Find a polynomial F(x) of degree 3 with real coefficients and the following zeros. 4,1-i help fast please?

f(x) = x³ - 6x² + 10x - 8

Step-by-step explanation:

Complex zeros occur as conjugate pairs, thus

Given x = 1 - i is a zero then x = 1 + i is also a zero

Given the zeros are

x = 4, x = 1 - i, x = 1 + i then the corresponding factors are

(x - 4), (x - (1 - i) ), (x - (1 + i ) ), that is

(x + 4), ( x - 1 + i), (x - 1 - i)

f(x) is then the product of the factors, that is

f(x) = (x - 4)((x - 1) + i)((x - 1) - i ) ← distribute the complex factors

     = (x - 4)((x - 1)²  - i² ) → i² = - 1

     = (x - 4)(x² - 2x + 1 + 1)

     = (x - 4)(x² - 2x + 2) ← distribute

     = x³ - 2x² + 2x - 4x² + 8x - 8 ← collect like term

f(x) = x³ - 6x² + 10x - 8

step-by-step explanation:

let

we know that

the y-intercept is the y-coordinate of the point when the value of x is equal to zero

so

the y-coordinate of the point a is the y-intercept

the answer part a) is

the value of y-intercept is

part b) find the slope m

we know that

the value of the slope is equal to

substitute the values

therefore

the answer part b) is

the value of the slope is equal to

the general answer is

the slope is

the value of y-intercept is

it is not that hard all u have to do is plot the graph

step-by-step explanation: