Add the complex numbers: (2 + 10i)+(7 - 15i)
9 - 5i
Describe the end behavior of the function: y = x2 - 3x
Rises to right and left
Factor the polynomial: y = x3 + 2x2 - 9x - 18
y= (x + 3)(x - 3)(x + 2)
Find all real and imaginary solutions of the polynomial equation: x2 + 16 = 0
x = 4i and -4i
Multiply the complex numbers: (3 - 2i)(3 + 4i)
17 + 6i
Describe the end behavior of the function: y = -x3
Falls to the right and rises to the left
Write the linear factorization of the polynomial: x4 + 7x3 + 10x2
x(x)(x + 5)(x + 2)
Find all real and imaginary solutions of the polynomial equation: 3x2 + 4x = 7
x = 1 and -7/3
Subtract the complex numbers: (6 + 5i) - (1 - 8i)
5 + 13i
If a polynomial has 4 turning points, what is the least possible degree?
5
Convert the polynomial below to intercept/factored form.
f(x) = 2(x - 1)2 - 8
f(x) = 2(x - 3)(x + 1)
Which zero of the polynomial equation would have a multiplicity of two?
x4 - 3x3 - 3x2 + 11x - 6 = 0
x = 1
Simplify and write in the standard form of a complex number.
(-20 + √-50)/-10
2 - √2i/2
Describe the behavior of a zero if it has an odd multiplicity.
It crosses the x-axis
Write as a product of linear factors, given that (x - 3) is a factor:
f(x)=2x4 - x3 + 35x2 - 25x - 375
f(x) = (x - 3)(x + 5i)(x - 5i)(2x + 5)
Ms. Hanley solved a 7th degree polynomial and found that it had 5 real zeros. Could this be true for this polynomial? Why or why not?
Yes! This would mean you have 2 imaginary zeros. These zeros would be a conjugate pair.
Divide: (2x4 + 3x3 - 5x2 + 4x + 7) ÷ (x2 + 2x - 1)
2x2 - x - 1 + (5x + 6)/(x2 + 2x - 1)
Divide:
(-3 + 5i)/(-3 + i)
7/5 - 6/5i
Describe the end behavior of the function and find the y-intercept: y = 7x - 5x2 + 8x
Falls to the right and left; (0,0)
Write a polynomial function in standard form with the given zeros: 1, -1, -2i
y = (x - 1)(x + 1)(x - 2i)(x + 2i)
y = x4 + 3x2 - 4
Find all real solutions of the polynomial equation:
x3 - 4x2 - 6x = 0
x = 0, 2 + √10, 2 - √10
Divide: (3x5 - 2x3 + 4x - 7) ÷ (x2 - 3)
3x3 + 7x + (25x - 7)/(x2 - 3)