Respuesta :
Answer:
The roots of f(x) are: -2, 3, (1+3i) and (1-3i)
Step-by-step explanation:
We are given an expression:
[tex]f(x)=x^4-3x^3+6x^2+2x-60[/tex]
(1+3i) is a root of f(x)
We have to find the remaining roots of f(x).
Since, (1+3i) is a root of f(x),
[tex]x-(1+3i)[/tex]
is a factor of given expression.
Now, we check if (1 - 3i) is a root of given function.
[tex]f(x)=x^4-3x^3+6x^2+2x-60\\f(1-3i)=(1-3i)^4-3(1-3i)^3+6(1-3i)^2+2(1-3i)-60\\= (28+96i)-3(-26+18i)+6(-8-6i)+2(1-3i)-60 = 0[/tex]
Thus, (1-3i) is also a root of given function.
Since, (1-3i) is a root of f(x),
[tex]x-(1-3i)[/tex]
is a factor of given expression.
Thus, we can write:
[tex](x-(1+3i))(x-(1-3i))\text{ is a factor of f(x)}\\x^2-x(1-3i)+x(1+3i)+(1-3i)(1+3i)\text{ is a factor of f(x)}\\x^2-2x+10\text{ is a factor of f(x)}[/tex]
Dividing f(x) by above expression:
[tex]\displaystyle\frac{x^4-3x^3+6x^2+2x-60}{x^2-2x+10} = x^2-x-6[/tex]
To find the root, we equate it to zero:
[tex]x^2-x-6 = 0\\x^2 - 3x+2x-6=0\\x(x-3)+2(x-3)=0\\(x+2)(x-3) = 0\\x = -2, x = 3[/tex]
Thus, the roots of f(x) are: -2, 3, (1+3i) and (1-3i)
