $x=3$x=3. Solve for the other zeros of $P\left(x\right)$P(x).

Question 16

The polynomial $P\left(x\right)=x^4+8x^2-9$P(x)=x4+8x2−9 has a zero at $x=3i$x=3i. Solve for the real zeros of $P\left(x\right)$P(x).

Question 17

Factor $P\left(x\right)=x^4-3x^3+6x^2-48x-160$P(x)=x4−3x3+6x2−48x−160 into linear factors given that $4i$4i is a zero of $P\left(x\right)$P(x).

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