Suppose that a polynomial function of degree 5 with rational coefficients has 0 (with multiplicity 2), 3, and 1 –2i as zeros. Find the remaining zero.
A. –2
B. –1 – 2i
C. 0
D. 1 + 2i
So the funtion is:
x^2(x - 7)(x + 3 - 5i)(x - 3 + 5i)
For it to have rational coefficients, you must get rid of the -3 + 5i term.
Therefore the remaining zero is:
3 - 5i
so your answer would be (C)
How can the answer be 0 ? I don't get it, I figured to be -1-2i Choice B. What did i do wrong? Thanks!
There is a little rule of thumb that you can count on for finding zeroes of polynomials with complex roots.
Complex roots always come in pairs. Each root of the pair is the complex conjugate of the other.
For example, if you have a root as 4+3i, the other root must be 4-3i. If another complex root is -2-i, then its conjugate is -2+i, etc.
To get the correct answer to the given problem, you only need to choose the complex conjugate of the given complex root.
The function would be
f(x)=x²(x-3)(x-1+2i)(x-1-2i)
= x5-5x4+11x³-15x²
7+what equal 125
To find the remaining zero, we know that a polynomial function with rational coefficients has complex roots in conjugate pairs. Since we are given that 1 - 2i is a zero, we know its conjugate, 1 + 2i, must also be a zero.
Therefore, the remaining zero is D. 1 + 2i.