#1. a zero at x=h means that (x-h) is a factor of f(x). So, a simple function which satisfies the conditions is
f(x) = (x+1)^2 (x-6)
#2. Since x^3-3x^2+x+5 = (x+1)(x^2-4x+5)
it has one real root at x = -1
Since the quadratic has no real roots (the discriminant is negative), -1 is the only root.
#3. f(x) = (x-1)(x^2+6x+2)
work it the same way as #2.
#4. f(x) = (x+2)(x^2-5x+4)
so see what you can do with that.
I need help with a few questions, please explain.
1. Write a polynomial function in standard form with zeros -1, -1, 6.
2. Find the roots of the equation
x^3 – 3x^2 + x + 5
3. Describe the number and type of roots for the polynomial (how many real and complex?). x^3 + 5x^2 – 4x – 2 = 0
4. One zero of x^3 – 3x^2 – 6x + 8 = 0 is -2. What are the other zeros of the function?
1 answer
0 answers