# SOLUTION: I have a question on polynomials with prescribed zeros. The problem is find polynomial f(x) of degree 4 that has real coefficients and zeros 2+i and -3i. I get most of the problem

Algebra ->  Algebra  -> Polynomials-and-rational-expressions -> SOLUTION: I have a question on polynomials with prescribed zeros. The problem is find polynomial f(x) of degree 4 that has real coefficients and zeros 2+i and -3i. I get most of the problem       Log On

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 Click here to see ALL problems on Polynomials-and-rational-expressions Question 75778: I have a question on polynomials with prescribed zeros. The problem is find polynomial f(x) of degree 4 that has real coefficients and zeros 2+i and -3i. I get most of the problem but what i dont understand is how they got from (x-2+i)(x-2-i)(x+3i)(x-3i) to (x^2-4x + 5)(x^2 +9) and then to x^4-4x^3+14x^2-36x +45. ALso is it the same steps to get from (x^3+1)(x^2-4) to (x+1)(x^2-x+1)(x+2)(x-2). Thank YouAnswer by ankor@dixie-net.com(15647)   (Show Source): You can put this solution on YOUR website!The problem is find polynomial f(x) of degree 4 that has real coefficients and zeros 2+i and -3i. : Think of it as the reverse of completing the square> : x = 2 + i : x - 2 = i: subtracted 2 from both sides : (x-2)^2 = i^2; squared both sides : x^2 - 4x + 4 = -1: FOILed (x-2)(x-2) : x^2 - 4x + 4 + 1 = 0; added 1 to both sides : x^2 - 4x + 5 = 0 : and : x = -3i : x^2 = 9(-1)squared both sides : x^2 = -9 : x^2 + 9 = 0; add 9 to both sides : Multiply (x^2 + 9) to (x^2 - 4x + 5) and you will get: x^4 - 4x^3 + 14x^2 - 36x + 45 : Did this help?