Construct a polynomial function with the stated properties. Reduce all fractions to lowest terms. Third-degree, with zeros of 2−i, 2+i, and 3, and a leading coefficient of −7.

Question

Can anybody help me out im really stuck and can use the help thank you

Answer 1

We could start with these roots and write these as factors.

x = 2-i x = 2 + i x = 3

put everything on the left, so that each of these =0, making them the factors, instead of the zeros.

x-2+i = 0 x-2-i = 0 x-3=0

Since these are factors, we can put the together to find the polynomial:

(x-2+i)(x-2-i)(x-3)=0

if we re-write the first two as a difference of squares, factored

[ (x-2) - i ] [ (x-2) + i ] (x-3)

the first two become (x-2)² - (i)² = (x² - 4x + 4 - (-1) ) = x² - 4x + 5

now the first 2 terms have been multiplied, still need to multiply that result with (x-3)

(x² - 4x + 5)(x - 3) = x³ - 3x² - 4x² + 12x + 5x - 15

combine like terms and get f(x) = x³ - 7x² + 17x - 15

So this takes care of our zeros. Still need a leading coefficient of -7.

Final f(x) = -7(x³ - 7x² + 17x - 15) = -7x³ + 49x² - 119x + 105

Answer 2

f(x) = -7(x - 3)[x - (2+i)][x - (2-i)]

= -7(x - 3)[(x-2) - i][(x-2) + i]

= -7(x - 3)[(x-2)² - i²]

= -7(x - 3)(x² - 4x + 5)

= -7(x³ - 4x² + 5x - 3x² + 12x - 15)

= -7x³ + 49x² - 119x + 105