Consider the given function.
f(x)=x^5-8x^3-9x
We see the degree of the polynomial is 5. According to the Fundamental Theorem of Algebra, the function has five zeros. This is including any repeated zeros.
Step 2
Let's now determine the type of zeros. To do so, we will examine the number of sign changes for f(x).
There is one sign change for the coefficients of f(x). According to Descartes' Rule of Signs, the function has one positive real zero. Next, let's write and simplify f(- x).
f(- x)= (- x)^5-8(- x)^3-9(- x)
⇕
f(- x)=- x^5+8x^3+9x
As we did for f(x), we will examine the number of sign changes for f(- x).
There is only one sign change for the coefficients of f(- x). Once again, according to Descartes' Rule of Signs, the function has one negative real zero. Therefore, there is only one option for the types of zeros of f(x). Remember, we know that there are five zeros in total.
Zeros
Real Zeros
1 positive
1 negative
Imaginary Zeros
3
Number of Zeros
1+ 1+3=5 total
Step 3
Let's now determine the real zeros by making a table of values.
x
x^5-8x^3-9x
f(x)=x^5-8x^3-9x
- 2
( - 2)^5-8( - 2)^3-9( - 2)
50
- 1
( - 1)^5-8( - 1)^3-9( - 1)
16
0
0^5-8( 0)^3-9( 0)
0
1
1^5-8( 1)^3-9( 1)
-16
2
2^5-8( 2)^3-9( 2)
-50
3
3^5-8( 3)^3-9( 3)
0
We see above that a zero occurs at x=0. Let's factor out x to depress the polynomial.
f(x)=x^5-8x^3-9x
⇕
f(x) = x(x^4-8x^2-9)
From the table, we know that another zero occurs at x=3. Thus, (x-3) is a factor of f(x). Let's use synthetic division to rewrite the depressed polynomial.
The above are the coefficients of the depressed polynomial. Moreover, since 3 is a root of f(x)=0, (x-3) is a factor of f(x).
f(x)=x(x^4-8x^2-9)
⇕
f(x)=x(x-3)( x^3+3x^2+x+3 )
Let's make a table of values of the depressed polynomial function x^3+3x^2+x+3 to find the remaining negative zero.
x
x^3+3x^2+x+3
P(x)=x^3+3x^2+x+3
- 4
( - 4)^3+3( - 4)^2+( - 4)+3
-17
- 3
( - 3)^3+3( - 3)^2+( - 3)+3
0
- 2
( - 2)^3+3( - 2)^2+( - 2)+3
5
- 1
( - 1)^3+3( - 1)^2+( - 1)+3
4
We see above that a zero of the depressed polynomial occurs at x=- 3. Thus, (x+3) is a factor of f(x). We can use synthetic division again to rewrite the depressed polynomial. If you want to see how to use synthetic division with the depressed polynomial, please see the end of this exercise.
f(x)=x(x-3)( x^3+3x^2+x+3 )
⇕
f(x)=x(x-3)(x+3)( x^2+1 )
Now the depressed is a quadratic polynomial. We can find its zeros by taking the square roots. Remember to consider the positive and negative solutions.
Since x=i and x=- i are zeros of the depressed polynomial, we can rewrite them as (x-i)(x+i).
f(x)=x(x-3)(x+3)( x^2+1 )
⇕
f(x)=x(x-3)(x+3)(x-i)(x+i)
The five zeros are 0, 3, -3, i, and - i.
Showing Our Work
Synthetic division
Let's use synthetic division to divide x^3+3x^2+x+3 by (x+3).
