Find all zeros of the function f(x) = (x - 4)(x + 3)(x + (4 + 3i))(x - (4 - 3i)).
Is it x = 4, -3, -4 - 3i, and 4 - 3i ?
Yes they are the zeros :)
if we put x=4 the 1st factor becomes 0, making the whole function = 0.
Hoped it helped!
While your answer is correct, I suspect a typo in the problem, since the expanded polynomial does not have real coefficients:
x^4 - (1-6i)x^3 - (37+6i)x^2 + (25-72i)x + 300
To find the zeros of the function f(x) = (x - 4)(x + 3)(x + (4 + 3i))(x - (4 - 3i)), we need to set f(x) equal to zero and solve for x.
Setting f(x) = 0, we have:
(x - 4)(x + 3)(x + (4 + 3i))(x - (4 - 3i)) = 0
Since multiplication is commutative, we can rearrange the factors:
(x + 3)(x - 4)(x + (4 + 3i))(x - (4 - 3i)) = 0
Now, we can apply the zero-product property, which states that if the product of several factors is zero, then at least one of the factors must be zero.
So we set each factor equal to zero and solve for x:
x + 3 = 0 => x = -3
x - 4 = 0 => x = 4
x + (4 + 3i) = 0 => x = -4 - 3i
x - (4 - 3i) = 0 => x = 4 - 3i
Therefore, the zeros of the function f(x) are x = 4, -3, -4 - 3i, and 4 - 3i, as you mentioned.
To find the zeros of the function f(x) = (x - 4)(x + 3)(x + (4 + 3i))(x - (4 - 3i)), we set f(x) equal to zero and solve for x.
Step 1: Set f(x) = 0
0 = (x - 4)(x + 3)(x + (4 + 3i))(x - (4 - 3i))
Step 2: Set each factor equal to zero and solve for x separately:
x - 4 = 0
x = 4
x + 3 = 0
x = -3
x + (4 + 3i) = 0
x + 4 + 3i = 0
x = -4 - 3i
x - (4 - 3i) = 0
x - 4 + 3i = 0
x = 4 - 3i
So the zeros of the function f(x) are x = 4, -3, -4 - 3i, and 4 - 3i. Therefore, your initial statement is correct.