Respuesta :
Answer:
1/2, 3
Step-by-step explanation:
This is a pretty involved problem, so I'm going to start by laying out two facts that our going to help us get there.
- The Fundamental Theorem of Algebra tells us that any polynomial has as many zeroes as its degree. Our function f(x) has a degree of 4, so we'll have 4 zeroes. Also,
- Complex zeroes come in pairs. Specifically, they come in conjugate pairs. If -2i is a zero, 2i must be a zero, too. The "why" is beyond the scope of this response, but this result is called the "complex conjugate root theorem".
In 2., I mentioned that both -2i and 2i must be zeroes of f(x). This means that both [tex]x-2i[/tex] and [tex]x+2i[/tex] are factors of f(x), and furthermore, their product, [tex]x^2+4[/tex], is also a factor. To see what's left after we factor out that product, we can use polynomial long division to find that
[tex]2x^4+5x^3+5x^2+20x-12=(x^2+4)(2x^2+5x-3)[/tex]
I'll go through to steps to factor that second expression below:
[tex]2x^2+5x-3=2x^2+6x-x-3\\=2x(x+3)-(x+3)\\=(2x-1)(x+3)[/tex]
Solving both of the expressions when f(x) = 0 gets us our final two zeroes:
[tex]2x-1=0\\2x=1\\x=1/2[/tex]
[tex]x+3=0\\x=-3[/tex]
So, the remaining zeroes are 1/2 and 3.
Answer:
The roots are -2i, 2i, 1/2 and -3.
Step-by-step explanation:
The complex roots exist as conjugate pairs so if one root is -2i the other is 2i. So a factor of f(x) is (x - 2i)(2 + 2i). = x^2 + 4.
So dividing:
2x^2 + 5x - 3 <---- Quotient.
--------------------------------------------
x^2 + 0x + 4 ) 2x^4 + 5x^3 + 5x^2 + 20x - 12
2x^4 + 0x^3 + 8x^2
5x^3 - 3x^2 + 20x
5x^3 + 0x^2 + 20x
-3x^2 - 12
-3x^2 - 12.
Finding the other 2 zeroes:
2x^2 + 5x - 3 = 0
(2x - 1) ( x + 3) = 0
x = 1/2, -3.
