Quest:
show that i (imaginary) is a zero of f (x)= x^3+ (3-i)x^2 - 4ix -1
Plug in i for x and add up the terms. You should get zero. Remember that i^2 = -1
To show that i is a zero of the function f(x) = x^3 + (3-i)x^2 - 4ix - 1, we need to substitute i for x in the function and demonstrate that the resulting expression evaluates to zero.
Let's start by substituting i for x in the function:
f(i) = i^3 + (3-i)i^2 - 4i(i) - 1
Now, let's simplify this expression step by step:
i^3 is equal to i * i^2. Since i^2 is -1, we have i * (-1) = -i.
(i) squared is equal to i * i = i^2, which is -1.
Substituting these values into the expression, we get:
f(i) = -i + (3-i)(-1) - 4i(-1) - 1
Next, we simplify further:
-1 multiplied by (3-i) gives us -3 + i.
-4i multiplied by -1 gives us 4i.
Now, we can rewrite the expression:
f(i) = -i - 3 + i - 4i - 1
Combining like terms, we have:
f(i) = -3 - 1
Simplifying further:
f(i) = -4
Since f(i) is equal to -4, and not zero, we have shown that i is not a zero of the function f(x) = x^3 + (3-i)x^2 - 4ix - 1.
If you were asked to show that i (imaginary) is a zero of the function, the result should be zero when evaluating f(i). However, in this case, it is not zero, so i is not a zero of the given function.