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Need help with this question ​-example-1
User Nate Dudek
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13 votes

Answer:


f(x) = a(x^2+4)(x-3)^2

Explanation:

So I'm assuming it means that one of the zeros is at x=3 with a multiplicity of 2 and it has an imaginary solution of 4i. Anyways, imaginary solutions come in conjugate pairs meaning if you have a complex solution of
a-bi there is another complex solution which is the conjugate of that which is
a+bi but since the imaginary solution is 4i, the complex number is just
0+4i so the conjugate is
0-4i or
-4i. So since you have x=3 as a zero that can be represented as
(x-3)^2 since 3 would make it 0 and it has a multiplicity of 2. As for the other factors, you won't just have (
(x-4i)(x+4i), you'll have a factor that is set up in a way that the solutions are:
x=\pm√(-4). That means it'll be a quadratic. So it'll be in the form
x^2+b. Since you're moving b to the other side and it's negative. that means it has to be positive and since the value is 4, you'll have the factor
(x^2+4) which when set equal to zero has the solutions sqrt(-4). So this gives you the equation


f(x) = a(x^2+4)(x-3)^2

User Johnymachine
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