Question

Find an equation with real coefficients of a polynomial function f that has the given characteristics.
Degree: 3
Zeros: 1, 2-3i
y-intercept: -26

Answer 1

x = 1, x = 2-3i, x = 2+3i


x - 1 = 0, x - (2 -3i) = 0, x - (2 + 3i) = 0


a(x-1)( x - (2 -3i) )( x - (2 + 3i) ) = 0


a(x-1)( x - 2 +3i )( x - 2 - 3i ) = 0


a(x-1)( (x - 2) +3i )( (x - 2) - 3i ) = 0


a(x-1)( (x - 2)^2 -(3i)^2 ) = 0


a(x-1)( x^2-4x+4+9 ) = 0


a(x-1)( x^2-4x+13) = 0


a[x( x^2-4x+13 )-1( x^2-4x+13 )] = 0


a[ x^3-4x^2+13x-x^2+4x-13] = 0


a[ x^3-5x^2+17x-13] = 0


ax^3-5ax^2+17ax-13a = 0


So the function is f(x) = ax^3-5ax^2+17ax-13a


But we know that the y-intercept is (0, -26)


So f(0) = -26


This means that

a(0)^3-5a(0)^2+17a(0)-13a = -26


-13a = -26


a = -26/(-13)


a = 2


So the function is now


f(x) = ax^3-5ax^2+17ax-13a


f(x) = 2x^3-5(2)x^2+17(2)x-13(2)


f(x) = 2x^3-10x^2+34x-26


So the answer is f(x) = 2x^3-10x^2+34x-26