Consider the given function.
f(x)=x^3+7x^2+4x-12
We see the degree of the polynomial is 3. According to the Fundamental Theorem of Algebra, the function has three 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)^3+7(- x)^2+4(- x)-12
⇕
f(- x)=- x^3 + 7x^2 - 4x - 12
As we did for f(x), we will examine the number of sign changes for f(- x).
There are two sign changes for the coefficients of f(- x). Once again, according to Descartes' Rule of Signs, the function has zero or two negative real zeros. Therefore, there are two options for the types of zeros of f(x). Remember, we know that there are three zeros in total.
Option 1
Option 2
Real Zeros
1 positive
Real Zeros
1 positive
0 negative
2 negative
Imaginary Zeros
2
Imaginary Zeros
Number of Zeros
1+ 0+2=3 total
Number of Zeros
1+ 2+ =3 total
Step 3
Let's now determine the real zeros by making a table of values.
x
x^3+7x^2+4x-12
f(x)=x^3+7x^2+4x-12
- 2
( -2)^3+7( -2)^2+4( -2)-12
0
- 1
( -1)^3+7( -1)^2+4( -1)-12
-10
0
0^3+7( 0)^2+4( 0)-12
- 12
1
1^3+7( 1)^2+4( 1)-12
0
2
2^3+7( 2)^2+4( 2)-12
32
We see above that a zero occurs at x=- 2. Let's use synthetic division to depress the polynomial.
rl IR-0.15cm r - 2 & |rr 1 & 7 & 4 & -12
Bring down the first coefficient
rl IR-0.15cm r - 2 & |rr 1 & 7 & 4 & -12 & c 1 & & &
The above are the coefficients of the depressed polynomial. Moreover, since - 2 is a zero of f(x)=0, (x+2) is a factor of f(x).
f(x)=x^3+7x^2+4x-12
⇕
f(x)=(x+2) ( x^2+5x-6 )
Now the depressed is a quadratic polynomial. We can find its zeros by factoring.
Since x=1 and x=-6 are zeros of the depressed polynomial, we can rewrite them as (x-1)(x+6).
f(x)=(x+2) ( x^2+5x-6 )
⇕
f(x)=(x+2)(x-1)(x+6)
The three zeros are - 2, 1, and -6.
