Question

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

Answer 1

Answer:

f(x)=x3−x2−6x2+16x−10

Step-by-step explanation:

If 3−i is a zero, then 3+i

must be a zero as they are conjugates:

f( x ) = ( x − 1 ) ( x − ( 3 − i )) ( x − ( 3 + i )) f ( x ) = ( x − 1 ) ( x2 − x ( 3 + i ) − x ( 3 − i ) + 9 + 1 ) f ( x )) ( x − 1 ) ( x2 − 6x + 10 ) f ( x ) = x3 − x 2 − 6 x 2 + 16x − 10

Answer 2

Answer:

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