Use synthetic division to factor the following polynomials completely by given factors or roots and find all the zeros; when you reach quadratic equation, performance regular factoring or quadratic formula.

, as factors and -3 as a root

The factors are (x - (1 - i√3)/2) , (x - (1 + i√3)/2) , (x - 3) , (x + 3) , (x - 1)

The zeroes are 1 , 3 , -3 , (1 - i√3)/2 , (1 + i√3)/2

Step-by-step explanation:

÷ (x - 3) =

÷ (x - 3) =

÷ (x - 3) =

÷ (x - 3) =

÷ (x - 1) =

x³ + (4x³ + 0 - x - 3) ÷ (x - 1) =

x³ + 4x² + (4x² - x - 3) ÷ (x - 1) =

x³ + 4x² + 4x + (3x - 3) ÷ (x - 1) =

x³ + 4x² + 4x + 3

∵ -3 is a root ⇒ (x + 3) is a factor

(x³ + 4x² + 4x + 3) ÷ (x + 3) =

x² + (x² + 4x + 3) ÷ (x + 3) =

x² + x + (x + 3) ÷ (x + 3) =

x² + x + 1 ⇒ use the formula to find the factors of this quadratic

∵ a = 1 , b = 1 and c = 1

∴

∴

∴ The factors are (x - (1 - i√3)/2) , (x - (1 + i√3)/2) , (x - 3) , (x + 3) , (x - 1)

The zeroes:

x - 3 = 0 ⇒ x = 3

x + 3 = 0 ⇒ x = -3

x - 1 = 0 ⇒ x = 1

x - (1 - i√3)/2 = 0 ⇒ x = (1 - i√3)/2

x - (1 + i√3)/2 = 0 ⇒ x = (1 + i√3)/2

The zeroes are 1 , 3 , -3 , (1 - i√3)/2 , (1 + i√3)/2

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.