Utah Math 3 - 2020 Edition 3.06 Roots and zeros
Lesson

Recall the relationship between the roots, or zeros, of a polynomial function. The roots, or zeros, of a function are the solutions to the polynomial equation, $P\left(x\right)=0$P(x)=0. It can also be viewed as the value of $x$x that makes the polynomial equal to $0$0, which are the $x$x-intercepts.

## Using the fundamental theorem of algebra

The fundamental theorem of algebra states that the number of roots a polynomial function has is equal to the degree of the function.  The theorem forms the foundation for solving polynomials equations.

Fundamental theorem of algebra

If $f\left(x\right)$f(x) is a polynomial of degree $n>0$n>0, and $a$a is a non-zero real number, then $f\left(x\right)$f(x) has exactly $n$n linear factors

$f\left(x\right)=a\left(x-c_1\right)\left(x-c_2\right)$f(x)=a(xc1)(xc2)...$x-c_n$xcn

where $c_1$c1, $c_2$c2, ... $c_n$cn are complex numbers. Therefore, $f\left(x\right)$f(x) as $n$n roots if we allow for multiplicities (duplicates).

Remember!

Both real and imaginary numbers belong to the set of complex numbers.

#### Worked examples

##### Question 1

Consider the function $h\left(x\right)=x^3-2x^2+9x-18$h(x)=x32x2+9x18

a. How many solutions will the function have?

Think: Remember, the number of solutions is equal to the degree of the function.

Do: $f\left(x\right)$f(x) will have $3$3 solutions.

b. Factor $h\left(x\right)$h(x) by grouping:

Think: We need to first find the GCF of the first two terms and the last two terms. Then we should have a binomial common factor to factor fully.

Do:

 $h\left(x\right)$h(x) $=$= $x^3-2x^2+9x-18$x3−2x2+9x−18 $=$= $x^3-2x^2+9x-18$x3−2x2+9x−18 $=$= $x^2\left(x-2\right)+9\left(x-2\right)$x2(x−2)+9(x−2) $=$= $\left(x-2\right)\left(x^2+9\right)$(x−2)(x2+9)

$x^2+9$x2+9 cannot be factored any further, so we have factored fully. The factors of $h\left(x\right)$h(x) are $x-2$x2 and $x^2+9$x2+9.

c. What are the roots of $h\left(x\right)$h(x)?

Think: To find the roots, set each factor equal to $0$0.

Do:

factor: $x-2$x2

 $x-2$x−2 $=$= $0$0 $x$x $=$= $2$2

factor $x^2+9$x2+9

 $x^2+9$x2+9 $=$= $0$0 $x^2$x2 $=$= $-9$−9 $\sqrt{x^2}$√x2 $=$= $\sqrt{-9}$√−9 $x$x $=$= $\left\{3i,-3i\right\}${3i,−3i}

The roots are: $2$2, $3i$3i, $-3i$3i

##### Question 2

Below is a portion of the graph of a polynomial function with only real roots. What is the degree of the function? Think: The Fundamental Theorem of Algebra states that the degree of a function is equal to the number of roots the function has.  The graph shows $3$3 real roots; therefore the leading term for the polynomial function is $ax^3$ax3

Do: The polynomial is a $3^{rd}$3rd degree function.

## Using roots to find a polynomial function

If a polynomial function $f\left(x\right)$f(x) the real roots $x=m$x=m and $x=n$x=n, then it must be of the form (based on the Fundamental Theorem of Algebra)

$f\left(x\right)=a\left(x-m\right)\left(x-n\right)$f(x)=a(xm)(xn)

with the leading coefficient (GCF) determinable from additional information given. We can extend this beyond quadratics to any polynomial.

#### Worked examples

##### Question 3

Given the roots $x=0$x=0 and $x=8$x=8 , what is the function when it passes through the point $\left(2,36\right)$(2,36) .

Think: The Fundamental Theorem of Algebra states a polynomial function can be written as the product of its factors and a non-zero real number:

$g\left(x\right)=a\left(x-c_1\right)\left(x-c_2\right)$g(x)=a(xc1)(xc2)

Do: To write the polynomial function using the roots, the roots must be turned into the factors of the function. $x=0$x=0 $\to$ $x+0=x$x+0=x and $x=8$x=8  $\to$  $\left(x-8\right)$(x8)

The function is $g\left(x\right)=ax\left(x-8\right)$g(x)=ax(x8)

Notice, the function still has an $a$a in it. The $a$a is an unknown value. To find this value, substitute the given point $\left(2,36\right)$(2,36) , which the graph passes through, into the above function.

 $g\left(x\right)$g(x) $=$= $ax\left(x-8\right)$ax(x−8) $36$36 $=$= $a\times2\left(2-8\right)$a×2(2−8) $36$36 $=$= $2a\times\left(-6\right)$2a×(−6) $36$36 $=$= $-12a$−12a $a$a $=$= $-3$−3

So the function is identified as $g\left(x\right)=-3x\left(x-8\right)$g(x)=3x(x8) or $g\left(x\right)=-3x^2+24x$g(x)=3x2+24x

Reflect: Without that extra piece of information of the point the curve goes through we could not identify the specific function. The diagram below shows five other possible candidates that have the same two roots as the identified function. Did you know?

Note: When a roots has multiplicity of $2$2 (double root), it implies that the curve of the function has one of two basic shapes as shown here: ##### Question 4

Write the cubic function, when the roots of the function are $x=-1$x=1 with multiplicity $2$2 and $x=10$x=10 with multiplicity $1$1 and the $y$y-intercept is $-5$5

Think: The number of roots a function has is equal to the degree of the function; however this function only has two roots. How? The $x=-1$x=1 has multiplicity of $2$2, which means it is a double root. Fundamental Theorem of Algebra states a polynomial function can be written as the product of its factors and a non-zero real number:

$h\left(x\right)=a\left(x-c_1\right)\left(x-c_2\right)\left(x-c_3\right)$h(x)=a(xc1)(xc2)(xc3)

Do: To write the polynomial function using the roots, the roots must be turned into the factors of the function.

$x=-1$x=1 with multiplicity of  $2$2 $\to$  $\left(x+1\right)^2$(x+1)2 and $x=10$x=10 $\to$ $\left(x-10\right)$(x10)

The function is $h\left(x\right)=a\left(x+1\right)^2\left(x-10\right)$h(x)=a(x+1)2(x10)

Notice, the function still has an $a$a in it. The $a$a is an unknown value. The graph of the function has a $y$y-intercept of $-5$5, therefore the graph passes through $\left(0,-5\right)$(0,5) then, upon substitution, we would know

 $h\left(x\right)$h(x) $=$= $a\left(x+1\right)^2\left(x-10\right)$a(x+1)2(x−10) $-5$−5 $=$= $a\left(0+1\right)^2\left(0-10\right)$a(0+1)2(0−10) $-5$−5 $=$= $-10a$−10a $a$a $=$= $\frac{1}{2}$12​

So the function is identified as $h\left(x\right)=\frac{1}{2}\left(x+1\right)^2\left(x-10\right)$h(x)=12(x+1)2(x10)

The graph of the function is shown here: #### Practice questions

##### Question 5

Which of the following equations has a root of multiplicity $4$4?

1. $\left(x+8\right)^4\left(x+3\right)=0$(x+8)4(x+3)=0

A

$\left(x+8\right)\left(x+3\right)^3=0$(x+8)(x+3)3=0

B

$\left(x+8\right)\left(x+3\right)=0$(x+8)(x+3)=0

C

$\left(x+8\right)^3\left(x+3\right)=0$(x+8)3(x+3)=0

D

$\left(x+8\right)^4\left(x+3\right)=0$(x+8)4(x+3)=0

A

$\left(x+8\right)\left(x+3\right)^3=0$(x+8)(x+3)3=0

B

$\left(x+8\right)\left(x+3\right)=0$(x+8)(x+3)=0

C

$\left(x+8\right)^3\left(x+3\right)=0$(x+8)3(x+3)=0

D

##### Question 6

Consider the cubic function that has been graphed.

1. State the $x$x-intercepts of the function.

2. Which $x$x-intercept is a double root?

3. Which form of a cubic function indicates it has a double root? (Select all that apply)

$y=a\left(x-p\right)\left(x-q\right)\left(x-r\right)$y=a(xp)(xq)(xr)

A

$y=a\left(x-p\right)^3\left(x-q\right)$y=a(xp)3(xq)

B

$y=a\left(x-p\right)\left(x+q\right)^2$y=a(xp)(x+q)2

C

$y=a\left(x-p\right)^2\left(x-q\right)$y=a(xp)2(xq)

D

$y=a\left(x-p\right)\left(x-q\right)\left(x-r\right)$y=a(xp)(xq)(xr)

A

$y=a\left(x-p\right)^3\left(x-q\right)$y=a(xp)3(xq)

B

$y=a\left(x-p\right)\left(x+q\right)^2$y=a(xp)(x+q)2

C

$y=a\left(x-p\right)^2\left(x-q\right)$y=a(xp)2(xq)

D
4. State the equation of the cubic function.

### Outcomes

#### III.N.CN.9

Know the fundamental theorem of algebra; show that it is true for quadratic polynomials.

#### III.A.APR.3

Identify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function defined by the polynomial.

#### III.F.IF.7.c

Graph polynomial functions, identifying zeros when suitable factorizations are available, and showing end behavior.