I was wondering if anyone could please help me with an Algebra 2 problem involving roots. The roots are 1/2, 1, and 3+2i. I need help finding the equation of that and it passes through the point (-1,200).
I did (x-1/2)(x-1)(x-3+2i)(x-3-2i) as my first step
Is that right?
Then i simplified it but it does not pass through (-1,200)?
Please don't switch names
There is a constant factor. You function is really
y = a(x-1/2)(x-1)(x-3+2i)(x-3-2i)
So, if you find that using your function, f(-1) = 100, then just let a=2, so that it scales the y value to where it passes through the point you want.
To find the equation with the given roots, you can start by writing the factors of the polynomial.
You correctly started by using the linear factors of the roots: (x - 1/2)(x - 1)(x - 3 + 2i)(x - 3 - 2i). Great job!
Next, you need to simplify it correctly. Let's go through the steps.
Expanding (x - 3 + 2i)(x - 3 - 2i) using the difference of squares, we get:
(x - 3 + 2i)(x - 3 - 2i) = (x - 3)^2 - (2i)^2 = (x - 3)^2 - 4i^2 = (x - 3)^2 - 4(-1) = (x - 3)^2 + 4.
So, the simplified factors become (x - 1/2)(x - 1)((x - 3)^2 + 4).
Now, we need to determine the constant term in the quadratic expression. Since the polynomial passes through the point (-1, 200), we can substitute the x and y coordinates into the equation.
200 = (-1 - 3)^2 + 4
200 = 16 + 4
200 = 20
Here, we found an inconsistency. The equation 200 = 20 is not true. So, either there was a mistake in the given point or the necessary information is missing.
Please double-check the given point or provide any additional information, and I'll be happy to help you further!