Please help!! I do not understand any of this!!
Information is given about a polynomial f(x) whose coefficients are real numbers. Find the remaining zeros of f.
Degree 4; zeros: 8,-6-i
you know this much..
(x-8)(x+6+i) and coefficents with real numbers adds the complex conjugate
(x-8)(x+6+i)(x+6-i) that is degree three.
so to get any fourth degree equation with those roots, you can add any other root, or make the root at 8 a double root such as
f(x)=(x-8)^2 * (x+6+i)(x+6-i)
I can help you understand how to find the remaining zeros of a polynomial. To do this, we need the degree of the polynomial and any known zeros.
In your case, the polynomial is of degree 4, and you are given two zeros: 8 and -6-i.
Let's start with the zero 8. If a polynomial has a zero at a specific value, then it means that when you plug in that value for x, the result is zero. So for this polynomial, we have:
f(8) = 0
Now, let's move on to the zero -6-i. In general, for a polynomial with real coefficients, if a complex number is a zero, then its complex conjugate is also a zero. So, the complex conjugate of -6-i is -6+i. Therefore, we have two zeros here:
f(-6-i) = 0
f(-6+i) = 0
Now that we have the equations, we can solve them to find the remaining zeros.
To solve the equation f(8) = 0, you would substitute 8 for x in the polynomial and set it equal to zero. This will give you an equation to solve for the remaining zeros.
To solve the equations f(-6-i) = 0 and f(-6+i) = 0, you would substitute -6-i and -6+i, respectively, for x in the polynomial and set each equation equal to zero. Solve each equation separately to find the remaining zeros.
Remember that when solving equations, you may need to use algebraic techniques like factoring, the quadratic formula, or synthetic division to find the zeros.
Once you solve the equations, the solutions you find will be the remaining zeros of the polynomial f(x).