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Use the Fundamental Theorem of Algebra to find the number of zeros. Then, use Descartes' Rule of Signs to determine the possible number of positive and negative real zeros.
x=4, x=4, x=-2i, x=2i
To find the zeros of the given polynomial function, we will follow three steps.
Let's do it!
Consider the given function. f(x)=x^4-8x^3+20x^2-32x+64 We see the degree of the polynomial is 4. According to the Fundamental Theorem of Algebra, 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 four sign changes for the coefficients of f(x). According to Descartes' Rule of Signs, the function has zero, two, or four positive real zeros. Next, let's write and simplify f(- x). f(- x) = (- x)^4 - 8(- x)^3 + 20(- x)^2 - 32(- x) + 64 ⇕ f(- x)=x^4+8x^3+20x^2+32x+64 As we did for f(x), we will examine the number of sign changes for f(- x).
There are no sign changes for the coefficients of f(- x). Once again, according to Descartes' Rule of Signs, the function has zero negative real zeros. Therefore, there are three options for the types of zeros of f(x). Remember, we know that there are four zeros in total.
| Option 1 | Option 2 | Option 3 | ||||
|---|---|---|---|---|---|---|
| Real Zeros | 0 positive | Real Zeros | 2 positive | Real Zeros | 4 positive | |
| 0 negative | 0 negative | 0 negative | ||||
| Imaginary Zeros | 4 | Imaginary Zeros | 2 | Imaginary Zeros | ||
| Number of Zeros | 0+ 0+4=4 total | Number of Zeros | 2+ 0+2=4 total | Number of Zeros | 4+ 0+ =4 total | |
Let's now determine the real zeros by making a table of values.
| x | x^4-8x^3+20x^2-32x+64 | f(x)=x^4-8x^3+20x^2-32x+64 |
|---|---|---|
| 0 | 0^4-8( 0)^3+20( 0)^2-32( 0)+64 | 64 |
| 1 | 1^4-8( 1)^3+20( 1)^2-32( 1)+64 | 45 |
| 2 | 2^4-8( 2)^3+20( 2)^2-32( 2)+64 | 32 |
| 3 | 3^4-8( 3)^3+20( 3)^2-32( 3)+64 | 13 |
| 4 | 4^4-8( 4)^3+20( 4)^2-32( 4)+64 | 0 |
| 5 | 5^4-8( 5)^3+20( 5)^2-32( 5)+64 | 29 |
We see above that a zero occurs at x=4. Let's use synthetic division to depress the polynomial.
Bring down the first coefficient
Multiply the coefficient by the divisor
Add down
Multiply the coefficient by the divisor
Add down
Multiply the coefficient by the divisor
Add down
Multiply the coefficient by the divisor
Add down
The above are the coefficients of the depressed polynomial. Moreover, since 4 is a root of f(x)=0, (x-4) is a factor of f(x). f(x)=x^4-8x^3+20x^2-32x+64 ⇕ f(x)=(x-4) ( x^3-4x^2+4x-16 ) Since there is an even number of positive zeros of our polynomial, let's make a table of values of the depressed polynomial function x^3+3x^2+64x+192 to find the second positive zero.
| x | x^3-4x^2+4x-16 | f(x)=x^3-4x^2+4x-16 |
|---|---|---|
| 0 | 0^3-4( 0)^2+4( 0)-16 | -16 |
| 1 | 1^3-4( 1)^2+4( 1)-16 | -15 |
| 2 | 2^3-4( 2)^2+4( 2)-16 | -16 |
| 3 | 3^3-4( 3)^2+4( 3)-16 | -13 |
| 4 | 4^3-4( 4)^2+4( 4)-16 | 0 |
| 5 | 5^3-4( 5)^2+4( 5)-16 | 29 |
We see above that a zero of the depressed polynomial occurs at x=4. Thus, (x-4) 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-4) ( x^3-4x^2+4x-16 ) ⇕ f(x)=(x-4)(x-4) ( x^2+4 ) 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.
LHS-16=RHS-16
sqrt(LHS)=sqrt(RHS)
sqrt(- a)= sqrt(a)* i
Calculate root
Since x=2i and x=-2i are zeros of the depressed polynomial, we can rewrite them as (x-2i)(x+2i).
f(x)=(x-4)(x-4) ( x^2+4 )
⇕
f(x)=(x-4)(x-4)(x-2i)(x+2i)
The four zeros are 4, 4, 2i, and -2i.
Bring down the first coefficient
Multiply the coefficient by the divisor
Add down
Multiply the coefficient by the divisor
Add down
Multiply the coefficient by the divisor
Add down