information is given about a polynomial f(x) whose coeffiecients are real numbers. Find the remaining zeros of f.
degree4, zeros i, 9+i
**I am just not getting these coefficients. I know that there has to be an easier way to learn this. Please help!!
They want to know the:
remaining zeros of f
complex roots come in pairs, so the others must be -i and 9-i.
Algebra I, I believe.
The reason they specify real coefficeints is so that you know the complex roots come in conjugate pairs. Otherwise, it is not true, since
(x-3)(x+i) = x^2 - 3x + ix - 3i
= x^2 + (-3+i)x - 3i
which is a perfectly useful quadratic, but its coefficients are not real, so the roots do not have to come in conjugate pairs.
To find the remaining zeros of a polynomial, we need to know its degree and the zeros that are already given. In this case, you mentioned that the polynomial has a degree of 4, and the zeros that are given are i and 9 + i.
The zeros of a polynomial with real coefficients occur in conjugate pairs. This means that if a complex number is a zero of the polynomial, its conjugate is also a zero. Complex conjugates are numbers of the form a + bi, where a and b are real numbers, and the conjugate of a complex number is obtained by changing the sign of the imaginary part.
Since the zero i is given, its conjugate -i is also a zero of the polynomial. Therefore, we know that the polynomial has zeros i, -i, 9 + i, and their conjugates.
To write the polynomial in factored form, we can start by setting up the factors for the known zeros:
(x - i)(x + i) is one factor since i and -i are zeros.
(x - (9 + i))(x - (9 - i)) is another factor since 9 + i and 9 - i are zeros.
Multiplying these factors together, we can find the remaining zeros of the polynomial:
[(x - i)(x + i)][(x - (9 + i))(x - (9 - i))] = (x^2 + 1)(x - 9 - i)(x - 9 + i)
To simplify this expression, we can multiply the binomials inside:
(x^2 + 1)(x - 9 - i)(x - 9 + i) = (x^2 + 1)(x^2 - 18x + 81 - i^2)
Since i^2 = -1, we can substitute this into the expression:
(x^2 + 1)(x^2 - 18x + 81 - (-1)) = (x^2 + 1)(x^2 - 18x + 81 + 1)
= (x^2 + 1)(x^2 - 18x + 82)
So, the remaining zeros of the polynomial f(x) with the given zeros i and 9 + i are the zeros of the factor:
(x^2 - 18x + 82)
To find the remaining zeros, we can either factor this quadratic equation further or use the quadratic formula.