Find a real plynomial equation with real coefficients that has the given roots.
3-2i, 3+2i
Should I just make it [x- (3-2i)][x + (3-2i)] then multiply it out and there's my answer?
angela, thanks for your help how did you get the 56.80 from my problem. I've tried so many different ways.
Look at the original post. I explained it there.
get out...
To find a polynomial equation with the given roots, you can use the fact that complex roots come in conjugate pairs.
If you have roots 3-2i and 3+2i, you can set up the factors as (x - (3-2i))(x - (3+2i)) and simplify it to find the polynomial equation.
Let's go through the steps:
Step 1: Expand the factors:
(x - (3-2i))(x - (3+2i))
Step 2: Simplify inside the parentheses:
(x - 3 + 2i)(x - 3 - 2i)
Step 3: Use the FOIL method:
(x - 3)(x - 3) + (x - 3)(-2i) + (x - 3)(2i) + (2i)(-2i)
Step 4: Simplify further:
(x - 3)^2 + (-2i)(x - 3) + (2i)(x - 3) + (2i)(-2i)
Step 5: Simplify the terms:
(x^2 - 6x + 9) + (-2ix + 6i) + (2ix - 6i) + (-4i^2)
Step 6: Simplify i^2:
(x^2 - 6x + 9) + (-2ix + 6i) + (2ix - 6i) - 4
Step 7: Combine like terms:
x^2 - 6x + 9 - 2ix + 6i + 2ix - 6i - 4
Step 8: Simplify further:
x^2 - 6x + 5
Therefore, the polynomial equation with the given roots 3-2i and 3+2i is x^2 - 6x + 5.
As for your second question about finding the value of 56.80, I apologize, but I couldn't find any reference to that in the previous conversation. Could you please provide more context or clarify your question?