Consider the given function.
f(x)=x^4-6x^3+9x^2+4x-12
We see the of the is 4. According to the , the function has four zeros. This is including any repeated zeros.
Let's now determine the type of zeros. To do so, we will examine the number of sign changes for f(x).
There are three sign changes for the coefficients of f(x). According to , the function has one or three positive real zeros. Next, let's write and simplify f(- x).
f(- x)= (- x)^4-6(- x)^3+9(- x)^2+4(- x)-12
⇕
f(- x)=x^4+6x^3+9x^2-4x-12
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 are two options for the types of zeros of f(x). Remember, we know that there are four zeros in total.
| Option 1
|
Option 2
|
| Real Zeros
|
1 positive
|
Real Zeros
|
3 positive
|
| 1 negative
|
1 negative
|
| Imaginary Zeros
|
2
|
Imaginary Zeros
|
|
| Number of Zeros
|
1+1+2=4 total
|
Number of Zeros
|
3+1+ =4 total
|
Step 3
Let's now determine the real zeros by making a table of values.
| x
|
x^4-6x^3+9x^2+4x-12
|
f(x)=x^4-6x^3+9x^2+4x-12
|
| - 2
|
( - 2)^4-6( - 2)^3+9( - 2)^2+4( - 2)-12
|
80
|
| - 1
|
( - 1)^4-6( - 1)^3+9( - 1)^2+4( - 1)-12
|
0
|
| 0
|
0^4-6( 0)^3+9( 0)^2+4( 0)-12
|
- 12
|
| 1
|
1^4-6( 1)^3+9( 1)^2+4( 1)-12
|
- 4
|
| 2
|
2^4-6( 2)^3+9( 2)^2+4( 2)-12
|
0
|
We see above that a zero occurs at x=- 1. Let's use to depress the polynomial.
rl IR-0.15cm r - 1 & |rr 1 & - 6 & 9 & 4 & - 12
Bring down the first coefficient
rl IR-0.15cm r - 1 & |rr 1 & - 6 & 9 & 4 & - 12 & c 1 & & & &
Multiply the coefficient by the divisor
rl IR-0.15cm r - 1 & |rr 1 & - 6 & 9 & 4 & - 12 & - 1 & & & & c 1 & & & &
rl IR-0.15cm r - 1 & |rr 1 & - 6 & 9 & 4 & - 12 & - 1 & & & & c 1 & - 7 & & &
▼
Repeat the process for all the coefficients
Multiply the coefficient by the divisor
rl IR-0.15cm r - 1 & |rr 1 & - 6 & 9 & 4 & - 12 & - 1 & 7& & & c 1 & - 7 & & &
rl IR-0.15cm r - 1 & |rr 1 & - 6 & 9 & 4 & - 12 & - 1 & 7& & & c 1 & - 7 & 16 & &
Multiply the coefficient by the divisor
rl IR-0.15cm r - 1 & |rr 1 & - 6 & 9 & 4 & - 12 & - 1 & 7& - 16 & & c 1 & - 7 & 16 & &
rl IR-0.15cm r - 1 & |rr 1 & - 6 & 9 & 4 & - 12 & - 1 & 7& - 16 & & c 1 & - 7 & 16 & - 12 &
Multiply the coefficient by the divisor
rl IR-0.15cm r - 1 & |rr 1 & - 6 & 9 & 4 & - 12 & - 1 & 7& - 16 &12 & c 1 & - 7 & 16 & - 12 &
rl IR-0.15cm r - 1 & |rr 1 & - 6 & 9 & 4 & - 12 & - 1 & 7& - 16 &12 & c 1 & - 7 & 16 & - 12 & 0
x=- b±sqrt(b^2-4ac)/2a
x=- ( - 5)±sqrt(( - 5)^2-4( 1)( 6))/2( 1)
x=5±sqrt((- 5)^2-4(1)(6))/2(1)
x=5±sqrt(25-4(1)(6))/2(1)
x=5±sqrt(25-24)/2
x=5±sqrt(1)/2
x=5± 1/2
| x=5± 1/2
|
| x=5+ 1/2
|
x=5- 1/2
|
| x=6/2
|
x=4/2
|
| x=3
|
x=2
|
Since x=3 and x=2 are zeros of the depressed polynomial, we can rewrite them as (x-3)(x-2).
f(x)=(x+1) (x-2) ( x^2-5x+6 )
⇕
f(x)=(x+1)(x-2)(x-3)(x-2)
The four zeros are - 1, 3, and 2, which is a repeated zero.
Graphing the Polynomial
We will use the zeros and the points obtained in the table to graph the polynomial function. Consider also that this is an even-degree polynomial with a positive leading coefficient. This tells us about the of the function.
&f(x) → + ∞ as x → - ∞
&f(x) → + ∞ as x → + ∞
Let's draw the function.
Let's use synthetic division to divide x^3-7x^2+16x-12 by (x-2).
rl IR-0.15cm r 2 & |rr 1 & - 7 & 16 & - 12
Bring down the first coefficient
rl IR-0.15cm r 2 & |rr 1 & - 7 & 16 & - 12 & c 1 & & &
Multiply the coefficient by the divisor
rl IR-0.15cm r 2 & |rr 1 & - 7 & 16 & - 12 & 2 & & & c 1 & & &
rl IR-0.15cm r 2 & |rr 1 & - 7 & 16 & - 12 & 2 & & & c 1 & - 5 & &
▼
Repeat the process for all the coefficients
Multiply the coefficient by the divisor
rl IR-0.15cm r 2 & |rr 1 & - 7 & 16 & - 12 & 2 & - 10 & & c 1 & - 5 & &
rl IR-0.15cm r 2 & |rr 1 & - 7 & 16 & - 12 & 2 & - 10 & & c 1 & - 5 & 6 &
Multiply the coefficient by the divisor
rl IR-0.15cm r 2 & |rr 1 & - 7 & 16 & - 12 & 2 & - 10 & 12 & c 1 & - 5 & 6 &
rl IR-0.15cm r 2 & |rr 1 & - 7 & 16 & - 12 & 2 & - 10 & 12 & c 1 & - 5 & 6 & 0
