Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. write the polynomial in standard form. 4, -2, and -1 + 2i

This is going to be a fourth degree polynomial because if x = 4, one of the factors is x - 4; if x = -2, then one of the factors is x + 2; and by the conjugate root theorem, if x = -1 + 2i, then -1-2i HAS to also be a root. If x = -1+2i, then the factor is (x-(-1+2i)). Simplfying that gives us (x+1-2i). Likewise, if x = -1-2i, then the factor is (x-(-1-2i)). Simplifying that gives us (x+1+2i). We will FOIL those complex factors first. Doing that we have . Once we simplify that down it's much easier to deal with than what it looks like right now. . Since i^2 = -1, what we have in the end is . Now we will multiply that by x-4. . Now finally we will multiply in the last factor of x+2. is your final fourth degree polynomial.

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!