Write the simplest polynomial equation with real coefficients having roots -i and 4.
I have noooooo idea how to even start this one :( I thought you could start backwards with synthetic substitution but it didn't work. Any help on this is greatly appreciated!! :D
(x-4)(x+i) = x^2 -4x -4i
I know i isn't considered a real coefficient, but when you multiply it out shouldn't there be an ix as you're working it out?
yes, of course, I erred. Thanks. The statement with real coefficents puzzles me.
Yeah it confuses me too :( thanks for your help though, I appreciate it! :D
To find the simplest polynomial equation with real coefficients having roots -i and 4, we need to use the concept of the conjugate pairs.
First, let's consider the complex root -i. According to the conjugate pairs theorem, complex roots always come in conjugate pairs. So, if -i is a root, then its conjugate, i, will also be a root.
Now, we have two roots: -i and 4. To form a polynomial equation, we can start by writing two separate linear equations using these roots:
x - (-i) = 0 (For -i)
x - 4 = 0 (For 4)
To convert these equations into polynomial form, we need to eliminate the complex number. To do that, we multiply both sides of each equation by their corresponding conjugates:
(x - (-i))(x - i) = 0 (For -i)
(x - 4)(x - 4) = 0 (For 4)
Simplifying these equations:
(x + i)(x - i) = 0
(x - 4)(x - 4) = 0
Now, let's expand these equations:
(x^2 - i^2) = 0
(x - 4)(x - 4) = 0
Since i^2 is equal to -1, we simplify further:
(x^2 + 1) = 0
(x - 4)(x - 4) = 0
Finally, we can multiply these two equations together to obtain the simplest polynomial equation:
(x^2 + 1)(x - 4)(x - 4) = 0
Expanding this equation gives us the simplest polynomial equation with real coefficients having roots -i and 4:
x^4 - 4x^3 - 8x^2 + 32x + 49 = 0
So, the simplest polynomial equation with real coefficients having roots -i and 4 is x^4 - 4x^3 - 8x^2 + 32x + 49 = 0.