Consider the given function.
f(x)=4x^3+2x^2-4x-2
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 only 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)= 4(- x)^3+2(- x)^2-4(- x)-2
⇕
f(- x)=- 4x^3+2x^2+4x-2
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
4x^3+2x^2-4x-2
f(x)=4x^3+2x^2-4x-2
- 2
4( - 2)^3+2( - 2)^2-4( - 2)-2
-18
- 1
4( - 1)^3+2( - 1)^2-4( - 1)-2
0
0
4( 0)^3+2( 0)^2-4( 0)-2
-2
1
4( 1)^3+2( 1)^2-4( 1)-2
0
2
4( 2)^3+2( 2)^2-4( 2)-2
30
We see above that a zero occurs at x=- 1. Let's use synthetic division to depress the polynomial.
rl IR-0.15cm r - 1 & |rr 4 & 2 & -4 & -2
Bring down the first coefficient
rl IR-0.15cm r - 1 & |rr 4 & 2 & -4 & -2 & c 4 & & &
The above are the coefficients of the depressed polynomial. Moreover, since - 1 is a zero of f(x)=0, (x+1) is a factor of f(x).
f(x)=4x^3+2x^2-4x-2
⇕
f(x)=(x+1) ( 4x^2-2x-2 )
Now the depressed is a quadratic polynomial, where a= 4, b= - 2, and c= -2. This means we can use the Quadratic Formula to find its zeros.
Let's now find the second and third zeros by splitting the final fraction into the positive and negative cases.
x=2± 6/8
x=2+6/8
x=2-6/8
x=8/8
x=-4/8
x=1
x=-1/2
Since x=1 and x=- 12 are zeros of the depressed polynomial, we can rewrite the quadratic expression and apply the Commutative Property of Multiplication.
f(x)=(x+1) ( 4x^2-2x-2 )
⇕
f(x)=4(x+1)(x-1)(x+ 12)
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 odd-degree polynomial with a positive leading coefficient. This tells us about the end behavior of the function.
f(x) → - ∞ & as x → - ∞
f(x) → + ∞ & as x → + ∞
Let's draw the function.
