Question 376785: Verify that 2i is a zero of P(x)= x^4-2x^3-11x^2-8x-60 Also find the remaining zeros.
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! If 2i is a zero of P(x), then there should be a zero remainder when P(x) is divided by (x-2i). And the quotient will be the other factor.
The simplest way to do divide is to use synthetic division:
2i | 1 -2 -11 -8 -60
---- 2i -4-4i 8-30i 60
--------------------------------
1 -2+2i -15-4i -30i 0
Our remainder, in the lower right corner, is zero. So (x-2i) is a factor of P(x). Not only that but the rest of the bottom line shows us the quotient (which is the other factor. So
The next zero can be found quickly because for all polynomials with real coefficients, like P(x), complex zeros, if any, will as complex conjugates. Therefore, since 2i was a zero it conjugate, -2i, will also be a zero.
Important: Complex conjugates are generally not negatives of each other. For the general complex number a+bi the complex conjugate is a-bi. In your problem the real part, i.e. the "a", of 2i is zero. In other words, 2i, written as complex number with real and imaginary parts is 0 + 2i. Its conjugate is 0 - 2i or just -2i. In this case the conjugates turn out to be negatives of each other, 2i and -2i. Only when the "a" is zero does the conjugate work out to be the negative of the original number!
Since -2i will be a zero we can divide by (x - (-2i)) or (x + 2i) knowing that it will divide evenly also. We divide not the original P(x) but the second factor of P(x) above:
-2i | 1 -2+2i -15-4i -30i
----- -2i +4i 30i
-------------------------
1 -2 -15 0
As predicted, (x+2i) is also a factor. Now we have:
The last factor is quadratic trinomial and it factors easily:
From the last two factors we can find the remaining zeros:
x - 5 = 0 or x + 3 = 0
Solving these we get:
x = 5 or x = -3
So the four zeros of P(x) are 2i, -2i, 5 and -3.
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